Unit 5 Learning Objectives

Note: your goal in using these objectives is to be able to respond to each one with a brief explanation. All of the objectives are covered (and expanded upon) in the flashcards below. Or you can consult this expanded list on our AP Bio Exam Study Outline.

TOPIC 5.1: Meiosis

  1. Explain how meiosis transmits genetic material from one generation to the next.
  2. Compare and contrast diploid and haploid cells, and explain how these terms connect to somatic cells and germ cells.
  3. Compare and contrast mitosis and meiosis (the types of daughter cells, the number of cell divisions)

Topics 5.2 and 5.6. Meiosis, Chromosomal Inheritance, and Genetic Diversity

  1. Explain how meiosis generates genetic diversity.
  2. Define “homologous” chromosomes (their origin, their relationship in terms of genetic information) and explain what happens to homologous pairs during meiosis
  3. Explain what crossing over is, and how it generates genetic diversity.
  4. Explain fertilization (in terms of haploid and diploid chromosome numbers), as well as fertilization’s contribution to genetic diversity.
  5. Compare meiosis 1 and meiosis 2, and explain what happens during each process.
  6. Connect the events of meiosis and fertilization to how sexual reproduction creates variation.
  7. Connect the events of meiosis to Mendel’s laws of segregation and independent assortment; and recombination of linked alleles.
  8. Explain how certain aspects of human genetic variation (Down’s syndrome, etc.) can be explained by chromosomal changes resulting from meiosis (nondisjunction).

Topic 5.3. Mendelian Genetics

  1. Explain Mendel’s laws of segregation and independent assortment, and connect them to what happens during meiosis.
  2. Explain relevant rules of probability that apply to genetics.
  3. Be able to solve genetics problems involving
    1. Monohybrid and dihybrid crosses with autosomal genes;
    2. Multiple alleles, with blood type (A, B, O system)as an illustrative example. Note that while blood type isn’t explicitly in the College Board’s standards, it can show up in problems related to inheritance patterns that involve multiple alleles.

Topic 5.4. Non-Mendelian Genetics

  1. Explain the chromosomal basis of linkage and recombination. When given data about linkage, be able to determine the distance (in map units) between linked alleles.
  2. Explain the inheritance patterns of sex-linked genes, and be able to solve genetics problems involving sex linkage.
  3. Explain non-XY sex determination systems, such as the ZZ/ZW system in birds, haplodiploidy in bees, and temperature-dependent sex determination in certain reptilian clades.
  4. Define polygenic traits, and describe why these usually have a bell-curve-shaped distribution pattern.
  5. Explain non-nuclear inheritance

Topic 5.5. Environmental Effects on Phenotype

  1. Explain how the interaction between genotype and environment is a major determinant of phenotype.

[note: Topic 5.6 was covered above with topic 5.2].


2. Unit 5 Cumulative Flashcards

[qdeck style=”width: 550px !important; min-height: 400px !important;” bold_text=”false” scroll=”true” dataset=”Unit 5 Cumulative Flashcards (v2.0)” qrecord_id=”sciencemusicvideosMeister1961-Unit 5 Cumulative Flashcards (v2.0)” dataset_intro=”true”]

[h] Unit 5 Cumulative Flashcards

[i] [start]

[q json=”true” yy=”4″ unit=”5.Heredity” topic=”5.1-2.Meiosis_and_Genetic_Diversity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3eacd2ba922c” question_number=”1″] Define “homologous” chromosomes (their origin, their relationship in term of genetic information) and explain what happens to homologous pairs during meiosis

[a] Homologous pairs are the matched chromosomes inherited from each parent. They have the same genes (A through I in the diagram), but possibly different alleles (for example, “b” vs “B”). During meiosis 1, two things happen to homologous pairs. During prophase 1, homologous pairs synapse and exchange bits of DNA. This creates novel, recombinant DNA sequences. During metaphase 1 and anaphase 1, homologous pairs independently assort, creating haploid gametes with unique combinations of maternal and paternal chromosomes.

[q json=”true” yy=”4″ unit=”5.Heredity” topic=”5.1-2.Meiosis_and_Genetic_Diversity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3e90e22be22c” question_number=”2″] What are the major events that occur during meiosis 1 and meiosis II?

[a] During meiosis 1, the developing gametes go from diploid to haploid. Crossing over creates new recombinant chromosomes, and independent assortment mixes chromosomes that were originally maternal or paternal in origin.

During meiosis 2, independent assortment occurs again as sister chromatids are pulled apart, resulting in 4 genetically unique haploid gametes.

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3e74f19d322c” question_number=”3″ topic=”5.1-2.Meiosis_and_Genetic_Diversity”] On a big picture level, explain how meiosis differs from mitosis.

[a] Mitosis clones the entire genome, creating diploid daughter cells that are identical to the mother cell. Meiosis reduces the genome from diploid to haploid and creates gametes (sperm cells or egg cells) that are genetically unique.

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3e59010e822c” question_number=”4″ topic=”5.1-2.Meiosis_and_Genetic_Diversity”] Explain how diploid cells are different from haploid cells.

[a] With the exception of gametes, the cells in eukaryotic organisms are diploid, with two sets of chromosomes: one set of chromosomes from the mother, and one from the father. When germ cells create gametes during meiosis, the diploid set of chromosomes is reduced to a single set of chromosomes. A cell with a single set of chromosomes is haploid.

[q json=”true” yy=”4″ unit=”5.Heredity” topic=”5.1-2.Meiosis_and_Genetic_Diversity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3e3868680a2c” question_number=”5″] Explain the processes by which sexual reproduction creates genetic variation.

[a] Sexual reproduction involves meiosis, followed by fertilization. Together, these processes create offspring who are different from their parents, and from one another.

  • During crossing over, homologous chromosomes exchange bits of DNA, creating novel, never-before-seen DNA sequences which can code for novel phenotypes.
  • During independent assortment, maternal and paternal chromosomes are randomly shuffled together. This creates haploid gametes with unique combinations of the maternal and paternal chromosomes that an individual inherited from his or her own parents.
  • During fertilization, DNA from sperm and egg are brought together, creating individuals with a unique genome and phenotype.

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3e1327a9ca2c” question_number=”6″ topic=”5.1-2.Meiosis_and_Genetic_Diversity”] Describe what happens during meiosis I. End your description with Anaphase 1.

[a] Meiosis I starts with interphase which results in duplication of the chromosomes. In prophase 1, the chromosomes condense (as they do during mitosis). During synapsis, homologous chromosomes line up and through crossing over, exchange bits of DNA, creating novel sequences of DNA. During metaphase 1, homologous pairs are moved by the spindle to the cell equator. Because each pair moves independently of every other, this sets the stage for independent assortment, which mixes chromosomes that were originally of maternal and paternal origins. During anaphase 1 the homologous pairs are pulled apart and dragged to opposing ends of the cell.

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3deb92dfa62c” question_number=”7″ topic=”5.1-2.Meiosis_and_Genetic_Diversity”] Describe what happens during meiosis II.

[a] Meiosis 1 ends with two haploid daughter cells. But because meiosis begins with chromosome replication, the daughter cells have doubled chromosomes. During prophase 2, the chromosomes condense, and a spindle reforms. The chromosomes arrive in the middle of the cell during metaphase 2. During anaphase 2, sister chromatids are pulled apart (which involves another round of independent assortment). Telophase II follows, rebuilding the nucleus. Finally, cytokinesis 2 results in four haploid cells. In males, the result of meiosis is four sperm cells. In females, often three of the cells resulting from meiosis are sacrificed so that one large egg cell results.

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3dc3fe15822c” question_number=”8″ topic=”5.1-2.Meiosis_and_Genetic_Diversity”] Mathematically explain the relationship between independent assortment during metaphase 1 and the creation of diversity in gametes. In your answer, you can ignore crossing over, and the independent assortment that happens during meiosis II. 

[a] The number of combinations resulting from independent assortment during metaphase 1 can be quantified as two to the number of homologous pairs. A species with two homologous pairs can create 22 chromosomally distinct gametes. A species with 4 homologous pairs can create 24 chromosomally distinct gametes (which equals 16). Humans, with 23 homologous pairs, can create 223 combinations of maternal and paternal chromosomes, which amounts to over 8 million chromosomally distinct gametes.

[!]5.3.Mendelian Genetics[/!]

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3d9a153f7a2c” question_number=”9″ topic=”5.3.Mendelian_Genetics”] Describe Mendel’s principle of segregation. In your answer, cover the difference between the terms homozygous and heterozygous.

[a] Mendel’s first law of inheritance is the principle of segregation. Mendel discovered that each individual has two copies of each gene. Each copy is called an allele. These alleles can be identical, making the individual homozygous, or they can be different, making the individual heterozygous. When individuals create gametes (sex cells), they pass on only one of their two alleles. In the zygote (fertilized egg) that results, the alleles from each parent come together so that the resulting offspring also has two copies of each gene.

[q json=”true” yy=”4″ dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3d51e7cede2c” question_number=”10″ unit=”5.Heredity” topic=”5.3.Mendelian_Genetics”] Define the terms dominant and recessive.

[a] If an individual possesses a dominant allele, she or he will express the phenotype that this allele codes for. Recessive alleles will only be expressed in individuals who are homozygous (and have inherited two copies of that allele).

[q json=”true” yy=”4″ unit=”5.Heredity” topic=”5.3.Mendelian_Genetics” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3d27fef8d62c” question_number=”11″] Explain what a monohybrid cross is, and what the results of such a cross will be. In your response, create a Punnett square to demonstrate your understanding.

[a] A monohybrid cross is a cross between two heterozygotes, such as a cross between parents whose genotypes are Bb and Bb. The result of such a cross is offspring in a 3:1 ratio, with three individuals with the dominant phenotype to each one with the recessive phenotype. 1/4 of the offspring will be homozygous dominant, 2/4 will be heterozygous (and still have the dominant phenotype) and 1/4 will be homozygous recessive (and show the recessive phenotype).
Here’s a Punnet square that represents this monohybrid cross.

B b
B BB Bb
b Bb bb

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3d02be3a962c” question_number=”12″ topic=”5.3.Mendelian_Genetics”] Using a Mendelian example with garden peas in which T and t control height (tall v. dwarf) and P and p control flower color (purple v. white), define Mendel’s principle of independent assortment.

[a] Independent assortment means that genes carrying different traits are segregated and passed on independently from one another. To use a Mendelian example with garden peas, independent assortment means that the genes controlling height (T) and those controlling flower color (P) are passed onto the offspring independently.

In a dihybrid cross between TtGg and TtGg, the result of this cross will be the combined result of independent crosses between Tt and Tt and that between Gg and Gg. Since each cross would produce a 3:1 ratio of dominants to recessives, and because the genes are passed on independently, the combined result is (3:1)(3:1) or 9:3:3:1 . That means 9/16 of the offspring show both dominant traits, 3/16 show one dominant and one recessive trait, 3/16 show the other dominant and the other recessive trait, and 1/16 show both recessive traits.

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3cd68158aa2c” question_number=”13″ topic=”5.3.Mendelian_Genetics”] What’s the connection between Mendel’s laws of inheritance and what happens during meiosis?

[a] Mendel’s first law is the principle of segregation. According to this principle, parents possess two alleles for each trait but pass only one on to their offspring, which inherit their two alleles from their parents. This principle corresponds to the fact that sexually reproducing organisms have a diploid phase that produces haploid gametes. The diploid phase corresponds to the two alleles in each parent. The haploid phase corresponds to the one allele that each parent passes on to their offspring. In the zygote, the diploid condition is restored.

Mendel’s second law is the law of independent assortment: what happens to one gene pair is independent of every other gene pair (at least the ones that Mendel studied). Independent assortment of genes corresponds to the independent assortment of chromosomes that happens during metaphase 1 of meiosis.

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3caeec8e862c” question_number=”14″ topic=”5.3.Mendelian_Genetics”] What is the rule of multiplication? Demonstrate your understanding by explaining how to predict the probability of genotype aabbcc resulting from a trihybrid cross between AaBbCc x AaBbCc.

[a] The rule of multiplication is the idea that the probability of independent events occurring together is the product of their individual probabilities.

The probability of genotype aabbcc resulting from a trihybrid cross (AaBbCc x AaBbCc) can be solved using the rule of multiplication as follows:  The chance of Aa x Aa producing aa is 1/4. The chance of Bb x Bb producing bb is also 1/4, and that’s also the probability of an offspring having the genotype cc. Because these alleles assort independently, you can use the rule of multiplication: 1/4 x 1/4 x 1/4 = 1/64.

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3c82afac9a2c” question_number=”15″ topic=”5.3.Mendelian_Genetics”] What’s the probability that in a family with five offspring, all of the children will be girls?

[a] Because each birth is independent, you use the rule of multiplication. The probability of each child being a girl is 1/2. The probability of 5 children all being girls is 1/2 x 1/2 x 1/2 x 1/2 x 1/2 (the product of all of these individual probabilities). In this case, the product is 1/32.

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3c5672caae2c” question_number=”16″ topic=”5.3.Mendelian_Genetics”] In a cross between parents whose genotypes are AaBbCCddEe and aaBbCcDdee, what’s the probability of a child being AabbCCddEe?

[a] To solve this, use the rule of multiplication.  The parental genotypes are AaBbCCddEe and aaBbCcDdee. You’re determining the probability of AabbCCddEe. To do this, determine each probability individually. Because these are independent events, multiply them together to determine the overall probability.

In this example, the probability of Aa and aa producing aa is 1/2. The probability of Bb x Bb producing bb is 1/4. Continue this with each allele. So, multiply 1/2 x 1/4 x 1/2 x 1/2 x 1/2. The overall probability is 1/64.

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3c27e1dcde2c” question_number=”17″ topic=”5.3.Mendelian_Genetics”] Explain the genetic and molecular basis of Tay-Sachs disease.

[a] In Tay-Sachs disease, a genetic mutation in an enzyme called beta-hexosaminidase A results in the build-up of a molecule called GM2 ganglioside in the brain. GM2 ganglioside buildup is toxic and ultimately fatal. Victims usually die before the age of five, and there is no treatment.

The disease is autosomal and recessive. In certain populations (Jewish people of Eastern European descent and Louisiana Cajuns) the frequency of the Tay-Sachs allele is significantly higher than in the surrounding populations. There has been speculation that there might be some heterozygote advantage at work (as with the sickle cell allele), possibly in relation to tuberculosis, but this advantage has not been substantiated.

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3bfdf906d62c” question_number=”18″ topic=”5.3.Mendelian_Genetics”] Describe the symptom and genetics of Huntington’s disease.

[a] Huntington’s disease is a neurological disorder caused by an autosomal dominant allele. A homozygote will pass the condition on to 100% of his or her offspring; a heterozygote will pass it on to half of his or her offspring.

The disease strikes individuals in middle age and causes damage to tissue in the brain that controls movement, leading to symptoms such as personality changes, unsteady gait, involuntary movements, and slurred speech. Most victims die from complications relating to the disease.

Because of the late onset of symptoms, most people with Huntington’s have, throughout history, reproduced before they knew they had the disease.

[!]5.4.Non-Mendelian Genetics[/!]

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3bcac0013e2c” question_number=”19″ topic=”5.4.Non-Mendelian_Genetics”] What are linked genes? Describe their inheritance pattern, and explain which of the Mendelian rules that linked genes don’t follow.

[a] Linked genes are genes that are on the same chromosome. For example, genes T and A in cell “2” are linked. Because they’re on the same chromosome, linked genes don’t independently assort (but they can recombine due to crossing over during meiosis).

[q json=”true” yy=”4″ unit=”5.Heredity” topic=”5.4.Non-Mendelian_Genetics” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3b72463d662c” question_number=”20″] Imagine a test cross involving genes AaBb x aabb. How would the inheritance pattern of these genes work if the genes were on different chromosomes? How would the inheritance pattern be different if the genes were linked?

[a] The test cross is between AaBb x aabb. If the genes weren’t linked, you’d expect a 1:1:1:1 ratio of phenotypes in the offspring (Dominant-dominant; Dominant Recessive; Recessive-Dominant; recessive-recessive). If the genes were linked, then most of the offspring would have one of the parental phenotypes (dominant-dominant, recessive-recessive), and a smaller percentage will have a recombinant phenotype (dominant-recessive or recessive-dominant).

[q json=”true” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3b485d675e2c” question_number=”21″ unit=”5.Heredity” topic=”5.4.Non-Mendelian_Genetics”] What’s the relationship between the percentage of recombination and the distance between any two linked genes?

[a] The further apart any genes are on the chromosome, the higher the percentage of recombinant gametes. For example, in the simple chromosome map below, genes A and E will recombine the most (because they’re the furthest apart) and B and C will recombine the least (because they’re the closest together). This percentage of recombination can be used to calculate the map distance between two alleles (measured in centiMorgans).

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3b17786daa2c” question_number=”22″ topic=”5.4.Non-Mendelian_Genetics”] What are sex-linked genes? Explain their inheritance pattern in males.

[a]   Sex-linked genes are genes that are located on the X chromosome. As a result, males can’t be heterozygous. They either have a sex-linked allele, or they don’t. Males who have a sex-linked allele will pass it on to their daughters, but never to their sons.

[q json=”true” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3afddbeade2c” question_number=”23″ unit=”5.Heredity” topic=”5.4.Non-Mendelian_Genetics”] Explain how the inheritance of a recessive sex-linked allele works in females.

[a] If a sex-linked allele is recessive, then female heterozygotes can be carriers, but won’t express the associated phenotype. For a female to express the phenotype of a recessive sex-linked trait, her mother has to be a carrier, and her father has to possess the allele (and thus the associated condition).

As mothers, female heterozygotes have a 1 in 2 chance of passing the allele on to their male offspring.

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3ae1eb5c2e2c” question_number=”24″ topic=”5.4.Non-Mendelian_Genetics”] What are polygenic traits, and what is their inheritance pattern?

[a] Polygenic traits are traits that are controlled by more than one gene. Examples in humans include the genes that control most traits, including phenotypes like height, eye color skin color, etc. This is also true in other organisms: in maize, kernel color is controlled by at least three genes.

Traits that are under the control of multiple genes do not show a binary distribution, as do traits controlled by a single gene (think of cystic fibrosis: you either have it, or you don’t). Rather, traits under the control of multiple genes tend to have what’s called a normal distribution, with an array of phenotypes. Using height as an example, most individuals are of average height: some individuals are very tall, and others are very short, with heights arrayed along a bell-shaped curve.

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3ac3a6c19a2c” question_number=”25″ topic=”5.4.Non-Mendelian_Genetics”] What is non-nuclear inheritance?

[a] Non-nuclear inheritance is the inheritance of genes that are not on nuclear chromosomes but in the chromosome of a mitochondrion or a chloroplast. Because these organelles are only passed on to offspring through the female gamete, their inheritance doesn’t follow Mendelian patterns. In the case of mitochondrial inheritance in animals, all of one’s mitochondria are inherited from one’s mother. In plants, only the ovule (and not the pollen) passes chloroplasts and mitochondria onto the offspring.

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3a99bdeb922c” question_number=”26″ topic=”5.4.Non-Mendelian_Genetics”] Describe the genetics and underlying biology of hemophilia.

[a] Hemophilia is caused by a mutation in a clotting factor gene. These genes code for proteins that are part of the clotting cascade that happens whenever a person gets a cut or an internal injury. The most common mutations are in the genes for the proteins called “clotting factor 8” or “clotting factor 9.”

Hemophilia is a sex-linked, recessive condition. Because it’s sex-linked and recessive, a hemophiliac male can only pass the allele on to his daughters. Daughters can be homozygous normal, heterozygous, or homozygous recessive (which is the only way a female can be a hemophiliac). A heterozygous mother is known as a carrier, and she has a 50% chance of passing the allele on to her son, who, if he inherits the allele, will express it.

The disease has been a target of gene therapy clinical trials.

[!]5.6.Chromosomal Inheritance[/!]

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3a747d2d522c” question_number=”27″ topic=”5.6.Chromosomal_Inheritance”] How is sex determined in mammals?

[a] Mammals have an XX / XY sex-determination system. Offspring that inherit an X and a Y chromosome will be male. Offspring that inherit two X chromosomes will be female.

Because males are XY, when they go through meiosis, half of their sperm will carry an X chromosome, and half will carry a Y. By contrast, a female can only pass on an X chromosome. As a result, it’s the sperm that determines the sex of the offspring, and any offspring will have a 50% chance of being male and a 50% chance of being female.

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3a4a94574a2c” question_number=”28″ topic=”5.6.Chromosomal_Inheritance”] Describe the ZW system of sex determination in birds, and the system of haplodiploidy in certain insects.

[a] In birds the sex chromosomes are known as Z and W. Male birds have two Z chromosomes (ZZ). Female birds have one Z chromosome and one W chromosome (ZW).

[q json=”true” yy=”4″ unit=”5.Heredity” topic=”5.6.Chromosomal_Inheritance” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3a22ff8d262c” question_number=”29″] Describe the system of haplodiploid sex determination in certain insects.

[a] In haplodiploid systems the males are haploid, and the females are diploid. Males develop from unfertilized eggs, while females develop from fertilized eggs.

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e39f46e9f562c” question_number=”30″ topic=”5.6.Chromosomal_Inheritance”] What is nondisjunction? What are its consequences? List some examples of conditions in humans caused by nondisjunction.

[a]  Nondisjunction means “failing to separate,” and it refers to a failure of chromosome separation during meiosis.  A nondisjunction during meiosis I occurs when homologous pairs fail to separate. Nondisjunction during meiosis II means the sister chromatids fail to separate. The result is a gamete (sex cell) that either has an extra chromosome or a missing chromosome.

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e39c831bd6a2c” question_number=”31″ topic=”5.6.Chromosomal_Inheritance”] What are the consequences of nondisjunction? List some examples of conditions in humans caused by nondisjunction.

[a] Nondisjunction results in gametes with an extra chromosome or missing a chromosome. If these gametes fuse with other gametes during fertilization, the result can be a zygote with an abnormal number of chromosomes. This can make the zygote unviable. Alternatively, developmental problems can arise in the multicellular organism that arises from this zygote. In humans, examples of conditions associated with nondisjunction include Turner Syndrome (females lacking an X chromosome), Klinefelter syndrome (males with an extra X chromosome), or Down Syndrome/trisomy 21.

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e3994f8b7d22c” question_number=”32″ topic=”5.6.Chromosomal_Inheritance”] In relationship to the events of meiosis, explain how nondisjunction occurs.

[a] Nondisjunction can occur during meiosis I or meiosis II. If homologous pairs fail to separate during anaphase 1 (“b”), or if sister chromatids fail to separate during anaphase 2 (“h”), then the resulting gametes will either have an extra chromosome or a missing chromosome (as shown in “e,” “f,” “j,” or “k”). If these processes occur together, the result can be gametes with two or three extra chromosomes. That level of chromosome duplication is usually only survivable if it involves the X chromosome.

[q json=”true” yy=”4″ unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Flashcards (v2.0)|e395f6ba6562c” question_number=”33″ topic=”5.6.Chromosomal_Inheritance”] Describe how trisomy 21/Down syndrome occurs.

[a] The cause of Down Syndrome is nondisjunction of the 21st chromosome during meiosis, generally in the egg, resulting in eggs with an extra copy of chromosome 21. After fertilization, the zygote has three copies of chromosome 21. Eggs are suspected as the source of the nondisjunction because of the correlation between maternal age and the incidence of Down syndrome. It’s thought that older eggs are more prone to nondisjunction than younger eggs and, in women, eggs are as old as the woman who carries them (unlike sperm which are continuously produced by men). At the same time, because birth by younger women is more common, most children with Down Syndrome are born to women under 35 years old.

[/qdeck]


3. Unit 5 Cumulative Multiple Choice Quiz 1

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[h] Unit 5 Cumulative Multiple Choice Quiz 1

[i]

Heredity

[q json=”true” multiple_choice=”true” unit=”5.Heredity” topic=”5.1.Meiosis” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|23562c1f6a6db5″ question_number=”1″] In the diagram below, the cell on the top row represents the chromosomal condition of a germ cell before DNA replication.

If this germ cell were to undergo meiosis, which of the cells on the bottom row could represent a viable daughter cell after meiosis I?

[c]IEEg[Qq][c]IE Ig[Qq][c]IEMg[Qq][c]IEQ=

Cg==[Qq]

[f]IE5vLiBUaGUgY2VsbCBvbiB0b3AgaXMgYSBkaXBsb2lkIGNlbGwgd2l0aCBmb3VyIGNocm9tb3NvbWVzLiBNZWlvc2lzIDEgaW52b2x2ZXMgdGhlIHJlcGxpY2F0aW9uIG9mIHRoZSBETkEgYW5kIGNoYW5nZXMgYSBjZWxsIGZyb20gZGlwbG9pZCB0byBoYXBsb2lkLiBGaW5kIGEgZGF1Z2h0ZXIgY2VsbCB3aGVyZSBlYWNoIGNocm9tb3NvbWUgY29uc2lzdHMgb2YgdHdvIHNpc3RlciBjaHJvbWF0aWRzLCBhbmQgdGhlIGNocm9tb3NvbWUgbnVtYmVyIGlzIGhhbGYgb2YgdGhhdCBvZiB0aGUgZ2VybSBjZWxsIHRoYXQgc3RhcnRlZCB0aGUgcHJvY2Vzcy4=[Qq]

[f]IEV4Y2VsbGVudC4gJiM4MjIwO0ImIzgyMjE7IHNob3dzIGEgZGF1Z2h0ZXIgY2VsbCB0aGF0JiM4MjE3O3MgaGFwbG9pZCwgYW5kIGluIHdoaWNoIGVhY2ggY2hyb21vc29tZSBjb25zaXN0cyBvZiB0d28gc2lzdGVyIGNocm9tYXRpZHMuIFRoYXQmIzgyMTc7cyB3aGF0IHlvdSYjODIxNztkIHNlZSBhZnRlciBtZWlvc2lzIEku[Qq]

[f]IE5vLiBZb3UmIzgyMTc7dmUgaWRlbnRpZmllZCB3aGF0IGEgZGF1Z2h0ZXIgY2VsbCBtaWdodCBsb29rIGxpa2UgYWZ0ZXIgbWVpb3NpcyAyLiBGaW5kIGEgZGF1Z2h0ZXIgY2VsbCB3aGVyZSB0aGUgY2VsbCBpcyBoYXBsb2lkLCBhbmQgaW4gd2hpY2ggZWFjaCBjaHJvbW9zb21lIGNvbnNpc3RzIG9mIHR3byBzaXN0ZXIgY2hyb21hdGlkcw==[Qq]

[f]IE5vLiBUaGUgcHJvYmxlbSB3aXRoIGNob2ljZSBEIGlzIHRoYXQgdGhlIGNlbGwgaGFzIHR3byBjb3BpZXMgb2YgdGhlIGxhcmdlciBjaHJvbW9zb21lLCBhbmQgaXMgbGFja2luZyB0aGUgc21hbGxlciBjaHJvbW9zb21lLiBGaW5kIGEgZGF1Z2h0ZXIgY2VsbCB3aGVyZSBib3RoIGNocm9tb3NvbWVzIGFyZSByZXByZXNlbnRlZCBpbiB0aGUgZGF1Z2h0ZXIgY2VsbHMsIHdoZXJlIGVhY2ggY2hyb21vc29tZSBjb25zaXN0cyBvZiB0d28gc2lzdGVyIGNocm9tYXRpZHMsIGFuZCB3aGVyZSB0aGUgY2hyb21vc29tZSBudW1iZXIgaXMgaGFsZiBvZiB0aGF0IG9mIHRoZSBnZXJtIGNlbGwgdGhhdCBzdGFydGVkIHRoZSBwcm9jZXNzLg==

Cg==

[Qq]

[q json=”true” multiple_choice=”true” unit=”5.Heredity” topic=”5.1.Meiosis” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|2355cca982e9b5″ question_number=”2″] In the diagram below, the cell on the top row represents the chromosomal condition of a germ cell before DNA replication.

If this germ cell were to undergo meiosis, which of the cells on the bottom row could represent a viable daughter cell after meiosis II?

[c]IEEg[Qq][c]IEIg[Qq][c]IE Mg[Qq][c]IEQ=

Cg==[Qq]

[f]IE5vLiBUaGF0IHJlcHJlc2VudHMgYSBwb3NzaWJsZSBvdXRjb21lIG9mIG1pdG9zaXMsIG5vdCBtZWlvc2lzLiBUaGUgY2VsbCBvbiB0b3AgaXMgYSBkaXBsb2lkIGNlbGwgd2l0aCBmb3VyIGNocm9tb3NvbWVzLiBNZWlvc2lzIGNvbnNpc3RzIG9mIHR3byBjZWxsIGRpdmlzaW9ucyBhbmQgcmVzdWx0cyBpbiBoYXBsb2lkIGRhdWdodGVyIGNlbGxzIHdpdGggc2luZ2xlIChub3QgZG91YmxlZCkgY2hyb21vc29tZXMu[Qq]

[f]IE5vLiAmIzgyMjA7QiYjODIyMTsgc2hvd3MgYSBkYXVnaHRlciBjZWxsIHRoYXQmIzgyMTc7cyBoYXBsb2lkLCBhbmQgaW4gd2hpY2ggZWFjaCBjaHJvbW9zb21lIGNvbnNpc3RzIG9mIHR3byBzaXN0ZXIgY2hyb21hdGlkcy4gVGhhdCYjODIxNztzIHdoYXQgeW91JiM4MjE3O2Qgc2VlIGFmdGVyIG1laW9zaXMgSS4gTWVpb3NpcyBJSSBjb25zaXN0cyBvZiBhbm90aGVyIGNlbGwgZGl2aXNpb24gaW4gd2hpY2ggdGhlIHNpc3RlciBjaHJvbWF0aWRzIGFyZSBwdWxsZWQgYXBhcnQu[Qq]

[f]IE5pY2UhIFlvdSYjODIxNzt2ZSBpZGVudGlmaWVkIHdoYXQgYSBkYXVnaHRlciBjZWxsIG1pZ2h0IGxvb2sgbGlrZSBhZnRlciBtZWlvc2lzIDIu[Qq]

[f]IE5vLCBmb3IgdHdvIHJlYXNvbnMuIEZpcnN0LMKgIGluIGNob2ljZSBEIGlzIHRoYXQgdGhlIGNlbGwgaGFzIHR3byBjb3BpZXMgb2YgdGhlIGxhcmdlciBjaHJvbW9zb21lLCBhbmQgaXMgbGFja2luZyB0aGUgc21hbGxlciBjaHJvbW9zb21lLiBTZWNvbmQsIGJ5IG1laW9zaXMgMiwgdGhlIGRvdWJsZWQgY2hyb21vc29tZXMgKHdoaWNoIGxvb2sgbGlrZSBhbiAmIzgyMjA7WCYjODIyMTspIGdldCBwdWxsZWQgYXBhcnQu

Cg==

RmluZCBhIGRhdWdodGVyIGNlbGwgd2hlcmUgYm90aCBjaHJvbW9zb21lcyBhcmUgcmVwcmVzZW50ZWQgaW4gdGhlIGRhdWdodGVyIGNlbGxzLCB3aGVyZSBlYWNoIGNocm9tb3NvbWUgY29uc2lzdHMgb2YganVzdCBvbmUgc2lzdGVyIGNocm9tYXRpZCwgYW5kIHdoZXJlIHRoZSBjaHJvbW9zb21lIG51bWJlciBpcyBoYWxmIG9mIHRoYXQgb2YgdGhlIGdlcm0gY2VsbCB0aGF0IHN0YXJ0ZWQgdGhlIHByb2Nlc3Mu

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|2355b5610c01b5″ question_number=”3″ topic=”5.1.Meiosis”] Diagrams A, B, and C below represent stages of cell division from the same organism. One of the three shows the first division of meiosis. Another one shows the second division of meiosis, and one shows division by mitosis.

What is the diploid number of chromosomes for this organism?

[c]IDIg[Qq][c]ID Qg[Qq][c]IDY=

Cg==[Qq]

[f]IE5vLiBEaXBsb2lkIG1lYW5zICYjODIyMDt0d28gc2V0cy4mIzgyMjE7IFRoZSBkaXBsb2lkIG51bWJlciBjYW4mIzgyMTc7dCBiZSB0d28sIGJlY2F1c2UgdGhlcmUgYXJlIEZPVVIgY2hyb21vc29tZXMgaW4gZGlhZ3JhbXMmIzgyMjE7QSYjODIyMTsgYW5kICYjODIyMDtDJiM4MjIxOyAod2hpY2ggaXMgYSBoaW50IGFzIHRvIHdoYXQgdGhlIGFuc3dlciBpcyku[Qq]

[f]IFllcy4gRGlwbG9pZCBtZWFucyAmIzgyMjA7dHdvIHNldHMsJiM4MjIxOyBhbmQgeW91IGNhbiBzZWUgdHdvIHNldHMgb2YgdHdvIGNocm9tb3NvbWVzIGluIGRpYWdyYW0gQy4gQWx0ZXJuYXRpdmVseSwgeW91IGNhbiBzaW1wbHkgY291bnQgdGhlIGZvdXIgY2hyb21vc29tZXMgaW4gZGlhZ3JhbSBBLg==[Qq]

[f]IE5vLiBOb25lIG9mIHRoZSBjZWxscyBhYm92ZSBoYXMgc2l4IGNocm9tb3NvbWVzLiBSZW1lbWJlciB0aGF0IGRpcGxvaWQgbWVhbnMgJiM4MjIwO3R3byBzZXRzLiYjODIyMTsgVGFrZSBhIGdvb2QgbG9vayBhdCB0aGUgZGlhZ3JhbSwgZmluZCBhIGNlbGwgd2l0aCB0d28gc2V0cyBvZiBjaHJvbW9zb21lcywgZG8gc29tZSBjb3VudGluZywgYW5kIHlvdSYjODIxNztsbCBoYXZlIHlvdXIgYW5zd2VyLg==

Cg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” topic=”5.2.Meiosis_and_Genetic_Diversity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|23559bc48935b5″ question_number=”4″] A diploid cell has three pairs of homologous chromosomes.  If the cell undergoes meiosis, how many possible combinations of chromosomes are there in the resulting haploid nuclei?

[c]IDEg[Qq][c]IDMg[Qq][c]IDQg[Qq][c]ID g=

Cg==[Qq]

[f]IE5vLiBJbiBtZWlvc2lzLCB0aGUgY2hyb21vc29tZSBudW1iZXIgaXMgcmVkdWNlZCBmcm9tIGRpcGxvaWQgKHR3byBzZXRzKSB0byBoYXBsb2lkIChvbmUgc2V0KS4gRWFjaCBnYW1ldGUgcmVjZWl2ZXMgb25lIG1lbWJlciBvZiBlYWNoIGhvbW9sb2dvdXMgcGFpci4gRm9yIGVhY2ggcGFpciwgdGhlcmUgYXJlIHR3byBwb3NzaWJpbGl0aWVzIGZvciB3aGljaCBjaHJvbW9zb21lIGdvZXMgdG8gdGhlIGdhbWV0ZSwgYW5kIGVhY2ggcGFpciBpcyBpbmRlcGVuZGVudCBvZiBldmVyeSBvdGhlciBwYWlyLiBUaGUgZ2VuZXJhbCBmb3JtdWxhIGlzIDI=bg==LCB3aGVyZSAmIzgyMjA7biYjODIyMTsgaXMgdGhlIG51bWJlciBvZiBob21vbG9nb3VzIHBhaXJzLiBXaGF0JiM4MjE3O3MgMg==Mw==Pw==[Qq]

[f]IE5vLiBJbiBtZWlvc2lzLCB0aGUgY2hyb21vc29tZSBudW1iZXIgaXMgcmVkdWNlZCBmcm9tIGRpcGxvaWQgKHR3byBzZXRzKSB0byBoYXBsb2lkIChvbmUgc2V0KS4gRWFjaCBnYW1ldGUgcmVjZWl2ZXMgb25lIG1lbWJlciBvZiBlYWNoIGhvbW9sb2dvdXMgcGFpci4gRm9yIGVhY2ggcGFpciwgdGhlcmUgYXJlIHR3byBwb3NzaWJpbGl0aWVzIGZvciB3aGljaCBjaHJvbW9zb21lIGdvZXMgdG8gdGhlIGdhbWV0ZSwgYW5kIGVhY2ggcGFpciBpcyBpbmRlcGVuZGVudCBvZiBldmVyeSBvdGhlciBwYWlyLiBUaGUgZ2VuZXJhbCBmb3JtdWxhIGlzIDI=bg==LCB3aGVyZSAmIzgyMjA7biYjODIyMTsgaXMgdGhlIG51bWJlciBvZiBob21vbG9nb3VzIHBhaXJzLiBXaGF0JiM4MjE3O3MgMg==Mw==Pw==[Qq]

[f]IE5vLiBJbiBtZWlvc2lzLCB0aGUgY2hyb21vc29tZSBudW1iZXIgaXMgcmVkdWNlZCBmcm9tIGRpcGxvaWQgKHR3byBzZXRzKSB0byBoYXBsb2lkIChvbmUgc2V0KS4gRWFjaCBnYW1ldGUgcmVjZWl2ZXMgb25lIG1lbWJlciBvZiBlYWNoIGhvbW9sb2dvdXMgcGFpci4gRm9yIGVhY2ggcGFpciwgdGhlcmUgYXJlIHR3byBwb3NzaWJpbGl0aWVzIGZvciB3aGljaCBjaHJvbW9zb21lIGdvZXMgdG8gdGhlIGdhbWV0ZSwgYW5kIGVhY2ggcGFpciBpcyBpbmRlcGVuZGVudCBvZiBldmVyeSBvdGhlciBwYWlyLiBUaGUgZ2VuZXJhbCBmb3JtdWxhIGlzIDI=bg==LCB3aGVyZSAmIzgyMjA7biYjODIyMTsgaXMgdGhlIG51bWJlciBvZiBob21vbG9nb3VzIHBhaXJzLiBXaGF0JiM4MjE3O3MgMg==Mw==Pw==[Qq]

[f]IEV4YWN0bHkuIEluIG1laW9zaXMsIHRoZSBjaHJvbW9zb21lIG51bWJlciBpcyByZWR1Y2VkIGZyb20gZGlwbG9pZCAodHdvIHNldHMpIHRvIGhhcGxvaWQgKG9uZSBzZXQpLiBFYWNoIGdhbWV0ZSByZWNlaXZlcyBvbmUgbWVtYmVyIG9mIGVhY2ggaG9tb2xvZ291cyBwYWlyLCBzbyBmb3IgZWFjaCBwYWlyLCB0aGVyZSBhcmUgdHdvIHBvc3NpYmlsaXRpZXMgZm9yIHdoaWNoIGNocm9tb3NvbWUgZ29lcyB0byB0aGUgZ2FtZXRlLCBhbmQgZWFjaCBwYWlyIGlzIGluZGVwZW5kZW50IG9mIGV2ZXJ5IG90aGVyIHBhaXIuIEluIHRoaXMgY2FzZSwgdGhlcmUgYXJlIHRocmVlIGhvbW9sb2dvdXMgcGFpcnMsIHNvIMKgMg==MyA=PSA4Lg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” topic=”5.2.Meiosis_and_Genetic_Diversity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|23557fd3fa85b5″ question_number=”5″] A diploid cell has three pairs of homologous chromosomes.  If the cell undergoes mitosis, how many possible combinations of chromosomes are there in the resulting diploid nuclei?

[c]ID Eg[Qq][c]IDMg[Qq][c]IDQg[Qq][c]IDg=

Cg==[Qq]

[f]IFllcy4gTWl0b3NpcyByZXN1bHRzIGluIGFuIGV4YWN0IGNvcHkgb2YgdGhlIHBhcmVudCBjZWxsLiBFYWNoIGRhdWdodGVyIGNlbGwgcmVjZWl2ZXMgb25lIGFuZCBvbmx5IG9uZSBzaXN0ZXIgY2hyb21hdGlkIGZyb20gZWFjaCBwYXJlbnRhbCBjaHJvbW9zb21lLCBhbmQgbm8gbmV3IGNocm9tb3NvbWFsIGNvbWJpbmF0aW9ucyBhcmUgY3JlYXRlZC4=

Cg==

SGVyZSYjODIxNztzIHdoYXQgdGhpcyBsb29rcyBsaWtlIGluIGEgY2VsbCB3aXRoIG9uZSBob21vbG9nb3VzIHBhaXIuIE5vdGUgdGhhdCB0aGUgZGF1Z2h0ZXIgY2VsbHMgYXJlIGEgY2xvbmUgb2YgdGhlIHBhcmVudCBjZWxsLCB3aXRoIG5vIG5ldyBjaHJvbW9zb21hbCBjb21iaW5hdGlvbnMgY3JlYXRlZC4=

Cg==

[Qq]

[f]IE5vLiBNaXRvc2lzIHJlc3VsdHMgaW4gYW4gZXhhY3QgY29weSBvZiB0aGUgcGFyZW50IGNlbGwuIEVhY2ggZGF1Z2h0ZXIgY2VsbCByZWNlaXZlcyBvbmUgYW5kIG9ubHkgb25lIHNpc3RlciBjaHJvbWF0aWQgZnJvbSBlYWNoIHBhcmVudGFsIGNocm9tb3NvbWUsIGFuZCBubyBuZXcgY2hyb21vc29tYWwgY29tYmluYXRpb25zIGFyZSBjcmVhdGVkLg==

Cg==

SGVyZSYjODIxNztzIHdoYXQgdGhpcyBsb29rcyBsaWtlIGluIGEgY2VsbCB3aXRoIG9uZSBob21vbG9nb3VzIHBhaXIuIE5vdGUgdGhhdCB0aGUgZGF1Z2h0ZXIgY2VsbHMgYXJlIGEgY2xvbmUgb2YgdGhlIHBhcmVudCBjZWxsLCB3aXRoIG5vIG5ldyBjaHJvbW9zb21hbCBjb21iaW5hdGlvbnMgY3JlYXRlZC4=

Cg==

[Qq]

[f]IE5vLiBNaXRvc2lzIHJlc3VsdHMgaW4gYW4gZXhhY3QgY29weSBvZiB0aGUgcGFyZW50IGNlbGwuIEVhY2ggZGF1Z2h0ZXIgY2VsbCByZWNlaXZlcyBvbmUgYW5kIG9ubHkgb25lIHNpc3RlciBjaHJvbWF0aWQgZnJvbSBlYWNoIHBhcmVudGFsIGNocm9tb3NvbWUsIGFuZCBubyBuZXcgY2hyb21vc29tYWwgY29tYmluYXRpb25zIGFyZSBjcmVhdGVkLg==

Cg==

SGVyZSYjODIxNztzIHdoYXQgdGhpcyBsb29rcyBsaWtlIGluIGEgY2VsbCB3aXRoIG9uZSBob21vbG9nb3VzIHBhaXIuIE5vdGUgdGhhdCB0aGUgZGF1Z2h0ZXIgY2VsbHMgYXJlIGEgY2xvbmUgb2YgdGhlIHBhcmVudCBjZWxsLCB3aXRoIG5vIG5ldyBjaHJvbW9zb21hbCBjb21iaW5hdGlvbnMgY3JlYXRlZC4=

Cg==

[Qq]

[f]IE5vLiBQZXJoYXBzIHlvdSBjb25mdXNlZCAmIzgyMjA7bWl0b3NpcyYjODIyMTsgZm9yICYjODIyMDttZWlvc2lzJiM4MjIxOyAoaW4gd2hpY2ggY2FzZSA4IHdvdWxkIGhhdmUgYmVlbiB0aGUgY29ycmVjdCBhbnN3ZXIpLg==

Cg==

TWl0b3NpcyByZXN1bHRzIGluIGFuIGV4YWN0IGNvcHkgb2YgdGhlIHBhcmVudCBjZWxsLiBFYWNoIGRhdWdodGVyIGNlbGwgcmVjZWl2ZXMgb25lIGFuZCBvbmx5IG9uZSBzaXN0ZXIgY2hyb21hdGlkIGZyb20gZWFjaCBwYXJlbnRhbCBjaHJvbW9zb21lLCBhbmQgbm8gbmV3IGNocm9tb3NvbWFsIGNvbWJpbmF0aW9ucyBhcmUgY3JlYXRlZC4=

Cg==

SGVyZSYjODIxNztzIHdoYXQgdGhpcyBsb29rcyBsaWtlIGluIGEgY2VsbCB3aXRoIG9uZSBob21vbG9nb3VzIHBhaXIuIE5vdGUgdGhhdCB0aGUgZGF1Z2h0ZXIgY2VsbHMgYXJlIGEgY2xvbmUgb2YgdGhlIHBhcmVudCBjZWxsLCB3aXRoIG5vIG5ldyBjaHJvbW9zb21hbCBjb21iaW5hdGlvbnMgY3JlYXRlZC4=

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|2355663777b9b5″ question_number=”6″ topic=”5.2.Meiosis_and_Genetic_Diversity”] Diagrams A, B and C below represent stages of cell division from the same organism. One of the three shows the first division of meiosis. Another one shows the second division of meiosis, and one shows division by mitosis.

Which diagram shows a type of cell division that, in a multicellular organism, would be used for growth and repair.

[c]IE Eg[Qq][c]IEIg[Qq][c]IEM=

Cg==[Qq]

[f]IFllcyEgJiM4MjIwO0EmIzgyMjE7IGlzIG1pdG9zaXMgKG1ldGFwaGFzZSwgdG8gYmUgcHJlY2lzZSksIHdoaWNoIGlzIHVzZWQgZm9yIGdyb3d0aCBhbmQgcmVwYWlyIGluIGV1a2FyeW90aWMgb3JnYW5pc21zLg==[Qq]

[f]IE5vLiBBbGwgb2YgdGhlc2UgY2VsbHMgYXJlIGZyb20gdGhlIHNhbWUgb3JnYW5pc20uIEluIGNlbGwgJiM4MjIwO0IsJiM4MjIxOyB0aGVyZSBhcmUgb25seSBoYWxmIG9mIHRoZSBjaHJvbW9zb21lcyBwcmVzZW50ICh0d28gYXMgb3Bwb3NlZCB0byBmb3VyKS4gVGhhdCBtYWtlcyBCIGxvb2sgbGlrZSBtZWlvc2lzLCB3aGljaCBpcyB1c2VkIGZvciByZXByb2R1Y3Rpb24sIG5vdCBncm93dGggYW5kIHJlcGFpci4=[Qq]

[f]IE5vLiBJbiBjZWxsIEIsIGl0IGxvb2tzIGxpa2UgaG9tb2xvZ291cyBwYWlycyBhcmUgbGluaW5nIHVwIGFuZCBhcmUgYWJvdXQgdG8gYmUgc3BsaXQgYXBhcnQuIFRoYXQmIzgyMTc7cyBzb21ldGhpbmcgdGhhdCBoYXBwZW5zIGluIG1laW9zaXMsIHdoaWNoIGlzIHVzZWQgZm9yIHJlcHJvZHVjdGlvbiwgbm90IGdyb3d0aCBhbmQgcmVwYWlyLg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|23554a46e909b5″ question_number=”7″ topic=”5.3.Mendelian_Genetics”] In cats, the pattern of tabby strips can be vertical (referred to as “mackerel tabby”) or blotched (referred to as “blotched tabby”), as shown below.

When true-breeding mackerel tabbies are bred with true-breeding blotched tabbies, all of the F1 offspring are mackerel tabbies.

Two F1 mackerel tabbies are crossed with one another. What’s the probability that a kitten resulting from this mating will be a mackerel tabby?

[c]IDAlIA==[Qq][c]IDI1JSA=[Qq][c]IDUwJSA=[Qq][c]IDc1 JQ==

Cg==[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcHJvYmxlbS4gWW91IGtub3cgdGhhdCB0aGUgbWFja2VyZWwgdGFiYnkgdHJhaXQgaXMgZG9taW5hbnQgKGJlY2F1c2UgYWxsIG9mIHRoZSBGMQ==IG9mZnNwcmluZyBhcmUgbWFja2VyZWwpLiBBbmQgeW91IGFsc28ga25vdyB0aGF0IHRoZSBGMQ==cyBhcmUgaGV0ZXJvenlnb3Rlcy4gSWYgeW91IGhhdmUgdGhlIGxldHRlciA=[Qq]M represent the allele for “mackerel” and m represent the allele for “blotched, then you’re crossing Mm x Mm, and you’d set up the Punnett square as follows:

M m
M
m

Now you draw it out, and figure out what percentage of the offspring will have genotype MM or Mm.

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcHJvYmxlbS4gWW91IGtub3cgdGhhdCB0aGUgbWFja2VyZWwgdGFiYnkgdHJhaXQgaXMgZG9taW5hbnQgKGJlY2F1c2UgYWxsIG9mIHRoZSBGMQ==IG9mZnNwcmluZyBhcmUgbWFja2VyZWwpLiBBbmQgeW91IGFsc28ga25vdyB0aGF0IHRoZSBGMQ==cyBhcmUgaGV0ZXJvenlnb3Rlcy4gSWYgeW91IGhhdmUgdGhlIGxldHRlciA=[Qq]M represent the allele for “mackerel” and m represent the allele for “blotched, then you’re crossing Mm x Mm, and you’d set up the Punnett square as follows:

M m
M
m

Now you draw it out and figure out what percentage of the offspring will have genotype MM or Mm.

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcHJvYmxlbS4gWW91IGtub3cgdGhhdCB0aGUgbWFja2VyZWwgdGFiYnkgdHJhaXQgaXMgZG9taW5hbnQgKGJlY2F1c2UgYWxsIG9mIHRoZSBGMQ==IG9mZnNwcmluZyBhcmUgbWFja2VyZWwpLiBBbmQgeW91IGFsc28ga25vdyB0aGF0IHRoZSBGMQ==cyBhcmUgaGV0ZXJvenlnb3Rlcy4gSWYgeW91IGhhdmUgdGhlIGxldHRlciA=[Qq]M represent the allele for “mackerel” and m represent the allele for “blotched, then you’re crossing Mm x Mm, and you’d set up the Punnett square as follows:

M m
M
m

Now you draw it out, and figure out what percentage of the offspring will have genotype MM or Mm.

[f]IFdheSB0byBnbyEgVGhlIHBlcmNlbnRhZ2Ugb2Ygb2Zmc3ByaW5nIHRoYXQgd2lsbCBzaG93IHRoZSBkb21pbmFudCB0cmFpdCB3aWxsIGJlIDc1JSAoMjUlIGhvbW96eWdvdXMgZG9taW5hbnQsIG9yIE1NLCBhbmQgNTAlIGhldGVyb3p5Z290ZXMsIG9yIE1tKS4gSGVyZSYjODIxNztzIHRoZSBQdW5uZXR0IFNxdWFyZQ==

Cg==Cg==Cg==Cg==
[Qq] M m
M MM Mm
m Mm mm

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|235530aa663db5″ question_number=”8″ topic=”5.3.Mendelian_Genetics”] The expression of fruit pigment in a newly developed variety of apples is under genetic control. The apples can have three phenotypes: no fruit pigment, yellow pigment, or red pigment. If there’s no fruit pigment, then the fruit is green. Two genes are involved in the control of pigment production. As shown in the table below, each of the genes has two alleles.

Gene 1 Gene 2
I: no fruit pigment Y: yellow fruit
i: fruit pigment y: red fruit

These two genes are not linked. This means that

[c]IHRoZSBnZW5lcyBhc3Nvcn QgaW5kZXBlbmRlbnRseS4=[Qq]

[f]IEV4YWN0bHkuIElmIHRoZSBnZW5lcyBhcmUgbm90IGxpbmtlZCwgdGhlbiB0aGV5JiM4MjE3O2xsIGFzc29ydCBpbmRlcGVuZGVudGx5Lg==[Qq]

[c]IHRoZSBnZW5lcyBhcmUgY2xvc2UgdG9nZXRoZXIgb24gdGhlIHNhbWUgY2hyb21vc29tZS4=[Qq]

[f]IE5vLiBJZiB0aGUgZ2VuZXMgYXJlIG5vdCBsaW5rZWQsIHRoZW4gdGhleSYjODIxNztyZSA=bm90IG9uIHRoZSBzYW1lIGNocm9tb3NvbWUuIFJlbWVtYmVyIHRoYXQgdGhlIG5leHQgdGltZSB5b3Ugc2VlIHRoaXMgcXVlc3Rpb24gYW5kIGNob29zZSBhbm90aGVyIGFuc3dlci4=[Qq]

[c]IHRoZXJlIGlzIHJlY29tYmluYXRpb24gdGhyb3VnaCBjcm9zc2luZyBvdmVyIGJldHdlZW4gdGhlIHR3byBnZW5lcyBkdXJpbmcgbWVpb3Npcy4=[Qq]

[f]IE5vLiBSZWNvbWJpbmF0aW9uIHRocm91Z2ggY3Jvc3Npbmcgb3ZlciBvbmx5IG9jY3VycyBpbiBnZW5lcyB0aGF0IGFyZSBsaW5rZWQgb24gdGhlIHNhbWUgY2hyb21vc29tZS4gVGhlc2UgZ2VuZXMgYXJlIG5vdCBsaW5rZWQuIFJlbWVtYmVyIHRoYXQgdGhlIG5leHQgdGltZSB5b3Ugc2VlIHRoaXMgcXVlc3Rpb24sIGFuZCBjaG9vc2UgYW5vdGhlciBhbnN3ZXIu[Qq]

[c]IHBhcnRpY3VsYXIgY29tYmluYXRpb25zIG9mIGFsbGVsZXMgb2YgdGhlc2UgZ2VuZXMgYXJlIGFsd2F5cyBpbmhlcml0ZWQgdG9nZXRoZXIu[Qq]

[f]IE5vLiBUaGUga2V5IGlkZWEgaXMgdGhhdCB0aGVzZSBnZW5lcyAoYW5kIHRoZWlyIGFsbGVsZXMpIGFyZSBub3QgbGlua2VkLiBBcyBhIHJlc3VsdCwgdGhlcmUgd29uJiM4MjE3O3QgYmUgcGFydGljdWxhciBjb21iaW5hdGlvbnMgb2YgYWxsZWxlcyB0aGF0IGFyZSBhbHdheXMgaW5oZXJpdGVkIHRvZ2V0aGVyLiBJbnN0ZWFkLCB0aGV5JiM4MjE3O2xsIGJlIGluaGVyaXRlZCBpbmRlcGVuZGVudGx5LiBSZW1lbWJlciB0aGF0IHRoZSBuZXh0IHRpbWUgeW91JiM4MjE3O2xsIHNlZSB0aGlzIHF1ZXN0aW9uIGFuZCBjaG9vc2UgYW5vdGhlciBhbnN3ZXIu

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|23551265cba9b5″ question_number=”9″ topic=”5.3.Mendelian_Genetics”] The pattern of tabby stripes in a cat’s fur is controlled by an autosomal gene with two alleles:

  • T: vertical colored stripes (called mackerel tabby)
  • t: swirly colored stripes (called blotched tabby)

The patterns are illustrated as follows.

A second autosomal gene (not linked to the tabby-striped gene)is the Agouti gene. This gene has two alleles:

  • A: tabby fur pattern
  • a: no tabby pattern.

A cat that is aa for the Agouti locus fails to show any stripes, regardless of its genotype at the tabby stripes locus. These cats are solid black.

A cross was carried out between cats of the following genotypes.

TTAa X ttAa

The cats had two kittens. Which of the images below shows phenotypes that would be possible? Note that not all the kittens in a litter would have to have the phenotype shown in any one of the choices below.

[c]IE Eg[Qq][c]IEIg[Qq][c]IEMg[Qq][c]IEQ=

Cg==[Qq]

[f]IEV4Y2VsbGVudCEgVGhpcyB0eXBlIG9mIGNyb3NzIGNvdWxkIG9ubHkgcHJvZHVjZSBraXR0ZW5zIHRoYXQgd2VyZSBibGFjaywgb3Igd2hpY2ggd2VyZSBtYWNrZXJlbCB0YWJieSAodGhlIGtpbmQgd2l0aCB2ZXJ0aWNhbCBzdHJpcGVzKS4gQmxvdGNoZWQgdGFiYnkgKHdoaWNoIGFwcGVhcnMgaW4gY2hvaWNlcyBCLCBDLCBhbmQgRCkgaXMgbm90IHBvc3NpYmxlLg==

Cg==

SGVyZSYjODIxNztzIHRoZSBQdW5uZXR0IHNxdWFyZQ==

Cg==Cg==[Qq]
tA ta
TA TtAA TtAa
Ta TtAa Ttaa

[f]IE5vLiBUaGUgY3Jvc3MgaXMgYmV0d2VlbiA=VFRBYQ==IGFuZCA=dHRBYQ==

[Qq]

Here’s the Punnett square

tA ta
TA TtAA TtAa
Ta TtAa Ttaa

The genotypes TtAA and TtAa result in mackerel tabby. Ttaa results in black. But this choice has some cats that are blotched tabby. Is that possible? What genotype would a blotched tabby have?

[f]IE5vLiBUaGUgY3Jvc3MgaXMgYmV0d2VlbiA=VFRBYQ==IGFuZCA=dHRBYQ==

[Qq]

Here’s the Punnett square

tA ta
TA TtAA TtAa
Ta TtAa Ttaa

The genotypes TtAA and TtAa result in mackerel tabby. Ttaa results in black. But this choice has cats that are blotched tabby. Is that possible? What genotype would a blotched tabby have?

[f]IEV4Y2VsbGVudCEgVGhpcyB0eXBlIG9mIGNyb3NzIGNvdWxkIG9ubHkgcHJvZHVjZSBraXR0ZW5zIHRoYXQgd2VyZSBibGFjaywgb3Igd2hpY2ggd2VyZSBtYWNrZXJlbCB0YWJieSAodGhlIGtpbmQgd2l0aCB2ZXJ0aWNhbCBzdHJpcGVzKS4gQmxvdGNoZWQgdGFiYnkgKHdoaWNoIGFwcGVhcnMgaW4gY2hvaWNlcyBCLCBDLCBhbmQgRCkgaXMgbm90IHBvc3NpYmxlLg==

Cg==

SGVyZSYjODIxNztzIHRoZSBQdW5uZXR0IHNxdWFyZQ==

[Qq]
tA ta
TA TtAA TtAa
Ta TtAa Ttaa

The genotypes TtAA and TtAa result in mackerel tabby. Ttaa results in black.

[q json=”true” xyz=”2″ multiple_choice=”true” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|2354f4213115b5″ question_number=”10″ unit=”5.Heredity” topic=”5.3.Mendelian_Genetics”] The pedigree below shows the inheritance of Tay-Sachs disease, an autosomal recessive neurological disorder. If individuals III-1 and III-2 have a fourth child, what is the probability that it is male and affected with Tay Sachs disease?

[c]IDMvNA==[Qq]

[f]IE5vLiBUYXkgU2FjaHMgaXMgY2F1c2VkIGJ5IGFuIGF1dG9zb21hbCByZWNlc3NpdmUgYWxsZWxlLCBzbyB3ZSBjYW4gcmVwcmVzZW50IHRoYXQgYWxsZWxlIHdpdGggYSBsb3dlcmNhc2UgJiM4MjIwO3QuJiM4MjIxOyBUaGUgbm9ybWFsIGFsbGVsZSB3ZSBjYW4gcmVwcmVzZW50IGJ5ICYjODIyMDtULiYjODIyMTsgSUlJLTEgYW5kIElJSS0yIGJvdGggaGF2ZSB0byBiZSBoZXRlcm96eWdvdGVzIChvdGhlcndpc2UsIHRoZXJlJiM4MjE3O3Mgbm8gd2F5IHRoYXQgdGhlc2UgdHdvIHVuYWZmZWN0ZWQgcGFyZW50cyB3b3VsZCBwYXNzIG9uIHRoZSBhbGxlbGUgdG8gb25lIG9mIHRoZWlyIGNoaWxkcmVuLCBzdWNoIGFzIElWLTEpLiBTbyB3ZSBjYW4gcmVwcmVzZW50IGJvdGggcGFyZW50cyYjODIxNzsgZ2Vub3R5cGVzIGFzIFR0LiBJZiB5b3Ugc2V0IHVwIGEgUHVubmV0dCBzcXVhcmUsIGl0JiM4MjE3O2xsIGxvb2sgbGlrZSB0aGlzOg==

Cg==Cg==Cg==Cg==
[Qq] T t
T TT Tt
t Tt tt

Take a look, and this tells you the chance that a child will have the disease. Now, you have to multiply that probability by the chance that any one of their children will be a boy. Here’s the Punnett Square for that:

X Y
X XX XY
X XX XY

[c]IDEvMg==[Qq]

[f]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

Cg==Cg==Cg==Cg==
[Qq] T t
T TT Tt
t Tt tt

Take a look, and this tells you the chance that a child will have the disease. Now, you have to multiply that probability by the chance that any one of their children will be a boy. Here’s the Punnett Square for that:

X Y
X XX XY
X XX XY

[c]IDEvNA==[Qq]

[f]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

Cg==Cg==Cg==Cg==
[Qq] T t
T TT Tt
t Tt tt

Take a look, and this tells you the chance that a child will have the disease. Now, you have to multiply that probability by the chance that any one of their children will be a boy. Here’s the Punnett Square for that:

X Y
X XX XY
X XX XY

[c]IDEv OA==[Qq]

[f]IEV4Y2VsbGVudCEgVGhlcmUmIzgyMTc7cyBhIDEvNCBjaGFuY2UgdGhhdCB0aGVpciBjaGlsZCB3aWxsIGhhdmUgdGhlIGNvbmRpdGlvbiBhbmQgYSAxLzIgY2hhbmNlIHRoYXQgdGhlaXIgY2hpbGQgd2lsbCBiZSBhIGJveS4gTXVsdGlwbHkgdGhlIHR3byB0b2dldGhlciwgYW5kIHlvdSBnZXQgMS84Lg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|2354b9ec07d1b5″ question_number=”11″ unit=”5.Heredity” topic=”5.3.Mendelian_Genetics”] The pedigree chart below depicts a specific inherited trait over three generations. What is the mode of inheritance for this trait?

[c]IFgtbGlua2VkIGRvbWluYW50[Qq]

[f]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

Cg==Cg==Cg==Cg==
[Qq] XD Y
XD XDXD XDY
Xd XDXd XdY

Notice that in X-linked dominance, it’s impossible for the daughters of a father with the condition to not inherit the condition. That’s because daughters inherit an X chromosome from their Dad (that’s how they get to be daughters), and if he has the allele causing the condition, he’s definitely going to pass it on to them. If the allele is dominant, the daughters will express the allele.

[c]IFgtbGlua2VkIHJlY2Vzc2l2ZQ==[Qq]

[f]IE5vLiBYLWxpbmtlZCByZWNlc3NpdmUgKGEgY29uZGl0aW9uIGxpa2UgaGVtb3BoaWxpYSkgY2FuJiM4MjE3O3QgZXhwbGFpbiB0aGlzIHBlZGlncmVlLiBXZSYjODIxNztsbCB1c2UgJiM4MjIwO2gmIzgyMjE7IHRvIHJlcHJlc2VudCB0aGUgcmVjZXNzaXZlIGFsbGVsZSwgYW5kICYjODIyMDtIJiM4MjIxOyB0byByZXByZXNlbnQgaXRzIGRvbWluYW50IGNvdW50ZXJwYXJ0LiBMb29rIGF0IHRoZSBjcm9zcyBiZXR3ZWVuIElJLTUgYW5kIElJLTYuIElmIHRoaXMgY29uZGl0aW9uIHdlcmUgWC1saW5rZWQgcmVjZXNzaXZlLCB0aGVuIHRoZSBmYXRoZXIgd291bGQgaGF2ZSB0byBiZSBYaA==WSBhbmQgdGhlIG1vdGhlciBYRA==WA==[Qq]d. The Punnett square would be

Xh Y
Xh XhXh XhY
Xh XhXh XhY

As you can see, all of the offspring would have to have this condition but, looking at the pedigree, the daughters don’t.

[c]IEF1dG9zb21hbC Bkb21pbmFudA==[Qq]

[f]IEV4Y2VsbGVudC4gQXV0b3NvbWFsIGRvbWluYW50IGluaGVyaXRhbmNlIGFjY291bnRzIGZvciBhbGwgdGhlIGNyb3NzZXMgaW4gdGhpcyBwZWRpZ3JlZS4=[Qq]

[c]IEF1dG9zb21hbCByZWNlc3NpdmU=[Qq]

[f]IE5vLiBUaGUgYXV0b3NvbWFsIGluaGVyaXRhbmNlIG1vZGVsLCBmb3IgdGhpcyBwZWRpZ3JlZSwgZmFpbHMgdG8gZXhwbGFpbiB0aGUgY3Jvc3MgYmV0d2VlbiBJSS01IGFuZCBJSS02LiBJZiB0aGlzIHdlcmUgYXV0b3NvbWFsIGluaGVyaXRhbmNlLCB0aGVuIGJvdGggb2YgdGhlc2UgaW5kaXZpZHVhbHMgd291bGQgaGF2ZSB0byBiZSBob21venlnb3VzIHJlY2Vzc2l2ZSwgYW5kIGFsbCBvZiB0aGVpciBjaGlsZHJlbiB3b3VsZCBoYXZlIHRvIGluaGVyaXQgdGhlIGNvbmRpdGlvbi4gRm9yIGV4YW1wbGUsIHVzaW5nICYjODIyMDthJiM4MjIxOyB0byByZXByZXNlbnQgdGhlIHJlY2Vzc2l2ZSBhbGxlbGUsIHRoZSBjcm9zcyBpcyBiZXR3ZWVuIA==YWE=IGFuZCA=YWE=LiBBbGwgdGhlIG9mZnNwcmluZyBoYXZlIHRvIGJlIGFhLg==

[Qq]
a a
a aa aa
a aa aa

 

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|23549dfb7921b5″ question_number=”12″ topic=”5.4.Non-Mendelian_Genetics”] (Note: This question is part of a series. Feel free to skip this introduction if you’ve already read it).

INTRODUCTION: Domestic chickens have been bred for many years to increase the number of eggs laid by females. Chick breeders need to be able to identify female chicks on the day after hatching, as only the females need to be kept for laying eggs.

Unlike mammals, which use a sex determination system with X and Y chromosomes, in chickens the sex chromosomes are known as Z and W. Male chickens have two Z chromosomes (ZZ). Female chickens have one Z chromosome and one W chromosome (ZW).

Some genes for feather color and pattern in chickens are carried on the Z chromosome but not on the W chromosome. One such example is the gene for striped feathers (barring).

The phenotypes associated with the two alleles of the barring gene are shown in the table below

Allele Adult Phenotype Day old chick phenotype
B Barred (black feathers striped with white bars) Black body with a white spot on the head
b Non-barred (black feathers) Black body and head

The type of inheritance associated with barring is

[c]IGF1dG9zb21hbCByZWNlc3NpdmU=[Qq]

[f]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[Qq]

[c]IGF1dG9zb21hbCBkb21pbmFudA==[Qq]

[f]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[Qq]

[c]IHNleC1saW5rZWQgcmVjZXNzaXZl[Qq]

[f]IE5vLCBidXQgeW91IGNob3NlIHRoZSBzZWNvbmQtYmVzdCBhbnN3ZXIuIEp1c3QgcmUtcmVhZCB0aGUgcXVlc3Rpb24gY2FyZWZ1bGx5LCBsb29raW5nIHNwZWNpZmljYWxseSBmb3IgYSBkZXNjcmlwdGlvbiBvZiB0aGUgYmVoYXZpb3Igb2YgdGhlIEJhcnJpbmcgYWxsZWxlLiBSZW1lbWJlciB3aGF0IHlvdSYjODIxNzt2ZSBsZWFybmVkLCBhbmQgdGhlIGFuc3dlciB3aWxsIGJlIG9idmlvdXMgdGhlIG5leHQgdGltZSB5b3Ugc2VlIHRoaXMgcXVlc3Rpb24u[Qq]

[c]IHNleC1saW5rZW QgZG9taW5hbnQ=[Qq]

[f]IFdheSB0byBnbyEgQmFycmluZyBpcyBzZXgtbGlua2VkIChpdCYjODIxNztzIGNhcnJpZWQgb24gdGhlIFogY2hyb21vc29tZSkgYW5kIGRvbWluYW50IChpdCYjODIxNztzIHJlcHJlc2VudGVkIGJ5IGEgY2FwaXRhbCBsZXR0ZXIpLg==[Qq]

[c]IG1pdG9jaG9uZHJpYWw=[Qq]

[f]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

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|2354845ef655b5″ question_number=”13″ topic=”5.4.Non-Mendelian_Genetics”] In a newly bred population of mice, coat color is under the influence of three alleles.

  • C: black coat
  • cR: brown coat
  • c: white coat

The black coat allele is dominant over both brown coat and white coat. The genotype cRc produces another coat color known as fawn.

Which of the following predictions is correct?

[c]IENjUg==IHggQ2M=Ug==OiBhbGwgYmxhY2sgbWljZS4=[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgdGhlIFB1bm5ldHQgc3F1YXJlIGZvciA=Q2M=Ug==IHggQ2M=Ug==[Qq]

C cR
C CC CcR
cR CcR cRcR

CC is black; CcR is black. But cRcR is brown.

[c]IGM=Ug==Yw==Ug==IHggYw==[Qq]Rc: all brown mice.

[f]IE5vLiBIZXJlJiM4MjE3O3MgdGhlIFB1bm5ldHQgc3F1YXJlIGZvciA=Yw==Ug==Yw==Ug==[Qq] x cRc.

cR cR
cR cRcR cRcR
c cRc cRc

cRcR is brown; cRc is fawn.

[c]IG M=Ug==Yw==Ug==IHggY2M6IGFsbCBmYXduIG1pY2Uu[Qq]

[f]IEZhYnVsb3VzLiBIZXJlJiM4MjE3O3MgdGhlIFB1bm5ldHQgc3F1YXJlIGZvciA=Yw==Ug==Yw==Ug==[Qq] x cc:

cR cR
c cRc cRc
c cRc cRc

All the offspring have the cRc genotype and the fawn phenotype

[c]IENjIHggY2M6IGFsbCB3aGl0ZSBtaWNlLg==[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgdGhlIFB1bm5ldHQgc3F1YXJlIGZvciBDYyB4IGNjLg==

Cg==Cg==[Qq]
C c
c Cc cc
c Cc cc

Cc is black; cc is white.

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|2354661a5bc1b5″ question_number=”14″ topic=”5.4.Non-Mendelian_Genetics”] The pedigree below shows inheritance of an X-linked recessive trait.

What is the probability that the child indicated by ? will be a boy who is affected by the trait?

[c]IDAlIA==[Qq][c]IDEyLj UlIA==[Qq][c]IDUwJSA=[Qq][c]IDc1JSA=[Qq][c]IDEwMCU=

Cg==[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcXVlc3Rpb24uIEltYWdpbmUgdGhhdCBpdCYjODIxNztzIGEgcmVjZXNzaXZlLCBzZXgtbGlua2VkIHRyYWl0LCBsaWtlIGhlbW9waGlsaWEuIEZvciBjbGFyaXR5LCBsZXQmIzgyMTc7cyBoYXZlICYjODIyMDtIJiM4MjIxOyBzdGFuZCBmb3IgdGhlIG5vcm1hbCBhbGxlbGUgYW5kICYjODIyMDtoJiM4MjIxOyBmb3IgdGhlIGFsbGVsZSB0aGF0IGNhdXNlcyB0aGUgcmVjZXNzaXZlIGNvbmRpdGlvbi4gVGhlIGNoaWxkIGF0IA==Pw==IGhhcyBhIGZhdGhlciB3aG8gaGFzIHRoZSB0cmFpdDogdGhlcmVmb3JlIHRoZSBmYXRoZXImIzgyMTc7cyBnZW5vdHlwZSBpcyBYaA==WS4gVGhlIGNoaWxkJiM4MjE3O3MgbW90aGVyIGhhZCBhIGJyb3RoZXIgd2hvIGhhZCB0aGUgcmVjZXNzaXZlIHRyYWl0LiBUaGVyZWZvcmUgPyYjODIxNztzIGdyYW5kbW90aGVyIGhhZCB0byBiZSBhIGNhcnJpZXIsIGFuZCA/JiM4MjE3O3MgbW90aGVyIGhhcyBhIDUwJSBjaGFuY2Ugb2YgYmVpbmcgYSBjYXJyaWVyLg==

[Qq]

Now use the rule of multiplication. There’s a 1/2 chance that ? will be a boy (true of any child, right?). There’s a 1/2 chance that the mother is a carrier. And, if she’s a carrier, there’s a 1/2 chance that she’ll pass on her Xh allele. Multiply all of these probabilities together, and you’ll have your answer.

[f]IEV4Y2VsbGVudC4gVGhlcmUmIzgyMTc7cyBhIDEvMiBjaGFuY2UgdGhhdCB0aGUgY2hpbGQgd2lsbCBiZSBhIGJveS4gVGhlcmUmIzgyMTc7cyBhIDEvMiBjaGFuY2UgdGhhdCB0aGUgbW9tIChpbmRpdmlkdWFsIDEpIGlzIGEgY2Fycmllci4gQW5kIHRoZXJlJiM4MjE3O3MgYSAxLzIgY2hhbmNlIHRoYXQgc2hlIChpbmRpdmlkdWFsIDEpIHdpbGwgcGFzcyBvbiBoZXIgcmVjZXNzaXZlIGFsbGVsZS4gMS8yIHggMS8yIHggMS8yID0gMS84LCBvciAxMi41JQ==[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcXVlc3Rpb24uIEltYWdpbmUgdGhhdCBpdCYjODIxNztzIGEgcmVjZXNzaXZlLCBzZXgtbGlua2VkIHRyYWl0LCBsaWtlIGhlbW9waGlsaWEuIEZvciBjbGFyaXR5LCBsZXQmIzgyMTc7cyBoYXZlICYjODIyMDtIJiM4MjIxOyBzdGFuZCBmb3IgdGhlIG5vcm1hbCBhbGxlbGUgYW5kICYjODIyMDtoJiM4MjIxOyBmb3IgdGhlIGFsbGVsZSB0aGF0IGNhdXNlcyB0aGUgcmVjZXNzaXZlIGNvbmRpdGlvbi4gVGhlIGNoaWxkIGF0IA==Pw==IGhhcyBhIGZhdGhlciB3aG8gaGFzIHRoZSB0cmFpdDogdGhlcmVmb3JlIHRoZSBmYXRoZXImIzgyMTc7cyBnZW5vdHlwZSBpcyBYaA==WS4gVGhlIGNoaWxkJiM4MjE3O3MgbW90aGVyIGhhZCBhIGJyb3RoZXIgd2hvIGhhZCB0aGUgcmVjZXNzaXZlIHRyYWl0LiBUaGVyZWZvcmUgPyYjODIxNztzIGdyYW5kbW90aGVyIGhhZCB0byBiZSBhIGNhcnJpZXIsIGFuZCA/JiM4MjE3O3MgbW90aGVyIGhhcyBhIDUwJSBjaGFuY2Ugb2YgYmVpbmcgYSBjYXJyaWVyLg==

[Qq]

Now use the rule of multiplication. There’s a 1/2 chance that “?” will be a boy (true of any child, right?). There’s a 1/2 chance that the mother is a carrier. And, if she’s a carrier, there’s a 1/2 chance that she’ll pass on her Xh allele. Multiply all of these probabilities together, and you’ll have your answer.

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcXVlc3Rpb24uIEltYWdpbmUgdGhhdCBpdCYjODIxNztzIGEgcmVjZXNzaXZlLCBzZXgtbGlua2VkIHRyYWl0LCBsaWtlIGhlbW9waGlsaWEuIEZvciBjbGFyaXR5LCBsZXQmIzgyMTc7cyBoYXZlICYjODIyMDtIJiM4MjIxOyBzdGFuZCBmb3IgdGhlIG5vcm1hbCBhbGxlbGUgYW5kICYjODIyMDtoJiM4MjIxOyBmb3IgdGhlIGFsbGVsZSB0aGF0IGNhdXNlcyB0aGUgcmVjZXNzaXZlIGNvbmRpdGlvbi4gVGhlIGNoaWxkIGF0IA==Pw==IGhhcyBhIGZhdGhlciB3aG8gaGFzIHRoZSB0cmFpdDogdGhlcmVmb3JlIHRoZSBmYXRoZXImIzgyMTc7cyBnZW5vdHlwZSBpcyBYaA==WS4gVGhlIGNoaWxkJiM4MjE3O3MgbW90aGVyIGhhZCBhIGJyb3RoZXIgd2hvIGhhZCB0aGUgcmVjZXNzaXZlIHRyYWl0LiBUaGVyZWZvcmUgPyYjODIxNztzIGdyYW5kbW90aGVyIGhhZCB0byBiZSBhIGNhcnJpZXIsIGFuZCA/JiM4MjE3O3MgbW90aGVyIGhhcyBhIDUwJSBjaGFuY2Ugb2YgYmVpbmcgYSBjYXJyaWVyLg==

[Qq]

Now use the rule of multiplication. There’s a 1/2 chance that “?” will be a boy (true of any child, right?). There’s a 1/2 chance that the mother is a carrier. And, if she’s a carrier, there’s a 1/2 chance that she’ll pass on her Xh allele. Multiply all of these probabilities together, and you’ll have your answer.

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcXVlc3Rpb24uIEltYWdpbmUgdGhhdCBpdCYjODIxNztzIGEgcmVjZXNzaXZlLCBzZXgtbGlua2VkIHRyYWl0LCBsaWtlIGhlbW9waGlsaWEuIEZvciBjbGFyaXR5LCBsZXQmIzgyMTc7cyBoYXZlICYjODIyMDtIJiM4MjIxOyBzdGFuZCBmb3IgdGhlIG5vcm1hbCBhbGxlbGUgYW5kICYjODIyMDtoJiM4MjIxOyBmb3IgdGhlIGFsbGVsZSB0aGF0IGNhdXNlcyB0aGUgcmVjZXNzaXZlIGNvbmRpdGlvbi4gVGhlIGNoaWxkIGF0IA==Pw==IGhhcyBhIGZhdGhlciB3aG8gaGFzIHRoZSB0cmFpdDogdGhlcmVmb3JlIHRoZSBmYXRoZXImIzgyMTc7cyBnZW5vdHlwZSBpcyBYaA==WS4gVGhlIGNoaWxkJiM4MjE3O3MgbW90aGVyIGhhZCBhIGJyb3RoZXIgd2hvIGhhZCB0aGUgcmVjZXNzaXZlIHRyYWl0LiBUaGVyZWZvcmUgPyYjODIxNztzIGdyYW5kbW90aGVyIGhhZCB0byBiZSBhIGNhcnJpZXIsIGFuZCA/JiM4MjE3O3MgbW90aGVyIGhhcyBhIDUwJSBjaGFuY2Ugb2YgYmVpbmcgYSBjYXJyaWVyLg==

[Qq]

Now use the rule of multiplication. There’s a 1/2 chance that “?” will be a boy (true of any child, right?). There’s a 1/2 chance that the mother is a carrier. And, if she’s a carrier, there’s a 1/2 chance that she’ll pass on her Xh allele. Multiply all of these probabilities together, and you’ll have your answer.

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|23544c7dd8f5b5″ question_number=”15″ topic=”5.4.Non-Mendelian_Genetics”] Hemophilia A is an X-linked recessive disorder in which the ability of the blood to clot is severely reduced. Hemophilia is caused by a mutation in the gene for the clotting component Factor VIII.

Jasmine’s brother has hemophilia A, but neither Jasmine nor anyone else in her family show symptoms of the disorder.

If Jasmine has a son, what is the probability that he will have hemophilia?

[c]IDAlIA==[Qq][c]IDI1 JSA=[Qq][c]IDUwJSA=[Qq][c]IDc1JSA=[Qq][c]IDEwMCU=

Cg==[Qq]

[f]IE5vLiBCZWNhdXNlIEphc21pbmUmIzgyMTc7cyBicm90aGVyIGlzIGEgaGVtb3BoaWxpYWMsIHdlIGtub3cgdGhhdCBoaXMgZ2Vub3R5cGUgaXMgWA==aA==WS4gQmVjYXVzZSB0aGUgaGVtb3BoaWxpYSBhbGxlbGUgaXMgb24gdGhlIFggY2hyb21vc29tZSwgdGhlbiBKYXNtaW5lJiM4MjE3O3MgYnJvdGhlciBoYWQgdG8gaGF2ZSBpbmhlcml0ZWQgdGhlIGNvbmRpdGlvbiBmcm9tIGhpcyBhbmQgSmFzbWluZSYjODIxNztzIG1vdGhlciwgd2hvIG11c3QgYmUgYSBjYXJyaWVyIChpdCYjODIxNztzIHN0YXRlZCBpbiB0aGUgcHJvYmxlbSB0aGF0IHNoZSBkb2VzbiYjODIxNzt0IGhhdmUgdGhlIGNvbmRpdGlvbiwgc28gc2hlIGNhbiBvbmx5IGJlIGEgY2Fycmllciku

Cg==

SWYgSmFzbWluZSYjODIxNztzIG1vdGhlciB3YXMgYSBjYXJyaWVyIGFuZCBoZXIgZmF0aGVyIHdhcyBub3QgYSBoZW1vcGhpbGlhYyAoYWxzbyBrbm93biBmcm9tIHRoZSBwcm9ibGVtKSwgdGhlbiBKYXNtaW5lIGhhcyBhIDEvMiBjaGFuY2Ugb2YgYmVpbmcgYSBjYXJyaWVyLiBBbmQgaWYgc2hlIHdlcmUgYSBjYXJyaWVyLCB0aGVuIHNoZSYjODIxNztkIGhhdmUgYSAxLzIgY2hhbmNlIG9mIHBhc3Npbmcgb24gdGhlIGFsbGVsZSB0byBoZXIgbWFsZSBjaGlsZC4gV2hhdCYjODIxNztzIDEvMiB0aW1lcyAxLzI/[Qq]

[f]IEV4Y2VsbGVudCEgVGhlcmUmIzgyMTc7cyBhIDUwJSBwcm9iYWJpbGl0eSB0aGF0IEphc21pbmUgaXMgYSBjYXJyaWVyLCBhbmQgdGhlcmUmIzgyMTc7cyBhIDUwJSBwcm9iYWJpbGl0eSB0aGF0IGEgY2FycmllciB3aWxsIHBhc3Mgb24gYW4gWCBjaHJvbW9zb21lIHdpdGggdGhlIGhlbW9waGlsaWEgYWxsZWxlIG9uIHRvIGhlciBzb24uIDUwJSBvZiA1MCUgaXMgMjUlLg==[Qq]

[f]IE5vLiBCZWNhdXNlIEphc21pbmUmIzgyMTc7cyBicm90aGVyIGlzIGEgaGVtb3BoaWxpYWMsIHdlIGtub3cgdGhhdCBoaXMgZ2Vub3R5cGUgaXMgWA==aA==WS4gQmVjYXVzZSB0aGUgaGVtb3BoaWxpYSBhbGxlbGUgaXMgb24gdGhlIFggY2hyb21vc29tZSwgdGhlbiBKYXNtaW5lJiM4MjE3O3MgYnJvdGhlciBoYWQgdG8gaGF2ZSBpbmhlcml0ZWQgdGhlIGNvbmRpdGlvbiBmcm9tIGhpcyBhbmQgSmFzbWluZSYjODIxNztzIG1vdGhlciwgd2hvIG11c3QgYmUgYSBjYXJyaWVyIChpdCYjODIxNztzIHN0YXRlZCBpbiB0aGUgcHJvYmxlbSB0aGF0IHNoZSBkb2VzbiYjODIxNzt0IGhhdmUgdGhlIGNvbmRpdGlvbiwgc28gc2hlIGNhbiBvbmx5IGJlIGEgY2Fycmllciku

Cg==

SWYgSmFzbWluZSYjODIxNztzIG1vdGhlciB3YXMgYSBjYXJyaWVyIGFuZCBoZXIgZmF0aGVyIHdhcyBub3QgYSBoZW1vcGhpbGlhYyAoYWxzbyBrbm93biBmcm9tIHRoZSBwcm9ibGVtKSwgdGhlbiBKYXNtaW5lIGhlcnNlbGYgaGFzIGEgMS8yIGNoYW5jZSBvZiBiZWluZyBhIGNhcnJpZXIuIEFuZCBpZiBzaGUgd2VyZSBhIGNhcnJpZXIsIHRoZW4gc2hlJiM4MjE3O2QgaGF2ZSBhIDEvMiBjaGFuY2Ugb2YgcGFzc2luZyBvbiB0aGUgYWxsZWxlIHRvIGhlciBtYWxlIGNoaWxkLiBXaGF0JiM4MjE3O3MgMS8yIHRpbWVzIDEvMj8=[Qq]

[f]IE5vLiBCZWNhdXNlIEphc21pbmUmIzgyMTc7cyBicm90aGVyIGlzIGEgaGVtb3BoaWxpYWMsIHdlIGtub3cgdGhhdCBoaXMgZ2Vub3R5cGUgaXMgWA==aA==WS4gQmVjYXVzZSB0aGUgaGVtb3BoaWxpYSBhbGxlbGUgaXMgb24gdGhlIFggY2hyb21vc29tZSwgdGhlbiBKYXNtaW5lJiM4MjE3O3MgYnJvdGhlciBoYWQgdG8gaGF2ZSBpbmhlcml0ZWQgdGhlIGNvbmRpdGlvbiBmcm9tIGhpcyBhbmQgSmFzbWluZSYjODIxNztzIG1vdGhlciwgd2hvIG11c3QgYmUgYSBjYXJyaWVyIChpdCYjODIxNztzIHN0YXRlZCBpbiB0aGUgcHJvYmxlbSB0aGF0IHNoZSBkb2VzbiYjODIxNzt0IGhhdmUgdGhlIGNvbmRpdGlvbiwgc28gc2hlIGNhbiBvbmx5IGJlIGEgY2Fycmllciku

Cg==

SWYgSmFzbWluZSYjODIxNztzIG1vdGhlciB3YXMgYSBjYXJyaWVyIGFuZCBoZXIgZmF0aGVyIHdhcyBub3QgYSBoZW1vcGhpbGlhYyAoYWxzbyBrbm93biBmcm9tIHRoZSBwcm9ibGVtKSwgdGhlbiBKYXNtaW5lIGhlcnNlbGYgaGFzIGEgMS8yIGNoYW5jZSBvZiBiZWluZyBhIGNhcnJpZXIuIEFuZCBpZiBzaGUgd2VyZSBhIGNhcnJpZXIsIHRoZW4gc2hlJiM4MjE3O2QgaGF2ZSBhIDEvMiBjaGFuY2Ugb2YgcGFzc2luZyBvbiB0aGUgYWxsZWxlIHRvIGhlciBtYWxlIGNoaWxkLiBXaGF0JiM4MjE3O3MgMS8yIHRpbWVzIDEvMj8=[Qq]

[f]IE5vLiBCZWNhdXNlIEphc21pbmUmIzgyMTc7cyBicm90aGVyIGlzIGEgaGVtb3BoaWxpYWMsIHdlIGtub3cgdGhhdCBoaXMgZ2Vub3R5cGUgaXMgWA==aA==WS4gQmVjYXVzZSB0aGUgaGVtb3BoaWxpYSBhbGxlbGUgaXMgb24gdGhlIFggY2hyb21vc29tZSwgdGhlbiBKYXNtaW5lJiM4MjE3O3MgYnJvdGhlciBoYWQgdG8gaGF2ZSBpbmhlcml0ZWQgdGhlIGNvbmRpdGlvbiBmcm9tIGhpcyBhbmQgSmFzbWluZSYjODIxNztzIG1vdGhlciwgd2hvIG11c3QgYmUgYSBjYXJyaWVyIChpdCYjODIxNztzIHN0YXRlZCBpbiB0aGUgcHJvYmxlbSB0aGF0IHNoZSBkb2VzbiYjODIxNzt0IGhhdmUgdGhlIGNvbmRpdGlvbiwgc28gc2hlIGNhbiBvbmx5IGJlIGEgY2Fycmllciku

Cg==

SWYgSmFzbWluZSYjODIxNztzIG1vdGhlciB3YXMgYSBjYXJyaWVyIGFuZCBoZXIgZmF0aGVyIHdhcyBub3QgYSBoZW1vcGhpbGlhYyAoYWxzbyBrbm93biBmcm9tIHRoZSBwcm9ibGVtKSwgdGhlbiBKYXNtaW5lIGhlcnNlbGYgaGFzIGEgMS8yIGNoYW5jZSBvZiBiZWluZyBhIGNhcnJpZXIuIEFuZCBpZiBzaGUgd2VyZSBhIGNhcnJpZXIsIHRoZW4gc2hlJiM4MjE3O2QgaGF2ZSBhIDEvMiBjaGFuY2Ugb2YgcGFzc2luZyBvbiB0aGUgYWxsZWxlIHRvIGhlciBtYWxlIGNoaWxkLiBXaGF0JiM4MjE3O3MgMS8yIHRpbWVzIDEvMj8=

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|2354273d1ab5b5″ question_number=”16″ unit=”5.Heredity” topic=”5.4.Non-Mendelian_Genetics”] Which of the following is true about mitochondrial DNA?

[c]IEl0IGlzIGluc2VydGVkIGludG8gdGhlIFggY2hyb21vc29tZS4=[Qq]

[f]IE5vLiBNaXRvY2hvbmRyaWFsIEROQSBpcyBleGFjdGx5IHdoYXQgaXQgc291bmRzIGxpa2U6IGEgbG9vcCBvZiBETkEgd2l0aGluIHRoZSBtaXRvY2hvbmRyaWEgb2YgYWxsIG9mIGFuIG9yZ2FuaXNtJiM4MjE3O3MgY2VsbHMuIEl0IHdvbiYjODIxNzt0IGJlIGluc2VydGVkIGludG8gdGhlIFggY2hyb21vc29tZSAob3IgYW55IG90aGVyIGNocm9tb3NvbWUpLg==[Qq]

[c]IEl0IGlzIGluaGVyaXRlZCBvbmx5IG Zyb20gdGhlIGZlbWFsZSBwYXJlbnQu[Qq]

[f]IENvcnJlY3QuIFdoZW4gYSBzcGVybSBmZXJ0aWxpemVzIGFuIGVnZywgb25seSB0aGUgc3Blcm0mIzgyMTc7cyBudWNsZXVzIGlzIGFsbG93ZWQgdG8gZW50ZXIuIFRoZSBzcGVybSYjODIxNztzIG1pdG9jaG9uZHJpYSwgbG9jYXRlZCBhdCB0aGUgYmFzZSBvZiB0aGUgZmxhZ2VsbHVtLCBhcmUgbGVmdCBiZWhpbmQuIENvbnNlcXVlbnRseSwgYWxsIG9mIGFuIGFuaW1hbCYjODIxNztzIG1pdG9jaG9uZHJpYSBhcmUgZnJvbSBpdHMgbW90aGVyLg==[Qq]

[c]IEl0IGV2b2x2ZXMgc2xvd2VyIHRoYW4gdGhlIGdlbmVzIGluIHRoZSBudWNsZXVzLg==[Qq]

[f]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[Qq]

[c]IEl0IHdhcyBkZXJpdmVkIGZyb20gdGhlIGdsb2JpbiBnZW5lcyBhcyBhbiBleHRyYSBjb3B5Lg==[Qq]

[f]IE5vLiBHbG9iaW5zIGFyZSBwcm90ZWlucyBpbnZvbHZlZCBpbiBiaW5kaW5nIHdpdGggb3IgdHJhbnNwb3J0aW5nIG94eWdlbi4gVGhlcmUgaGFzIGJlZW4gZHVwbGljYXRpb24gb2YgZ2xvYmluIGdlbmVzIHRocm91Z2hvdXQgb3VyIGdlbm9tZSwgcmVzdWx0aW5nIGluIHNldmVyYWwgdHlwZXMgb2YgaGVtb2dsb2JpbiwgYW5kIG90aGVyIHByb3RlaW5zIHN1Y2ggYXMgbXlvZ2xvYmluIGFuZCBuZXVyb2dsb2Jpbi4gQnV0IG5vbmUgb2YgdGhvc2UgcHJvdGVpbnMgYXJlIGNvbm5lY3RlZCB0byBtaXRvY2hvbmRyaWFsIEROQS4=[Qq]

[c]IEl0IGZpcnN0IGFwcGVhcmVkIGluIGh1bWFucyBhbmQgaXMgbm90IGZvdW5kIGluIG90aGVyIG1lbWJlcnMgb2YgdGhlIEFuaW1hbGlhIGtpbmdkb20u[Qq]

[f]IE5vLiBNaXRvY2hvbmRyaWFsIEROQSBpcyBub3Qgb25seSBwcmVzZW50IGluIGFsbCBhbmltYWxzLCBidXQgaXQmIzgyMTc7cyBwcmVzZW50IGluIGFueSBvcmdhbmlzbSB0aGF0IGhhcyBtaXRvY2hvbmRyaWEgKG1lYW5pbmcgYWxsIGV1a2FyeW90ZXMpLg==

Cg==

Cg==

[Qq]

[q json=”true” multiple_choice=”true” unit=”5.Heredity” topic=”5.5.Environmental_Effects_on_Phenotype” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|23540b4c8c05b5″ question_number=”17″] A marine biologist measured the average temperature of 18 loggerhead turtle nests and determined the sex ratio of the eggs that hatched. Which of the following is an accurate interpretation of the results?

[c]IEF0IGFueSB0ZW1wZXJhdHVyZSBhYm92ZSAzMCDCsEMgYWxsIHR1cnRsZXMgdGhhdCBoYXRjaCB3aWxsIGJlIG1hbGUu[Qq]

[f]IE5vLiBMb29rIGF0IHRoZSBZLWF4aXMsIHdoaWNoIGlzIHRoZSByYXRpbyBpbiAlIGZlbWFsZXMuIEFib3ZlIDMwIMKwQywgYWxtb3N0IGFsbCB0dXJ0bGUgaGF0Y2hsaW5ncyB3aWxsIGJlIGZlbWFsZS4=[Qq]

[c]IEF0IGFueSB0ZW1wZXJhdHVyZSBiZWxvdyAyNSDCsEMgYWxsIHR1cnRsZXMgdGhhdCBoYXRjaCB3aWxsIGJlIGZlbWFsZS4=[Qq]

[f]IE5vLiBMb29rIGF0IHRoZSBZLWF4aXMsIHdoaWNoIGlzIHRoZSByYXRpbyBpbiAlIGZlbWFsZXMuIEF0IDI1IMKwQywgbm9uZSBvZiB0aGUgdHVydGxlIGhhdGNobGluZ3Mgd2lsbCBiZSBmZW1hbGUu[Qq]

[c]IEF0IDI5IMKwQywgdGhlcmUgaXMgYSA1MCUgY2hhbmNlIHRoYXQg YW55IHR1cnRsZSB0aGF0IGhhdGNoZXMgd2lsbCBiZSBtYWxlLg==[Qq]

[f]IEV4Y2VsbGVudC4gU2V4IGRldGVybWluYXRpb24gaW4gbG9nZ2VyaGVhZHMgaXMgZGV0ZXJtaW5lZCBieSB0ZW1wZXJhdHVyZS4gQWJvdmUgMjkgwrBDLCBtb3N0IHR1cnRsZSBoYXRjaGxpbmdzIHdpbGwgYmUgZmVtYWxlLiBCZWxvdyB0aGF0IHRlbXBlcmF0dXJlLCBtb3N0IHdpbGwgYmUgbWFsZS4gQXQgMjkgwrBDLCB0aGVyZSYjODIxNztzIGEgNTAlIGNoYW5jZSBvZiBhIGhhdGNobGluZyBkZXZlbG9waW5nIGludG8gZWl0aGVyIHNleC4=[Qq]

[c]IFNleCBkZXRlcm1pbmF0aW9uIGluIExvZ2dlcmhlYWQgdHVydGxlcyBpcyBkZXRlcm1pbmVkIGdlbmV0aWNhbGx5LCBub3QgYnkgdGhlIGVudmlyb25tZW50Lg==[Qq]

[f]IE5vLiBJZiB0aGUgc2V4IHdlcmUgZGV0ZXJtaW5lZCBnZW5ldGljYWxseSAoYXMgd2l0aCB0aGUgWFgvWFkgc3lzdGVtIGluIG1hbW1hbHMgb3IgdGhlIFpaL1pXIHN5c3RlbSBpbiBiaXJkcyksIHRoZW4gdGhlIGVudmlyb25tZW50YWwgdGVtcGVyYXR1cmUgd291bGRuJiM4MjE3O3QgaGF2ZSBhbiBlZmZlY3Qgb24gdGhlIHNleCBvZiB0aGUgb2Zmc3ByaW5nLg==

Cg==

[Qq]

[q json=”true” multiple_choice=”true” unit=”5.Heredity” topic=”5.5.Environmental_Effects_on_Phenotype” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|2353b2d2c82db5″ question_number=”18″] A botanist observes two plants with the same flower shape and structure, but different colors.  One has blue petals and the other red.  The botanist hypothesizes that the color is due to the soil. Which of the following experiments would best test if environmental differences accounted for the difference in phenotype?

[c]IENvbGxlY3Qgc2VlZHMgZnJvbSBlYWNoIHBsYW50LsKgIFNvdyAxMCBzZWVkcyBmcm9tIGVhY2ggcGxhbnQgaW4gdGhlIHNvaWwgdGhlIHBhcmVudGFsIHBsYW50IHdhcyBOT1QgZ3Jvd2luZyBpbi7CoCBPbmNlIHRoZSBwbGFudHMgcmVhY2ggbWF0dXJpdHkgYW5kIGZsb3dlciwgc2NvcmUgdGhlIG51bWJlciBvZiByZWQgYW5kIGJsdWUgZmxvd2VycyB0aGF0IHJlc3VsdC4=[Qq]

[f]IE5vLiBXaGF0IHlvdSYjODIxNztyZSB0cnlpbmcgdG8gdGVzdCBpcyB0aGUgZWZmZWN0IG9mIHRoZSBlbnZpcm9ubWVudCBhbG9uZSBvbiB0aGUgcGhlbm90eXBlLiBUaGUgY3JlYXRpb24gb2Ygc2VlZHMgaW52b2x2ZXMgbWVpb3NpcyBhbmQgZmVydGlsaXphdGlvbiBhbmQgY3JlYXRlcyBvZmZzcHJpbmcgdGhhdCBhcmUgZ2VuZXRpY2FsbHkgZGlmZmVyZW50IGZyb20gdGhlaXIgcGFyZW50cy4gSGVyZSYjODIxNztzIGEgaGludCBmb3IgdGhlIG5leHQgdGltZSB5b3Ugc2VlIHRoaXMgcXVlc3Rpb24uIEhvdyBjb3VsZCB5b3UgZWxpbWluYXRlIGdlbmV0aWMgZGlmZmVyZW5jZXMgZnJvbSB5b3VyIGV4cGVyaW1lbnQ/[Qq]

[c]IENvbGxlY3Qgc2VlZHMgZnJvbSBlYWNoIHBsYW50IGFuZCBzb3cgMTAgc2VlZHMgZnJvbSBlYWNoIHBsYW50IGluIGNvbW1lcmNpYWwgcG90dGluZyBzb2lsLsKgIE9uY2UgdGhlIHBsYW50cyByZWFjaCBtYXR1cml0eSBhbmQgZmxvd2VyLCBzY29yZSB0aGUgbnVtYmVyIG9mIHJlZCBhbmQgYmx1ZSBmbG93ZXJzIHRoYXQgcmVzdWx0Lg==[Qq]

[f]IE5vLiBXaGF0IHlvdSYjODIxNztyZSB0cnlpbmcgdG8gdGVzdCBpcyB0aGUgZWZmZWN0IG9mIHRoZSBlbnZpcm9ubWVudCBhbG9uZSBvbiB0aGUgcGhlbm90eXBlLiBUaGUgY3JlYXRpb24gb2Ygc2VlZHMgaW52b2x2ZXMgbWVpb3NpcyBhbmQgZmVydGlsaXphdGlvbiBhbmQgY3JlYXRlcyBvZmZzcHJpbmcgdGhhdCBhcmUgZ2VuZXRpY2FsbHkgZGlmZmVyZW50IGZyb20gdGhlaXIgcGFyZW50cy4gSGVyZSYjODIxNztzIGEgaGludCBmb3IgdGhlIG5leHQgdGltZSB5b3Ugc2VlIHRoaXMgcXVlc3Rpb24uIEhvdyBjb3VsZCB5b3UgZWxpbWluYXRlIGdlbmV0aWMgZGlmZmVyZW5jZXMgZnJvbSB5b3VyIGV4cGVyaW1lbnQ/[Qq]

[c]IENsb25lIGVhY2ggcGxhbnQgYnkgdGFraW5nIHZlZ2V0YXRpdmUgY3V0dGluZ3MuwqAgUGxhbnQgdGhyZWUgY3V0dGluZ3MgZnJvbSBlYWNoIGFkdWx0IHBsYW50IGluIHRoZSBzb2lsIHdoZXJlIHRoZSByZWQgcG xhbnQgd2FzIGdyb3dpbmcgYW5kIHRocmVlIHdoZXJlIHRoZSBibHVlIHBsYW50IHdhcyBncm93aW5nLiBTY29yZSB0aGUgbnVtYmVyIG9mIHJlZCBhbmQgYmx1ZSBmbG93ZXJzIGluIGVhY2ggc29pbCB0eXBlLg==[Qq]

[f]IE5pY2Ugam9iLiBTaW5jZSB5b3UmIzgyMTc7cmUgdXNpbmcgdmVnZXRhdGl2ZSBjdXR0aW5ncywgeW91JiM4MjE3O3ZlIGtlcHQgdGhlIGdlbm90eXBlIG9mIGVhY2ggcGxhbnQgY29uc3RhbnQuIFRoZXJlZm9yZSwgdGhpcyBtZXRob2Qgd291bGQgYWxsb3cgeW91IHRvIHRlc3QgZm9yIHRoZSBlZmZlY3Qgb2YgdGhlIGVudmlyb25tZW50IGFsb25lIG9uIHRoZSBwaGVub3R5cGUu[Qq]

[c]IENsb25lIGVhY2ggcGxhbnQgYnkgdGFraW5nIHZlZ2V0YXRpdmUgY3V0dGluZ3MuwqAgUGxhbnQgdGhyZWUgY3V0dGluZ3MgZnJvbSBlYWNoIGFkdWx0IHBsYW50IGluIGNvbW1lcmNpYWwgcG90dGluZyBzb2lsIGFuZCBncm93IHRoZW0gaW4gdGhlIGdyZWVuaG91c2UuwqAgQ3Jvc3MgdGhlbSB0byBjcmVhdGUgYW4gRg==MQ==IGdlbmVyYXRpb24gYW5kIHRoZW4gYmFja2Nyb3NzIHRvIHRoZSBwYXJlbnRhbCBwbGFudHMuwqAgU2NvcmUgdGhlIG51bWJlciBvZiByZWQgYW5kIGJsdWUgZmxvd2VycyBpbiBlYWNoIGJhY2tjcm9zcy4=[Qq]

[f]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

Cg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|2353923a21b5b5″ question_number=”19″ topic=”5.6.Chromosomal_Inheritance”] Down syndrome and and Klinefelter’s syndrome have which of the following in common?

[c]IEJvdGggaW52b2x2ZSB0aGUgWCBvciBZIGNocm9tb3NvbWVz[Qq]

[f]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[Qq]

[c]IEJvdGggYXJlIHRoZSByZXN1bHQgb2YgcG9pbnQgbXV0YXRpb25zLg==[Qq]

[f]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[Qq]

[c]IEJvdGggYXJlIGFzc29jaWF0ZWQgd2l0aCBj aGFuZ2VzIGluIGNocm9tb3NvbWUgbnVtYmVy[Qq]

[f]IFllcy4gQm90aCBEb3duIHN5bmRyb21lIGFuZCBLbGluZWZlbHRlciYjODIxNztzIHN5bmRyb21lIGludm9sdmUgY2hhbmdlcyBpbiBjaHJvbW9zb21lIG51bWJlci4gRG93biBTeW5kcm9tZSBvY2N1cnMgd2hlbiBhIHBlcnNvbiByZWNlaXZlcyB0aHJlZSBjb3BpZXMgb2YgY2hyb21vc29tZSAyMSwgcmF0aGVyIHRoYW4gdHdvLiBLbGluZWZlbHRlciYjODIxNztzIHN5bmRyb21lIG9jY3VycyB3aGVuIGEgY2hyb21vc29tYWxseSBtYWxlIHBlcnNvbiBoYXMgYW4gZXh0cmEgWCBjaHJvbW9zb21lLg==[Qq]

[c]IEJvdGggY29uZGl0aW9ucyBhcmUgY2F1c2VkIGJ5IGF1dG9zb21hbCBkb21pbmFudCBhbGxlbGVz[Qq]

[f]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

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|2352a0154d15b5″ question_number=”20″ topic=”5.6.Chromosomal_Inheritance”] Assuming that the chromosomes shown in the germ cell on the top of the diagram represent one of the 22 homologous pairs of autosomes found in human beings, then which of the zygotes on the bottom row could develop into an individual with Down syndrome?

[c]IGEg[Qq][c]IGIg[Qq][c]IG Mg[Qq][c]IGQ=

Cg==[Qq]

[f]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[Qq]

[f]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[Qq]

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[x][restart]

[/qwiz]


4. Unit 5 Cumulative Multiple Choice Quiz 2

[qwiz style=”width: 550px !important; min-height: 400px !important;” dataset=”Unit 5 Cumulative MC Dataset 2″ qrecord_id=”sciencemusicvideosMeister1961-Unit 5 Cumulative MC Quiz 2″]

[h]Unit 5 MC Quiz 2

[i]

[q json=”true” multiple_choice=”true” unit=”5.Heredity” topic=”5.2.Meiosis_and_Genetic_Diversity” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a65c1a082b8d5″ question_number=”1″] Meiosis enhances genetic diversity by

[c]IHJlcGxpY2F0aW5nIGNocm9tb3NvbWFsIEROQSwgYWxsb3dpbmcgZm9yIGdlbmUgZHVwbGljYXRpb24gYW5kIGZvciBzdWJzZXF1ZW50IGdlbmV0aWMgZGl2ZXJzaWZpY2F0aW9u[Qq]

[f]IE5vLiBXaGlsZSBtZWlvc2lzIGRvZXMgaW52b2x2ZSB0aGUgcmVwbGljYXRpb24gb2YgRE5BLCBpdCBkb2VzbiYjODIxNzt0IG5vcm1hbGx5IGludm9sdmUgZ2VuZSBkdXBsaWNhdGlvbi4gSGVyZSYjODIxNztzIGEgaGludC4gV2hhdCYjODIxNztzIGhhcHBlbmluZyBpbiB0aGUgZGlhZ3JhbSBiZWxvdz8=

Cg==

[Qq]

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Cg==

Cg==

[Qq]Here’s a hint. What’s happening in the diagram below?

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[f]IE5vLiBUaGUgdHJhbnNsb2NhdGlvbiBvZiBjaHJvbW9zb21lIHNlZ21lbnRzIGlzIGEgdHlwZSBvZiBjaHJvbW9zb21hbCBtdXRhdGlvbi4=

Cg==

SGVyZSYjODIxNztzIGEgaGludC4gV2hhdCYjODIxNztzIGhhcHBlbmluZyBpbiB0aGUgZGlhZ3JhbSBiZWxvdz8=

Cg==

[Qq]

[c]IGV4Y2hhbmdpbmcgZ2VuZXMgYmV0d2VlbiBob21vbG9nb3VzIHNpc3RlciBjaH JvbWF0aWRzLCBjcmVhdGluZyBuZXcgY29tYmluYXRpb25zIG9mIGFsbGVsZXMu[Qq]

[f]IE5pY2Ugam9iLiBBcyB5b3UgY2FuIHNlZSBiZWxvdywgdGhhdCYjODIxNztzIGFuIGltcG9ydGFudCBwYXJ0IG9mIGhvdyBtZWlvc2lzIGNyZWF0ZXMgdmFyaWF0aW9uLg==

Cg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a65622a9b34d5″ question_number=”2″ topic=”5.1.Meiosis”] The diagram below shows the amount of DNA per cell in a dividing cell. Which roman numeral shows the amount of DNA in an unfertilized egg cell?

[c]IEkg[Qq][c]IElJSSA=[Qq][c]IElWIA==[Qq][c]IF ZJ

Cg==[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMuIFlvdSYjODIxNztyZSBpbnN0cnVjdGVkIHRvIGxvb2sgZm9yIHRoZSBhbW91bnQgb2YgRE5BIGluIGFuIA==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[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMuIFlvdSYjODIxNztyZSBpbnN0cnVjdGVkIHRvIGxvb2sgZm9yIHRoZSBhbW91bnQgb2YgRE5BIGluIGFuIA==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[Qq]

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[f]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

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a654ae2244cd5″ question_number=”3″ topic=”5.1.Meiosis”] In the diagram below, maternal chromosomes are colored red, and paternal chromosomes are colored blue.

If a cell with a diploid number of six is undergoing a type of cell division, then which of the diagrams below could represent the arrangement of chromosomes in one of the daughter cells at some stage of the cell division process?

[c]IEEg[Qq][c]IE Ig[Qq][c]IEMg[Qq][c]IEQ=

Cg==[Qq]

[f]IE5vLiBJZiB0aGUgZGlwbG9pZCBudW1iZXIgaXMgc2l4LCB0aGVuIHRoZSBjZWxsIHdvdWxkIG5ldmVyIGhhdmUgbW9yZSB0aGFuIHNpeCBjaHJvbW9zb21lcy4gSW4gdGhpcyBkaWFncmFtLCB5b3Ugc2VlIDEyLiBSZW1lbWJlciB0aGF0IHRoZSBuZXh0IHRpbWUgeW91IHNlZSB0aGlzIHF1ZXN0aW9uLg==[Qq]

[f]IE5pY2Ugam9iLiBJZiB0aGUgY2VsbCB3ZXJlIGdvaW5nIHRocm91Z2ggbWVpb3NpcywgdGhlbiB5b3UgbWlnaHQgc2VlIHRocmVlIGRvdWJsZWQgY2hyb21vc29tZXMgYWZ0ZXIgbWVpb3NpcyAxICh3aGljaCBpcyB3aGF0IHlvdSBzZWUgaGVyZSkuIEJlY2F1c2Ugb2YgaW5kZXBlbmRlbnQgYXNzb3J0bWVudCwgaXQmIzgyMTc7cyBjb21wbGV0ZWx5IHBvc3NpYmxlIHRoYXQgdHdvIG9mIHRoZSBjaHJvbW9zb21lcyB3b3VsZCBiZSBtYXRlcm5hbCwgYW5kIG9uZSB3b3VsZCBiZSBwYXRlcm5hbC4=[Qq]

[f]IE5vLiBJZiBhIGNlbGwgaGFzIGEgZGlwbG9pZCBudW1iZXIgb2Ygc2l4LCB0aGVuIHlvdSBjb3VsZCBzZWUgc2l4IGNocm9tb3NvbWVzIGxpbmluZyB1cCBkdXJpbmcgbWV0YXBoYXNlIDEgKHRoZSBhbGlnbm1lbnQgcGhhc2UpIG9mIG1pdG9zaXMuIFRoZSBwcm9ibGVtIGlzIHRoYXQgaGFsZiB0aGUgY2hyb21vc29tZXMgd291bGQgaGF2ZSB0byBiZSBwYXRlcm5hbCwgYW5kIGhhbGYgd291bGQgYmUgbWF0ZXJuYWwuIEhlcmUgeW91IGhhdmUgZm91ciBtYXRlcm5hbCBjaHJvbW9zb21lcyBhbmQgdHdvIHBhdGVybmFsIG9uZXMuIFJlbWVtYmVyIHRoYXQgdGhlIG5leHQgdGltZSB5b3Ugc2VlIHRoaXMgcXVlc3Rpb24u[Qq]

[f]IE5vLiBXaGlsZSBhIGRpcGxvaWQgY2VsbCB3b3VsZCBsaW5lIHVwIGl0cyBjaHJvbW9zb21lcyBpbiBhIHdheSB0aGF0JiM4MjE3O3Mgc2ltaWxhciB0byB0aGlzIGR1cmluZyBtZXRhcGhhc2UgMSBvZiBtZWlvc2lzIDEsIHRoZXJlJiM4MjE3O3Mgb25lIHByb2JsZW06IGZvdXIgb2YgdGhlIGNocm9tb3NvbWVzIGFyZSBwYXRlcm5hbCwgYW5kIHR3byBhcmUgbWF0ZXJuYWwuIEZvciB0aGlzIHRvIGJlIGNvcnJlY3QsIGhhbGYgd291bGQgaGF2ZSB0byBiZSBtYXRlcm5hbCwgYW5kIGhhbGYgcGF0ZXJuYWwuIFJlbWVtYmVyIHRoYXQgdGhlIG5leHQgdGltZSB5b3Ugc2VlIHRoaXMgcXVlc3Rpb24u

Cg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” topic=”5.2.Meiosis_and_Genetic_Diversity” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a653145a180d5″ question_number=”4″] A diploid cell has four pairs of homologous chromosomes.  If the cell undergoes meiosis, how many possible combinations of chromosomes are there in the resulting haploid nuclei?

[c]IDE2 IA==[Qq][c]IDMyIA==[Qq][c]IDgg[Qq][c]IDQ=

Cg==[Qq]

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[f]IE5vLiBJbiBtZWlvc2lzLCB0aGUgY2hyb21vc29tZSBudW1iZXIgaXMgcmVkdWNlZCBmcm9tIGRpcGxvaWQgKHR3byBzZXRzKSB0byBoYXBsb2lkIChvbmUgc2V0KS4gRWFjaCBnYW1ldGUgcmVjZWl2ZXMgb25lIG1lbWJlciBvZiBlYWNoIGhvbW9sb2dvdXMgcGFpci4gRm9yIGVhY2ggcGFpciwgdGhlcmUgYXJlIHR3byBwb3NzaWJpbGl0aWVzIGZvciB3aGljaCBjaHJvbW9zb21lIGdvZXMgdG8gdGhlIGdhbWV0ZSwgYW5kIGVhY2ggcGFpciBpcyBpbmRlcGVuZGVudCBvZiBldmVyeSBvdGhlciBwYWlyLiBUaGUgZ2VuZXJhbCBmb3JtdWxhIGlzIDI=bg==LCB3aGVyZSAmIzgyMjA7biYjODIyMTsgaXMgdGhlIG51bWJlciBvZiBob21vbG9nb3VzIHBhaXJzLiDCoFdoYXQmIzgyMTc7cyAyNCA=KG9yIDIgeCAyIHggMiB4IDIpPw==[Qq]

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[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a6517a91eb4d5″ question_number=”5″ topic=”5.2.Meiosis_and_Genetic_Diversity”] Assuming that the chromosomes shown in the germ cell on the top of this diagram represent X chromosomes, then which of the zygotes on the bottom row would develop into an individual with Turner syndrome?

[c]IGEg[Qq][c]IGIg[Qq][c]IGMg[Qq][c]IG Q=

Cg==[Qq]

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[f]IEV4Y2VsbGVudC4gVGhlIHp5Z290ZSBhdCAmIzgyMjA7ZCYjODIyMTsgaXMgbWlzc2luZyBhIGNocm9tb3NvbWUuIFNpbmNlIHBlb3BsZSB3aXRoIFR1cm5lciBzeW5kcm9tZSBhcmUgZmVtYWxlcyB3aG8gYXJlIG1pc3NpbmcgYW4gWCBjaHJvbW9zb21lLCB0aGVuIHRoYXQgY2VsbCBmaXRzIHRoZSBkZXNjcmlwdGlvbiBvZiBhIHp5Z290ZSB0aGF0IGNvdWxkIGxlYWQgdG8gYSBwZXJzb24gd2l0aCBUdXJuZXIgc3luZHJvbWUu

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a64f9648420d5″ question_number=”6″ topic=”5.2.Meiosis_and_Genetic_Diversity”] The giant Pacific Octopus can lay up to 100,000 eggs (each about the size of a grain of rice). All of the eggs are fertilized with sperm from the same father. From the list below, the process that does not contribute to genetic diversity in the offspring is

[c]IGNyb3NzaW5nLW92ZXIgZHVyaW5nIHByb3BoYXNlIDEu[Qq]

[f]IE5vLiBDcm9zc2luZyBvdmVyIG9jY3VycyBkdXJpbmcgbWVpb3Npcy4gSXQgaW52b2x2ZXMgbWl4aW5nIG9mIG1hdGVybmFsIGFuZCBwYXRlcm5hbCBhbGxlbGVzLCBhbmQgaXQmIzgyMTc7cyBhIG1ham9yIHNvdXJjZSBvZiB2YXJpYXRpb24uIFdoYXQmIzgyMTc7cyBvbiB0aGUgbGlzdCB0aGF0IHdvdWxkIA==ZGVjcmVhc2U=IHZhcmlhdGlvbj8=[Qq]

[c]IEROQSByZXBhaXIgZHVyaW5nIHRoZSBTIHBo YXNlIHRoYXQgcHJlY2VkZXMgbWVpb3Npcy4=[Qq]

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Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a64dd73f570d5″ question_number=”7″ topic=”5.3.Mendelian_Genetics”] In cats, the pattern of tabby strips can be vertical (referred to as “mackerel tabby”) or blotched (referred to as “blotched tabby”), as shown below.

When true-breeding mackerel tabbies are bred with true-breeding blotched tabbies, all of the F1 offspring are mackerel tabbies.

An F1 mackerel tabby is crossed with a blotched tabby. What’s the probability that a kitten resulting from this mating will be blotched tabby?

[c]IDAlIA==[Qq][c]IDI1JSA=[Qq][c]IDUw JSA=[Qq][c]IDc1JSA=[Qq][c]IDEwMCU=

Cg==[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcHJvYmxlbS4gWW91IGtub3cgdGhhdCB0aGUgbWFja2VyZWwgdGFiYnkgdHJhaXQgaXMgZG9taW5hbnQgKGJlY2F1c2UgYWxsIG9mIHRoZSBGMQ==IG9mZnNwcmluZyBhcmUgbWFja2VyZWwpLiBBbmQgeW91IGFsc28ga25vdyB0aGF0IHRoZSBGMQ==cyBhcmUgaGV0ZXJvenlnb3Rlcy4gQmVjYXVzZSB0aGUgYWxsZWxlIGZvciBibG90Y2hlZCBpcyByZWNlc3NpdmUsIGEgYmxvdGNoZWQgdGFiYnkgaGFzIHRvIGJlIGhvbW96eWdvdXMu

[Qq]

If you let the letter M represent the allele for “mackerel” and m represent the allele for “blotched, then you’re crossing Mm x mm, and you’d set up the Punnett square as follows:

M m
m
m

Now you draw it out, and figure out what percentage of the offspring will have genotype mm (the only genotype you can have if you’re blotched tabby).

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcHJvYmxlbS4gWW91IGtub3cgdGhhdCB0aGUgbWFja2VyZWwgdGFiYnkgdHJhaXQgaXMgZG9taW5hbnQgKGJlY2F1c2UgYWxsIG9mIHRoZSBGMQ==IG9mZnNwcmluZyBhcmUgbWFja2VyZWwpLiBBbmQgeW91IGFsc28ga25vdyB0aGF0IHRoZSBGMQ==cyBhcmUgaGV0ZXJvenlnb3Rlcy4gQmVjYXVzZSB0aGUgYWxsZWxlIGZvciBibG90Y2hlZCBpcyByZWNlc3NpdmUsIGEgYmxvdGNoZWQgdGFiYnkgaGFzIHRvIGJlIGhvbW96eWdvdXMu

[Qq]

If you let the letter M represent the allele for “mackerel” and m represent the allele for “blotched, then you’re crossing Mm x mm, and you’d set up the Punnett square as follows:

M m
m
m

Now you draw it out, and figure out what percentage of the offspring will have genotype mm (the only genotype you can have if you’re blotched tabby).

[f]IEZhYnVsb3VzLiBJZiB5b3UgbGV0IHRoZSBsZXR0ZXIgTQ==IHJlcHJlc2VudCB0aGUgYWxsZWxlIGZvciAmIzgyMjA7bWFja2VyZWwmIzgyMjE7IGFuZCA=bQ==IHJlcHJlc2VudCB0aGUgYWxsZWxlIGZvciAmIzgyMjA7YmxvdGNoZWQsIHRoZW4geW91JiM4MjE3O3JlIGNyb3NzaW5nIA==[Qq]Mm x mm. 50% of the offspring will be blotched tabby, with genotype mm. Here’s the Punnett square:

M m
m Mm mm
m Mm mm

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcHJvYmxlbS4gWW91IGtub3cgdGhhdCB0aGUgbWFja2VyZWwgdGFiYnkgdHJhaXQgaXMgZG9taW5hbnQgKGJlY2F1c2UgYWxsIG9mIHRoZSBGMQ==IG9mZnNwcmluZyBhcmUgbWFja2VyZWwpLiBBbmQgeW91IGFsc28ga25vdyB0aGF0IHRoZSBGMQ==cyBhcmUgaGV0ZXJvenlnb3Rlcy4gQmVjYXVzZSB0aGUgYWxsZWxlIGZvciBibG90Y2hlZCBpcyByZWNlc3NpdmUsIGEgYmxvdGNoZWQgdGFiYnkgaGFzIHRvIGJlIGhvbW96eWdvdXMu

[Qq]

If you let the letter M represent the allele for “mackerel” and m represent the allele for “blotched, then you’re crossing Mm x mm, and you’d set up the Punnett square as follows:

M m
m
m

Now you draw it out, and figure out what percentage of the offspring will have genotype mm (the only genotype you can have if you’re blotched tabby).

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcHJvYmxlbS4gWW91IGtub3cgdGhhdCB0aGUgbWFja2VyZWwgdGFiYnkgdHJhaXQgaXMgZG9taW5hbnQgKGJlY2F1c2UgYWxsIG9mIHRoZSBGMQ==IG9mZnNwcmluZyBhcmUgbWFja2VyZWwpLiBBbmQgeW91IGFsc28ga25vdyB0aGF0IHRoZSBGMQ==cyBhcmUgaGV0ZXJvenlnb3Rlcy4gQmVjYXVzZSB0aGUgYWxsZWxlIGZvciBibG90Y2hlZCBpcyByZWNlc3NpdmUsIGEgYmxvdGNoZWQgdGFiYnkgaGFzIHRvIGJlIGhvbW96eWdvdXMu

[Qq]

If you let the letter M represent the allele for “mackerel” and m represent the allele for “blotched, then you’re crossing Mm x mm, and you’d set up the Punnett square as follows:

M m
m
m

Now you draw it out, and figure out what percentage of the offspring will have genotype mm (the only genotype you can have if you’re blotched tabby).

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a64c18366c0d5″ question_number=”8″ topic=”5.3.Mendelian_Genetics”] A plant with genotype DdEeFf was crossed with a plant with genotype ddEEFf. What is the probability of obtaining DdEeFf progeny?

Note from Mr. W: the goal is to solve this problem without making an enormous Punnett square.

[c]IDEvNCA=[Qq][c]IDEv OCA=[Qq][c]IDEvMTYg[Qq][c]IDEvMzIg[Qq][c]IDEvNjQ=

Cg==[Qq]

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[f]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[Qq]

[f]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[Qq]

[f]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

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a64a7e6e3f4d5″ question_number=”9″ topic=”5.3.Mendelian_Genetics”] Two strains of mice that are heterozygous for four genes (AaBbCcDd)are crossed together. What proportion of the total offspring would be homozygous for all four genes? (Note from Mr. W: Do not solve this by creating a huge Punnett square. Use the rules of probability.)

[c]IDEvMjU2IA==[Qq][c]IDEvMzIg[Qq][c]IDEv MTYg[Qq][c]IDEvOCA=[Qq][c]IDEvNA==

Cg==[Qq]

[f]IE5vLiBUaGluayBhYm91dCBpdCB0aGlzIHdheS4gSW4gYSBtb25vaHlicmlkIGNyb3NzIGJldHdlZW4gQWEgYW5kIEFhLCB0aGUgb2Zmc3ByaW5nIGNhbiBiZSBob21venlnb3VzIGluIHR3byB3YXlzOiB0aGV5IGNhbiBiZSBhYSBvciBBQS4gSWYgeW91IGltYWdpbmUgdGhlIFB1bm5ldHQgc3F1YXJlIGZvciBBYSB4IEFhLCB5b3UmIzgyMTc7bGwgc2VlIHRoYXQgMS8yIG9mIHRoZSBzcXVhcmVzIGFyZSBmaWxsZWQgd2l0aCBhYSBvciBBQS4gVGhlIHNhbWUgaXMgdHJ1ZSBmb3IgQmIgeCBCYiwgQ2MgeCBDYywgYW5kIERkIHggRGQuIE5vdyB1c2UgdGhlIHJ1bGUgb2YgbXVsdGlwbGljYXRpb24sIGFuZCB5b3UmIzgyMTc7bGwgaGF2ZSB0aGUgYW5zd2VyLg==[Qq]

[f]IE5vLiBUaGluayBhYm91dCBpdCB0aGlzIHdheS4gSW4gYSBtb25vaHlicmlkIGNyb3NzIGJldHdlZW4gQWEgYW5kIEFhLCB0aGUgb2Zmc3ByaW5nIGNhbiBiZSBob21venlnb3VzIGluIHR3byB3YXlzOiB0aGV5IGNhbiBiZSBhYSBvciBBQS4gSWYgeW91IGltYWdpbmUgdGhlIFB1bm5ldHQgc3F1YXJlIGZvciBBYSB4IEFhLCB5b3UmIzgyMTc7bGwgc2VlIHRoYXQgMS8yIG9mIHRoZSBzcXVhcmVzIGFyZSBmaWxsZWQgd2l0aCBhYSBvciBBQS4gVGhlIHNhbWUgaXMgdHJ1ZSBmb3IgQmIgeCBCYiwgQ2MgeCBDYywgYW5kIERkIHggRGQuIE5vdyB1c2UgdGhlIHJ1bGUgb2YgbXVsdGlwbGljYXRpb24sIGFuZCB5b3UmIzgyMTc7bGwgaGF2ZSB0aGUgYW5zd2VyLg==[Qq]

[f]IEV4Y2VsbGVudCEgSW4gYSB0ZXRyYWh5YnJpZCBjcm9zcyAoQWFCYkNjRGQgeCBBYUJiQ2NEZCksIHRoZSBudW1iZXIgb2Ygb2Zmc3ByaW5nIHRoYXQgd2lsbCBiZSBob21venlnb3VzIGZvciBhbGwgZm91ciBnZW5lcyBpcyAxLzIgeCAxLzIgeCAxLzIgeCAxLzIsIG9yIDEvMTYu[Qq]

[f]IE5vLiBUaGluayBhYm91dCBpdCB0aGlzIHdheS4gSW4gYSBtb25vaHlicmlkIGNyb3NzIGJldHdlZW4gQWEgYW5kIEFhLCB0aGUgb2Zmc3ByaW5nIGNhbiBiZSBob21venlnb3VzIGluIHR3byB3YXlzOiB0aGV5IGNhbiBiZSBhYSBvciBBQS4gSWYgeW91IGltYWdpbmUgdGhlIFB1bm5ldHQgc3F1YXJlIGZvciBBYSB4IEFhLCB5b3UmIzgyMTc7bGwgc2VlIHRoYXQgMS8yIG9mIHRoZSBzcXVhcmVzIGFyZSBmaWxsZWQgd2l0aCBhYSBvciBBQS4gVGhlIHNhbWUgaXMgdHJ1ZSBmb3IgQmIgeCBCYiwgQ2MgeCBDYywgYW5kIERkIHggRGQuIE5vdyB1c2UgdGhlIHJ1bGUgb2YgbXVsdGlwbGljYXRpb24sIGFuZCB5b3UmIzgyMTc7bGwgaGF2ZSB0aGUgYW5zd2VyLg==[Qq]

[f]IE5vLiBUaGluayBhYm91dCBpdCB0aGlzIHdheS4gSW4gYSBtb25vaHlicmlkIGNyb3NzIGJldHdlZW4gQWEgYW5kIEFhLCB0aGUgb2Zmc3ByaW5nIGNhbiBiZSBob21venlnb3VzIGluIHR3byB3YXlzOiB0aGV5IGNhbiBiZSBhYSBvciBBQS4gSWYgeW91IGltYWdpbmUgdGhlIFB1bm5ldHQgc3F1YXJlIGZvciBBYSB4IEFhLCB5b3UmIzgyMTc7bGwgc2VlIHRoYXQgMS8yIG9mIHRoZSBzcXVhcmVzIGFyZSBmaWxsZWQgd2l0aCBhYSBvciBBQS4gVGhlIHNhbWUgaXMgdHJ1ZSBmb3IgQmIgeCBCYiwgQ2MgeCBDYywgYW5kIERkIHggRGQuIE5vdyB1c2UgdGhlIHJ1bGUgb2YgbXVsdGlwbGljYXRpb24sIGFuZCB5b3UmIzgyMTc7bGwgaGF2ZSB0aGUgYW5zd2VyLg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a648bf65544d5″ question_number=”10″ unit=”5.Heredity” topic=”5.3.Mendelian_Genetics”] Based on the pedigree chart below, what is the most likely mode of inheritance for the condition shown?

[c]IFgtbGlua2VkIGRvbWluYW50[Qq]

[f]IE5vLiBJZiB0aGUgY29uZGl0aW9uIHdlcmUgWC1saW5rZWQgZG9taW5hbnQsIHRoZW4gb25lIG9mIHRoZSBwYXJlbnRzIG9mIElWLTEgd291bGQgaGF2ZSB0byBoYXZlIHRoZSBjb25kaXRpb24u[Qq]

[c]IFgtbGlua2VkIHJlY2Vzc2l2ZQ==[Qq]

[f]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[Qq]

[c]IEF1dG9zb21hbCBkb21pbmFudA==[Qq]

[f]IE5vLiBJZiB0aGUgY29uZGl0aW9uIHdlcmUgZG9taW5hbnQsIHRoZW4gb25lIG9mIHRoZSBwYXJlbnRzIG9mIElWLTEgd291bGQgaGF2ZSB0byBoYXZlIHRoZSBjb25kaXRpb24u[Qq]

[c]IEF1dG9zb21hbC ByZWNlc3NpdmU=[Qq]

[f]IEZhYnVsb3VzISBBdXRvc29tYWwgcmVjZXNzaXZlIGlzIHRoZSBvbmx5IGluaGVyaXRhbmNlIHBhdHRlcm4gdGhhdCBmaXRzIHRoaXMgcGVkaWdyZWUu

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a647005c694d5″ question_number=”11″ unit=”5.Heredity” topic=”5.3.Mendelian_Genetics”] If two individuals who both had genotype PpQqRr mated, what is the expected frequency of PPQQRR offspring? Note: DON’T CREATE A HUGE PUNNETT SQUARE TO SOLVE THIS PROBLEM.

[c]IDEvNCA=[Qq][c]IDEvOCA=[Qq][c]IDEvMTYg[Qq][c]IDEvMzIg[Qq][c]IDEv NjQ=

Cg==[Qq]

[f]IE5vLiBVc2UgdGhlIHJ1bGUgb2YgbXVsdGlwbGljYXRpb24uIElmIHlvdSBjcm9zcyBQcCB4IFBwLCB0aGVuIHdoYXQgZnJhY3Rpb24gb2YgdGhlIG9mZnNwcmluZyB3aWxsIGhhdmUgZ2Vub3R5cGUgUFA/IEhlcmUmIzgyMTc7cyB0aGUgUHVubmV0dCBTcXVhcmUu

Cg==Cg==Cg==Cg==
[Qq] P p
P PP Pp
p Pp pp

Now take that probability (1/4), multiply it by the probability of producing QQ and RR offspring, and you’ll have your answer.

[f]IE5vLiBVc2UgdGhlIHJ1bGUgb2YgbXVsdGlwbGljYXRpb24uIElmIHlvdSBjcm9zcyBQcCB4IFBwLCB0aGVuIHdoYXQgZnJhY3Rpb24gb2YgdGhlIG9mZnNwcmluZyB3aWxsIGhhdmUgZ2Vub3R5cGUgUFA/IEhlcmUmIzgyMTc7cyB0aGUgUHVubmV0dCBTcXVhcmUu

Cg==Cg==Cg==Cg==
[Qq] P p
P PP Pp
p Pp pp

Now take that probability (1/4), multiply it by the probability of producing QQ and RR offspring, and you’ll have your answer.

[f]IE5vLiBVc2UgdGhlIHJ1bGUgb2YgbXVsdGlwbGljYXRpb24uIElmIHlvdSBjcm9zcyBQcCB4IFBwLCB0aGVuIHdoYXQgZnJhY3Rpb24gb2YgdGhlIG9mZnNwcmluZyB3aWxsIGhhdmUgZ2Vub3R5cGUgUFA/IEhlcmUmIzgyMTc7cyB0aGUgUHVubmV0dCBTcXVhcmUu

Cg==Cg==Cg==Cg==
[Qq] P p
P PP Pp
p Pp pp

Now take that probability (1/4), multiply it by the probability of producing QQ and RR offspring, and you’ll have your answer.

[f]IE5vLiBVc2UgdGhlIHJ1bGUgb2YgbXVsdGlwbGljYXRpb24uIElmIHlvdSBjcm9zcyBQcCB4IFBwLCB0aGVuIHdoYXQgZnJhY3Rpb24gb2YgdGhlIG9mZnNwcmluZyB3aWxsIGhhdmUgZ2Vub3R5cGUgUFA/

Cg==

SGVyZSYjODIxNztzIHRoZSBQdW5uZXR0IFNxdWFyZS4=

Cg==Cg==[Qq]
P p
P PP Pp
p Pp pp

Now take that probability (1/4), multiply it by the probability of producing QQ and RR offspring, and you’ll have your answer.

[f]IEZhYnVsb3VzLiBGb3IgYW55IG9uZSBvZiB0aGVzZSBnZW5lcywgdGhlIHByb2JhYmlsaXR5IG9mIHByb2R1Y2luZyBhIGhvbW96eWdvdXMgZG9taW5hbnQgaXMgMS80LiBVc2luZyB0aGUgcnVsZSBvZiBtdWx0aXBsaWNhdGlvbiwgeW91IGdldCAxLzQgeCAxLzQgeCAxLzQsIHdoaWNoIGVxdWFscyAxLzY0Lg==

Cg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a64541537e4d5″ question_number=”12″ topic=”5.4.Non-Mendelian_Genetics”] The pedigree below shows inheritance of an X-linked recessive trait.

What is the probability that the female designated as “1″ is a carrier for the trait?

[c]IDAlIA==[Qq][c]IDI1JSA=[Qq][c]IDUw JSA=[Qq][c]IDc1JSA=[Qq][c]IDEwMCU=

Cg==[Qq]

[f]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SA==WA==aA==KSwgYW5kIHRoZSBmYXRoZXIgd2FzIG5vcm1hbCAoWA==[Qq]HY).

If you can, your best move at this point is to make a Punnett square and figure out the answer. If you need to see the Punnett square that I made to solve this problem, read on…

Here’s the Punnett square:

XH Xh
XH XHXH XHXh
Y XHY XhY

Look at the two girls in the top row, and you should be able to figure out the probability that any one of the daughters will be a carrier.

[f]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SA==WA==aA==KSwgYW5kIHRoZSBmYXRoZXIgd2FzIG5vcm1hbCAoWA==[Qq]HY).

If you can, your best move at this point is to make a Punnett square and figure out the answer. If you need to see the Punnett square that I made to solve this problem, read on…

Here’s the Punnett square:

XH Xh
XH XHXH XHXh
Y XHY XhY

Look at the two girls in the top row, and you should be able to figure out the probability that any one of the daughters will be a carrier.

[f]IEV4Y2VsbGVudCEgVGhlIHByb2JhYmlsaXR5IHRoYXQgaW5kaXZpZHVhbCAxIChvbmUgb2YgdGhlIGRhdWdodGVycykgaXMgYSBjYXJyaWVyIGlzIDUwJSwgYXMgeW91IGNhbiBzZWUgYnkgdGhpcyBQdW5uZXR0IHNxdWFyZQ==

Cg==Cg==Cg==Cg==
[Qq] XH Xh
XH XHXH XHXh
Y XHY XhY

[f]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SA==WA==aA==KSwgYW5kIHRoZSBmYXRoZXIgd2FzIG5vcm1hbCAoWA==[Qq]HY).

If you can, your best move at this point is to make a Punnett square and figure out the answer. If you need to see the Punnett square that I made to solve this problem, read on…

Here’s the Punnett square:

XH Xh
XH XHXH XHXh
Y XHY XhY

Look at the two girls in the top row, and you should be able to figure out the probability that any one of the daughters will be a carrier.

[f]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SA==WA==aA==KSwgYW5kIHRoZSBmYXRoZXIgd2FzIG5vcm1hbCAoWA==[Qq]HY).

If you can, your best move at this point is to make a Punnett square and figure out the answer. If you need to see the Punnett square that I made to solve this problem, read on…

Here’s the Punnett square:

XH Xh
XH XHXH XHXh
Y XHY XhY

Look at the two girls in the top row, and you should be able to figure out the probability that any one of the daughters will be a carrier.

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a643a78b518d5″ question_number=”13″ topic=”5.4.Non-Mendelian_Genetics”]

(Note: this question is part of a series. Feel free to skip this introduction if you’ve already read it).

INTRODUCTION: Domestic chickens have been bred for many years to increase the number of eggs laid by the females. Chick breeders need to be able to identify female chicks on the day after hatching, as only the females need to be kept for laying eggs.

Unlike mammals, which use a sex determination system with X and Y chromosomes, in chickens the sex chromosomes are known as Z and W. Male chickens have two Z chromosomes (ZZ). Female chickens have one Z chromosome and one W chromosome (ZW).

Some genes for feather color and pattern in chickens are carried on the Z chromosome but not on the W chromosome. One such example is the gene for striped feathers (barring).

The phenotypes associated with the two alleles of the barring gene are shown in the table below

Allele Adult Phenotype Day old chick phenotype
B Barred (black feathers striped with white bars) Black body with a white spot on the head
b Non-barred (black feathers) Black body and head

An individual with genotype ZBZb will, as an adult, have which phenotype?

[c]IEJhcnJl ZCBtYWxl[Qq]

[f]IEV4Y2VsbGVudC4gQW4gaW5kaXZpZHVhbCB3aXRoIGdlbm90eXBlIA==Wg==Qg==Wg==Yg==[Qq] will be a barred male (ZZ = male, and B is the allele for barring)

[c]IEJhcnJlZCBmZW1hbGU=[Qq]

[f]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[Qq]

[c]IE5vbi1iYXJyZWQgZmVtYWxl[Qq]

[f]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[Qq]

[c]IE5vbi1iYXJyZWQgbWFsZQ==[Qq]

[f]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

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a6420dc324cd5″ question_number=”14″ topic=”5.4.Non-Mendelian_Genetics”] Hemophilia A is an X-linked recessive disorder in which the ability of the blood to clot is severely reduced. Hemophilia is caused by a mutation in the gene for the clotting component Factor VIII.

Jasmine’s brother has hemophilia A, but neither Jasmine nor anyone else in her family show symptoms of the disorder.

If Jasmine has a son, what is the probability that he will have hemophilia?

[c]IDAlIA==[Qq][c]IDI1 JSA=[Qq][c]IDUwJSA=[Qq][c]IDc1JSA=[Qq][c]IDEwMCU=

Cg==[Qq]

[f]IE5vLiBCZWNhdXNlIEphc21pbmUmIzgyMTc7cyBicm90aGVyIGlzIGEgaGVtb3BoaWxpYWMsIHdlIGtub3cgdGhhdCBoaXMgZ2Vub3R5cGUgaXMgWA==aA==WS4gQmVjYXVzZSB0aGUgaGVtb3BoaWxpYSBhbGxlbGUgaXMgb24gdGhlIFggY2hyb21vc29tZSwgdGhlbiBKYXNtaW5lJiM4MjE3O3MgYnJvdGhlciBoYWQgdG8gaGF2ZSBpbmhlcml0ZWQgdGhlIGNvbmRpdGlvbiBmcm9tIGhpcyBhbmQgSmFzbWluZSYjODIxNztzIG1vdGhlciwgd2hvIG11c3QgYmUgYSBjYXJyaWVyIChpdCYjODIxNztzIHN0YXRlZCBpbiB0aGUgcHJvYmxlbSB0aGF0IHNoZSBkb2VzbiYjODIxNzt0IGhhdmUgdGhlIGNvbmRpdGlvbiwgc28gc2hlIGNhbiBvbmx5IGJlIGEgY2Fycmllciku

Cg==

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[f]IEV4Y2VsbGVudCEgVGhlcmUmIzgyMTc7cyBhIDUwJSBwcm9iYWJpbGl0eSB0aGF0IEphc21pbmUgaXMgYSBjYXJyaWVyLCBhbmQgdGhlcmUmIzgyMTc7cyBhIDUwJSBwcm9iYWJpbGl0eSB0aGF0IGEgY2FycmllciB3aWxsIHBhc3Mgb24gYW4gWCBjaHJvbW9zb21lIHdpdGggdGhlIGhlbW9waGlsaWEgYWxsZWxlIG9uIHRvIGhlciBzb24uIDUwJSBvZiA1MCUgaXMgMjUlLg==[Qq]

[f]IE5vLiBCZWNhdXNlIEphc21pbmUmIzgyMTc7cyBicm90aGVyIGlzIGEgaGVtb3BoaWxpYWMsIHdlIGtub3cgdGhhdCBoaXMgZ2Vub3R5cGUgaXMgWA==aA==WS4gQmVjYXVzZSB0aGUgaGVtb3BoaWxpYSBhbGxlbGUgaXMgb24gdGhlIFggY2hyb21vc29tZSwgdGhlbiBKYXNtaW5lJiM4MjE3O3MgYnJvdGhlciBoYWQgdG8gaGF2ZSBpbmhlcml0ZWQgdGhlIGNvbmRpdGlvbiBmcm9tIGhpcyBhbmQgSmFzbWluZSYjODIxNztzIG1vdGhlciwgd2hvIG11c3QgYmUgYSBjYXJyaWVyIChpdCYjODIxNztzIHN0YXRlZCBpbiB0aGUgcHJvYmxlbSB0aGF0IHNoZSBkb2VzbiYjODIxNzt0IGhhdmUgdGhlIGNvbmRpdGlvbiwgc28gc2hlIGNhbiBvbmx5IGJlIGEgY2Fycmllciku

Cg==

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[f]IE5vLiBCZWNhdXNlIEphc21pbmUmIzgyMTc7cyBicm90aGVyIGlzIGEgaGVtb3BoaWxpYWMsIHdlIGtub3cgdGhhdCBoaXMgZ2Vub3R5cGUgaXMgWA==aA==WS4gQmVjYXVzZSB0aGUgaGVtb3BoaWxpYSBhbGxlbGUgaXMgb24gdGhlIFggY2hyb21vc29tZSwgdGhlbiBKYXNtaW5lJiM4MjE3O3MgYnJvdGhlciBoYWQgdG8gaGF2ZSBpbmhlcml0ZWQgdGhlIGNvbmRpdGlvbiBmcm9tIGhpcyBhbmQgSmFzbWluZSYjODIxNztzIG1vdGhlciwgd2hvIG11c3QgYmUgYSBjYXJyaWVyIChpdCYjODIxNztzIHN0YXRlZCBpbiB0aGUgcHJvYmxlbSB0aGF0IHNoZSBkb2VzbiYjODIxNzt0IGhhdmUgdGhlIGNvbmRpdGlvbiwgc28gc2hlIGNhbiBvbmx5IGJlIGEgY2Fycmllciku

Cg==

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[f]IE5vLiBCZWNhdXNlIEphc21pbmUmIzgyMTc7cyBicm90aGVyIGlzIGEgaGVtb3BoaWxpYWMsIHdlIGtub3cgdGhhdCBoaXMgZ2Vub3R5cGUgaXMgWA==aA==WS4gQmVjYXVzZSB0aGUgaGVtb3BoaWxpYSBhbGxlbGUgaXMgb24gdGhlIFggY2hyb21vc29tZSwgdGhlbiBKYXNtaW5lJiM4MjE3O3MgYnJvdGhlciBoYWQgdG8gaGF2ZSBpbmhlcml0ZWQgdGhlIGNvbmRpdGlvbiBmcm9tIGhpcyBhbmQgSmFzbWluZSYjODIxNztzIG1vdGhlciwgd2hvIG11c3QgYmUgYSBjYXJyaWVyIChpdCYjODIxNztzIHN0YXRlZCBpbiB0aGUgcHJvYmxlbSB0aGF0IHNoZSBkb2VzbiYjODIxNzt0IGhhdmUgdGhlIGNvbmRpdGlvbiwgc28gc2hlIGNhbiBvbmx5IGJlIGEgY2Fycmllciku

Cg==

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[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a64073faf80d5″ question_number=”15″ topic=”5.4.Non-Mendelian_Genetics”] Hemophilia A is an X-linked recessive disorder in which the ability of the blood to clot is severely reduced. Hemophilia is caused by a mutation in the gene for the clotting component Factor VIII.

Jasmine’s brother has hemophilia A, but neither Jasmine nor anyone else in her family show symptoms of the disorder.

If Jasmine’s husband had hemophilia, what would the probability be of their son being a hemophiliac?

[c]IDAlIA==[Qq][c]IDI1 JSA=[Qq][c]IDUwJSA=[Qq][c]IDc1JSA=[Qq][c]IDEwMCU=

Cg==[Qq]

[f]IE5vLiBCZWNhdXNlIEphc21pbmUmIzgyMTc7cyBicm90aGVyIGlzIGEgaGVtb3BoaWxpYWMsIHdlIGtub3cgdGhhdCBoaXMgZ2Vub3R5cGUgaXMgWA==aA==WS4gQmVjYXVzZSB0aGUgaGVtb3BoaWxpYSBhbGxlbGUgaXMgb24gdGhlIFggY2hyb21vc29tZSwgdGhlbiBKYXNtaW5lJiM4MjE3O3MgYnJvdGhlciBoYWQgdG8gaGF2ZSBpbmhlcml0ZWQgdGhlIGNvbmRpdGlvbiBmcm9tIGhpcyBhbmQgSmFzbWluZSYjODIxNztzIG1vdGhlciwgd2hvIG11c3QgYmUgYSBjYXJyaWVyIChpdCYjODIxNztzIHN0YXRlZCBpbiB0aGUgcHJvYmxlbSB0aGF0IHNoZSBkb2VzbiYjODIxNzt0IGhhdmUgdGhlIGNvbmRpdGlvbiwgc28gc2hlIGNhbiBvbmx5IGJlIGEgY2Fycmllciku

Cg==

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[f]IEV4Y2VsbGVudCEgVGhlcmUmIzgyMTc7cyBhIDUwJSBwcm9iYWJpbGl0eSB0aGF0IEphc21pbmUgaXMgYSBjYXJyaWVyLCBhbmQgdGhlcmUmIzgyMTc7cyBhIDUwJSBwcm9iYWJpbGl0eSB0aGF0IGEgY2FycmllciB3aWxsIHBhc3Mgb24gYW4gWCBjaHJvbW9zb21lIHdpdGggdGhlIGhlbW9waGlsaWEgYWxsZWxlIG9uIHRvIGhlciBzb24uIEFzIHlvdSBtdXN0IGhhdmUgdW5kZXJzdG9vZCwgdGhlIGZhdGhlciYjODIxNztzIHN0YXR1cyBhcyBhIGhlbW9waGlsaWFjIGlzIGlycmVsZXZhbnQgc2luY2UgaGUmIzgyMTc7bGwgb25seSBwYXNzIG9uIGhpcyBYIGNocm9tb3NvbWUgdG8gaGlzIGRhdWdodGVycy4gNTAlIG9mIDUwJSBpcyAyNSUu[Qq]

[f]IE5vLiBCZWNhdXNlIEphc21pbmUmIzgyMTc7cyBicm90aGVyIGlzIGEgaGVtb3BoaWxpYWMsIHdlIGtub3cgdGhhdCBoaXMgZ2Vub3R5cGUgaXMgWA==aA==WS4gQmVjYXVzZSB0aGUgaGVtb3BoaWxpYSBhbGxlbGUgaXMgb24gdGhlIFggY2hyb21vc29tZSwgdGhlbiBKYXNtaW5lJiM4MjE3O3MgYnJvdGhlciBoYWQgdG8gaGF2ZSBpbmhlcml0ZWQgdGhlIGNvbmRpdGlvbiBmcm9tIGhpcyBhbmQgSmFzbWluZSYjODIxNztzIG1vdGhlciwgd2hvIG11c3QgYmUgYSBjYXJyaWVyIChpdCYjODIxNztzIHN0YXRlZCBpbiB0aGUgcHJvYmxlbSB0aGF0IHNoZSBkb2VzbiYjODIxNzt0IGhhdmUgdGhlIGNvbmRpdGlvbiwgc28gc2hlIGNhbiBvbmx5IGJlIGEgY2Fycmllciku

Cg==

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[f]IE5vLiBCZWNhdXNlIEphc21pbmUmIzgyMTc7cyBicm90aGVyIGlzIGEgaGVtb3BoaWxpYWMsIHdlIGtub3cgdGhhdCBoaXMgZ2Vub3R5cGUgaXMgWA==aA==WS4gQmVjYXVzZSB0aGUgaGVtb3BoaWxpYSBhbGxlbGUgaXMgb24gdGhlIFggY2hyb21vc29tZSwgdGhlbiBKYXNtaW5lJiM4MjE3O3MgYnJvdGhlciBoYWQgdG8gaGF2ZSBpbmhlcml0ZWQgdGhlIGNvbmRpdGlvbiBmcm9tIGhpcyBhbmQgSmFzbWluZSYjODIxNztzIG1vdGhlciwgd2hvIG11c3QgYmUgYSBjYXJyaWVyIChpdCYjODIxNztzIHN0YXRlZCBpbiB0aGUgcHJvYmxlbSB0aGF0IHNoZSBkb2VzbiYjODIxNzt0IGhhdmUgdGhlIGNvbmRpdGlvbiwgc28gc2hlIGNhbiBvbmx5IGJlIGEgY2Fycmllciku

Cg==

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[f]IE5vLiBCZWNhdXNlIEphc21pbmUmIzgyMTc7cyBicm90aGVyIGlzIGEgaGVtb3BoaWxpYWMsIHdlIGtub3cgdGhhdCBoaXMgZ2Vub3R5cGUgaXMgWA==aA==WS4gQmVjYXVzZSB0aGUgaGVtb3BoaWxpYSBhbGxlbGUgaXMgb24gdGhlIFggY2hyb21vc29tZSwgdGhlbiBKYXNtaW5lJiM4MjE3O3MgYnJvdGhlciBoYWQgdG8gaGF2ZSBpbmhlcml0ZWQgdGhlIGNvbmRpdGlvbiBmcm9tIGhpcyBhbmQgSmFzbWluZSYjODIxNztzIG1vdGhlciwgd2hvIG11c3QgYmUgYSBjYXJyaWVyIChpdCYjODIxNztzIHN0YXRlZCBpbiB0aGUgcHJvYmxlbSB0aGF0IHNoZSBkb2VzbiYjODIxNzt0IGhhdmUgdGhlIGNvbmRpdGlvbiwgc28gc2hlIGNhbiBvbmx5IGJlIGEgY2Fycmllciku

Cg==

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

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a63e8fb14ecd5″ question_number=”16″ topic=”5.4.Non-Mendelian_Genetics”] Blood type inheritance involves three alleles. While there are several notation systems for representing these alleles, a common one has alleles IA and IB as codominant to one another, with each of these dominant over i. If a child has type O blood (genotype ii), which of the following blood types is impossible in either parent?

[c]IFR5cGUgTw==[Qq]

[f]IE5vLiBUaGUgYmVzdCB3YXkgdG8gdGhpbmsgYWJvdXQgdGhpcyBpcyB0byBkcmF3IGEgUHVubmV0dCBzcXVhcmUsIGFuZCBwdXQgdGhlIGdlbm90eXBlIG9mIHRoZSB0eXBlIE8gY2hpbGQgaW4gdGhlIGxvd2VyIHJpZ2h0IGJveCwgYXMgc2hvd24gYmVsb3cu

Cg==Cg==Cg==Cg==
[Qq]
ii

If the child is ii, then each of their parents must have given them an i allele, as shown in the slightly more complete Punnett square shown below.

___ i
___
i ii

Each parent is __i. It’s completely possible that the child could have a parent that’s type O (genotype ii). But what genotype is impossible for either of the parents to have?

[c]IFR5cGUgQg==[Qq]

[f]IE5vLiBUaGUgYmVzdCB3YXkgdG8gdGhpbmsgYWJvdXQgdGhpcyBpcyB0byBkcmF3IGEgUHVubmV0dCBzcXVhcmUsIGFuZCBwdXQgdGhlIGdlbm90eXBlIG9mIHRoZSB0eXBlIE8gY2hpbGQgaW4gdGhlIGxvd2VyIHJpZ2h0IGJveCwgYXMgc2hvd24gYmVsb3cu

Cg==Cg==Cg==Cg==
[Qq] ___ ___
___
___ ii

If the child is ii, then each of their parents must have given them an i allele, as shown in the slightly more complete Punnett square shown below.

___ i
___
i ii

Each parent is __i. It’s completely possible that the child could have a parent that’s type B (genotype IBi). But what genotype is impossible for either of the parents to have?

[c]IFR5cG UgQUI=[Qq]

[f]IFdheSB0byBnby7CoEEgY2hpbGQgd2l0aCB0eXBlIE8gYmxvb2QgbmVlZHMgZWFjaCBvZiB0aGVpciBwYXJlbnRzIHRvIGdpdmUgdGhlbSBhbiBpIGFsbGVsZS4gR2l2ZW4gdGhhdCwgdHlwZSBBQiBpcyBpbXBvc3NpYmxlIChiZWNhdXNlIGEgdHlwZSBBQiBwYXJlbnQgY291bGQgZ2l2ZSB0aGVpciBjaGlsZCBlaXRoZXIgSQ==QSA=b3IgSQ==Qg==LCBidXQgbm90IGFpLg==[Qq]

[c]IFR5cGUgQQ==[Qq]

[f]IE5vLiBUaGUgYmVzdCB3YXkgdG8gdGhpbmsgYWJvdXQgdGhpcyBpcyB0byBkcmF3IGEgUHVubmV0dCBzcXVhcmUsIGFuZCBwdXQgdGhlIGdlbm90eXBlIG9mIHRoZSB0eXBlIE8gY2hpbGQgaW4gdGhlIGxvd2VyIHJpZ2h0IGJveCwgYXMgc2hvd24gYmVsb3cu

Cg==Cg==Cg==Cg==
[Qq] ___ ___
___
___ ii

If the child is ii, then each of their parents must have given them an i allele, as shown in the slightly more complete Punnett square shown below.

___ i
___
i ii

Each parent is __i. It’s completely possible that the child could have a parent that’s type A (genotype IAi). But what genotype is impossible for either of the parents to have?

[q json=”true” multiple_choice=”true” unit=”5.Heredity” topic=”5.5.Environmental_Effects_on_Phenotype” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a63cf5e9220d5″ question_number=”17″] Cerateropteris richii is a fern that, in adult diploid form, can be male (producing only male gametes), or hermaphroditic (producing both male and female gametes).

Students sowed haploid spores of the fern C. richii on Petri plates containing growth media over a period of 5 days as shown in the diagram on the right.

After 14 days of incubation, the students scored the number of hermaphrodites and males in each sector of the plate.  Each of the 30 students scored 10 plants in each sector of each plate.  The combined data is shown below.

Which of the following is the most likely mechanism by which the ratio of males to hermaphrodites changes depending on the day of sowing?

[c]IEhpZ2hlciB0ZW1wZXJhdHVyZXMgaW4gdGhlIGdyb3d0aCBtZWRpdW0gZmF2b3IgdGhlIGRldmVsb3BtZW50IG9mIHNwb3JlcyBpbnRvIG1hbGVzLg==[Qq]

[f]IE5vLiBUaGVyZSBpcyBubyBkYXRhIGFib3V0IHRlbXBlcmF0dXJlLiBTZWUgaWYgeW91IGNhbiB0aGluayBvZiBzb21ldGhpbmcgZWxzZSB0aGF0JiM4MjE3O3MgZGV0ZXJtaW5pbmcgdGhlIGluY3JlYXNlZCBudW1iZXIgb2YgaGVybWFwaHJvZGl0ZXMu[Qq]

[c]IFNwb3JlcyBzb3duIGxhdGVyIHVuZGVyZ28gY3Jvc3Npbmcgb3ZlciB0byBtb3ZlIHRoZSBzZXgtZGV0ZXJtaW5pbmcgcmVnaW9uIG9mIHRoZSBZIGNocm9tb3NvbWUgdG8gYW5vdGhlciB0aGUgWCBjaHJvbW9zb21lIHdoZXJlIGl0IHdpbGwgbm90IGJlIGV4cHJlc3NlZC4=[Qq]

[f]IE5vLiBUaGUgc3BvcmVzIGFyZSBoYXBsb2lkLCBzbyB0aGVyZSBpcyBubyBwYWlyaW5nIG9mIGhvbW9sb2dvdXMgY2hyb21vc29tZXMgYW5kIG5vIGNyb3NzaW5nIG92ZXIuIFNlZSBpZiB5b3UgY2FuIHRoaW5rIG9mIHNvbWV0aGluZyBlbHNlIHRoYXQmIzgyMTc7cyBkZXRlcm1pbmluZyB0aGUgaW5jcmVhc2VkIG51bWJlciBvZiBoZXJtYXBocm9kaXRlcy4=[Qq]

[c]IFNwb3JlcyBzb3duIGxhdGVyIGFyZSBtb3JlIGxpa2VseSB0byBoYXZlIHRoZWlyIFggY2hyb21vc29tZSBtdXRhdGUgaW50byBhIFkgY2hyb21vc29tZS4=[Qq]

[f]IE5vLiBUaGVyZSYjODIxNztzIG5vIGV2aWRlbmNlIG9mIGFueXRoaW5nIHRoYXQgd291bGQgY2F1c2Ugc3VjaCBhIGhpZ2ggbXV0YXRpb24gcmF0ZS4gU2VlIGlmIHlvdSBjYW4gdGhpbmsgb2Ygc29tZXRoaW5nIGVsc2UgdGhhdCYjODIxNztzIGRldGVybWluaW5nIHRoZSBpbmNyZWFzZWQgbnVtYmVyIG9mIGhlcm1hcGhyb2RpdGVzLg==[Qq]

[c]IEhpZ2ggY29uY2VudHJhdGlvbnMgb2YgY2hlbWljYWwgc2lnbmFscyBzZWNyZXRlZCBieSBn ZXJtaW5hdGluZyBmZXJucyBjYXVzZSBzcG9yZXMgdG8gZGV2ZWxvcCBpbnRvIG1hbGVzLg==[Qq]

[f]IFllcy4gSWYgZWFybHktZ3Jvd2luZyBwbGFudHMgc2VjcmV0ZSBhIGNoZW1pY2FsIHNpZ25hbCB0aGF0IGluZHVjZXMgY2hhbmdlcyBpbiBzcG9yZXMgdGhhdCBhcmUgc293biBsYXRlciwgdGhlIGxhdGVyIHNwb3JlcyBjYW4gZGV2ZWxvcCBpbnRvIG1hbGVzLg==

Cg==

[Qq]

[q json=”true” multiple_choice=”true” unit=”5.Heredity” topic=”5.5.Environmental_Effects_on_Phenotype” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a63b5c20f54d5″ question_number=”18″] A marine biologist measured the average temperature of 18 loggerhead turtle nests and determined the sex ratio of the eggs that hatched. Which of the following is an accurate interpretation of the results?

[c]IEF0IGFueSB0ZW1wZXJhdHVyZSBhYm92ZSAzMCDCsEMgYWxsIHR1cnRsZXMgdGhhdCBoYXRjaCB3aWxsIGJlIG1hbGUu[Qq]

[f]IE5vLiBMb29rIGF0IHRoZSBZLWF4aXMsIHdoaWNoIGlzIHRoZSByYXRpbyBpbiAlIGZlbWFsZXMuIEFib3ZlIDMwIMKwQywgYWxtb3N0IGFsbCB0dXJ0bGUgaGF0Y2hsaW5ncyB3aWxsIGJlIGZlbWFsZS4=[Qq]

[c]IEF0IGFueSB0ZW1wZXJhdHVyZSBiZWxvdyAyNSDCsEMgYWxsIHR1cnRsZXMgdGhhdCBoYXRjaCB3aWxsIGJlIGZlbWFsZS4=[Qq]

[f]IE5vLiBMb29rIGF0IHRoZSBZLWF4aXMsIHdoaWNoIGlzIHRoZSByYXRpbyBpbiAlIGZlbWFsZXMuIEF0IDI1IMKwQywgbm9uZSBvZiB0aGUgdHVydGxlIGhhdGNobGluZ3Mgd2lsbCBiZSBmZW1hbGUu[Qq]

[c]IEF0IDI5IMKwQywgdGhlcmUgaXMgYSA1MCUgY2hhbmNlIHRoYXQg YW55IHR1cnRsZSB0aGF0IGhhdGNoZXMgd2lsbCBiZSBtYWxlLg==[Qq]

[f]IEV4Y2VsbGVudC4gU2V4IGRldGVybWluYXRpb24gaW4gbG9nZ2VyaGVhZHMgaXMgZGV0ZXJtaW5lZCBieSB0ZW1wZXJhdHVyZS4gQWJvdmUgMjkgwrBDLCBtb3N0IHR1cnRsZSBoYXRjaGxpbmdzIHdpbGwgYmUgZmVtYWxlLiBCZWxvdyB0aGF0IHRlbXBlcmF0dXJlLCBtb3N0IHdpbGwgYmUgbWFsZS4gQXQgMjkgwrBDLCB0aGVyZSYjODIxNztzIGEgNTAlIGNoYW5jZSBvZiBhIGhhdGNobGluZyBkZXZlbG9waW5nIGludG8gZWl0aGVyIHNleC4=[Qq]

[c]IFNleCBkZXRlcm1pbmF0aW9uIGluIExvZ2dlcmhlYWQgdHVydGxlcyBpcyBkZXRlcm1pbmVkIGdlbmV0aWNhbGx5LCBub3QgYnkgdGhlIGVudmlyb25tZW50Lg==[Qq]

[f]IE5vLiBJZiB0aGUgc2V4IHdlcmUgZGV0ZXJtaW5lZCBnZW5ldGljYWxseSAoYXMgd2l0aCB0aGUgWFgvWFkgc3lzdGVtIGluIG1hbW1hbHMgb3IgdGhlIFpaL1pXIHN5c3RlbSBpbiBiaXJkcyksIHRoZW4gdGhlIGVudmlyb25tZW50YWwgdGVtcGVyYXR1cmUgd291bGRuJiM4MjE3O3QgaGF2ZSBhbiBlZmZlY3Qgb24gdGhlIHNleCBvZiB0aGUgb2Zmc3ByaW5nLg==

Cg==

Cg==

[Qq]

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a639c258c88d5″ question_number=”19″ topic=”5.6.Chromosomal_Inheritance”] The karyotype shown below is an example of a

[c]IHRyaX NvbXku[Qq]

[f]IE5pY2Ugam9iISBBcyB5b3Ugb2JzZXJ2ZWQsIHRoZXJlIGFyZSB0aHJlZSBjb3BpZXMgb2YgY2hyb21vc29tZSAxNSwgY3JlYXRpbmcgYSB0cmlzb215Lg==[Qq]

[c]IG1vbm9zb215Lg==[Qq]

[f]IE5vLiBBIG1vbm9zb215IHdvdWxkIGhhdmUganVzdCBvbmUgY2hyb21vc29tZSwgaW5zdGVhZCBvZiBhIHBhaXIuIFlvdSBtaWdodCBoYXZlIGJlZW4gY29uZnVzZWQgYnkgdGhlIHNpbmdsZSBYIGFuZCBZIGNocm9tb3NvbWVzLCBidXQgdGhhdCYjODIxNztzIHRoZSBwYWlyIHRoYXQgY3JlYXRlcyBhIGNocm9tb3NvbWFsbHkgbWFsZSBtYW1tYWwuIFRha2UgYSBjbG9zZXIgbG9vayBhdCB0aGUgY2hyb21vc29tZXMgaW4gdGhlIGthcnlvdHlwZSwgYW5kIHNlZSBpZiB5b3UgY2FuIGZpbmQgYW5vdGhlciB0eXBlIG9mIGNocm9tb3NvbWFsIGRpZmZlcmVuY2UgaW4gdGhpcyBrYXJ5b3R5cGUu[Qq]

[c]IHBvbHlwbG9pZHku[Qq]

[f]IE5vLiBQb2x5cGxvaWR5IGludm9sdmVzIHRoZSBkdXBsaWNhdGlvbiBvZiBlbnRpcmUgY2hyb21vc29tZSBzZXRzLiBUaGlzIGlzIGEgc2luZ2xlIHNldC4gVGFrZSBhIGNsb3NlciBsb29rIGF0IHRoZSBjaHJvbW9zb21lcyBpbiB0aGUga2FyeW90eXBlLCBhbmQgc2VlIGlmIHlvdSBjYW4gZmluZCBzb21lIHR5cGUgb2YgY2hyb21vc29tYWwgZGlmZmVyZW5jZSB0aGF0IGNvcnJlc3BvbmRzIHRvIG9uZSBvZiB0aGUgYW5zd2Vycw==[Qq]

[c]IGNocm9tb3NvbWFsIGludmVyc2lvbi4=[Qq]

[f]IE5vLiBBIGNocm9tb3NvbWFsIGludmVyc2lvbiB3b3VsZG4mIzgyMTc7dCBiZSBwb3NzaWJsZSB0byBkaXNjZXJuIGluIHRoaXMgdHlwZSBvZiBibGFjay1hbmQtd2hpdGUga2FyeW90eXBlLiBBbiBpbnZlcnNpb24gcmVzdWx0cyB3aGVuIGNocm9tb3NvbWFsIHNlZ21lbnRzIGFyZSBmbGlwcGVkLCBhcyBpcyBzaG93biBpbiB0aGUgZGlhZ3JhbSBiZWxvdyAoZnJvbSBub2JlbHByaXplLm9yZyk=

Cg==

Cg==

[Qq]Take a closer look at the chromosomes in the karyotype, and see if you can find some type of chromosomal difference that corresponds to one of the other answers.

[q json=”true” xyz=”2″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative MC Dataset 2|1a637938da2cd5″ question_number=”20″ topic=”5.6.Chromosomal_Inheritance”] In an experimental organism, the alleles for height and body color are linked on the same chromosome. Allele “T” results in a tall phenotype, while “t” results in a short phenotype. “B” results in a black body, while “b” results in brown. A test cross resulted in four types of offspring. The alleles and their position on their chromosomes for the two recombinant offspring are shown below.

Which of the following is a test cross that could have produced these results? Note from Mr. W: this is a very difficult question.

[c]IEEg[Qq][c]IEIg[Qq][c]IE Mg[Qq][c]IEQ=

Cg==[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcXVlc3Rpb24uIEEgdGVzdCBjcm9zcw==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[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcXVlc3Rpb24uIEEgdGVzdCBjcm9zcw==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[Qq]

[f]IENvcnJlY3QhIEEgY3Jvc3MgYmV0d2VlbiBUQi8vdGIgYW5kIHRiLy90YiB3aWxsIHJlc3VsdCBpbiB0aGUgdHdvIHJlY29tYmluYW50IG9mZnNwcmluZyBzaG93biBhYm92ZSwgYXMgd2VsbCBhcyB0d28gcGFyZW50YWwtdHlwZSBvZmZzcHJpbmcgKFRCLy90YiBhbmQgdGIvL3RiKS4=[Qq]

[f]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[Qq]

[/qwiz]


5. Unit 5 Cumulative Free Response Questions

[qwiz style=”width: 550px !important; min-height: 400px !important;” dataset=”Unit 5 Cumulative FRQs” qrecord_id=”sciencemusicvideosMeister1961-Unit 5 Cumulative FRQs”]

[h]Unit 5 Cumulative FRQs

[i]

[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative FRQs|1a62b350e778d5″ question_number=”1″ topic=”5.1-5.2.Meiosis_and_Genetic_Diversity”] The diagrams below represent three reproductive strategies.

  • A is a multicellular organism about 1 centimeter in length
  • B is much smaller. The second image in row B shows individual cells
  • C is about 10 centimeters in length.

PART 1: Identify which of the strategies above will produce offspring with the greatest genetic variation.

PART 2: Justify your selection.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IFBBUlQgMTogU3RyYXRlZ3kgQiB3aWxsIHByb2R1Y2Ugb2Zmc3ByaW5nIHdpdGggdGhlIGdyZWF0ZXN0IGdlbmV0aWMgdmFyaWF0aW9uLg==

Cg==

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

Cg==

[Qq]

[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative FRQs|1a629c087090d5″ question_number=”2″ topic=”5.1-5.2.Meiosis_and_Genetic_Diversity”] The diagram below shows two cells, P and Q, that during sexual reproduction combine to form cell R. Cell R divides to produce an embryo.

PART 1: Label each of the following cells as haploid or diploid: Cell P, Cell Q, Cell R, Cells of the embryo

PART 2: Each cell of the embryo in the diagram has the potential to generate any human cell. Explain.

PART 3: For a zygote to be formed and for that zygote to develop into an embryo, two different types of cell division are involved. Define and describe each type of cell division, and explain its significance.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IFBBUlQgMTogQ2VsbCBQOiA=SGFwbG9pZA==LCBDZWxsIFE6IA==SGFwbG9pZA==[Qq], Cell R: Diploid, Cells of the embryo: Diploid

PART 2: Each cell in the embryo is a diploid cell that possesses the entire genome for the whole organism. Because the cells are early-stage embryonic cells, epigenetic changes (such as methylation) will not yet have occurred, allowing each cell to be the source of an entirely new organism.

PART 3: Zygote formation and embryonic development require meiosis and mitosis. Meiosis occurs during the formation of gametes, or sex cells. Sex cells are shown as P & Q in the diagram. Meiosis produces daughter cells that are haploid, with half the number of chromosomes and, because of the way that meiosis occurs, half of the genome. Meiosis sets the stage for the full diploid number to be restored at fertilization (at R) when two haploid gametes fuse together. During mitosis, daughter cells are produced that are genetic clones of the parent cell, containing the entire genome. At the same time, these daughter cells can differentiate into the specialized cells and tissues of a multicellular organism.

[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative FRQs|1a62826bedc4d5″ question_number=”3″ topic=”5.1-5.2.Meiosis_and_Genetic_Diversity”] Identical twins are genetic clones, formed when the cells in a dividing zygote (a fertilized egg) separate into two physically distinct individuals, instead of remaining connected within the same body.

Two identical twin brothers (David and John) married two identical twin sisters (Sarah and Anne). Each couple had a single child, as shown in the diagram below.

Estimate the percentage of alleles shared by the individuals listed in the table below. For each individual, justify your answer.

Individuals % of alleles shared Justification
David and John
Anne and Lisa
Sarah and Lisa

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]

Cg==Cg==Cg==Cg==
[Qq]Individuals % of alleles shared Justification
David and John 100% David and John arose from the same fertilized egg cell. All of their alleles will be the same (except for ones that change through mutation)
Anne and Lisa 50% Through meiosis, Anne produced the unfertilized egg that John’s sperm fertilized, producing the zygote that became Lisa. Meiosis involves passing on 50% of one’s alleles to one’s offspring (thus a 50% match)
Sarah and Lisa 50% Because Sarah is Anne’s identical twin, they both possess the same genes. During meiosis, they’ll randomly pass on half of their alleles to their egg cells. In the same way that Lisa has 50% of Anne’s alleles, she’ll also have 50% of Sarah’s alleles.

 

[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative FRQs|1a626b2376dcd5″ question_number=”4″ topic=”5.3.Mendelian_Genetics”] A team of geneticists is investigating an inherited condition found in three generations within a family. They create the pedigree chart below.

PART 1: Propose the most likely mode of inheritance for this condition.

PART 2: Justify your proposal with at least two pieces of evidence from the pedigree chart.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IFBBUlQgMTo=VGhlIGNvbmRpdGlvbiBpcyBtb3N0IGxpa2VseSBhdXRvc29tYWwgYW5kIGRvbWluYW50LsKg

Cg==

[Qq]PART 2:  We know that the condition is most likely dominant because two parents with the condition (such as parents II-5 and II-6 can have a child without the condition. That would be the case if both II-5 and II-6 were heterozygotes (Aa, with “A” being the allele that causes the condition). In that case, they’d have a 1/4 chance of having a child whose genotype is “aa” and would not have the condition (as we see in III-9). If the condition were recessive then the genotypes of II-5 and II-6 would have to be “aa,” and all of their children would have to be “aa.”

We know that the condition is autosomal because it appears equally in males and females. Additional proof that the condition is autosomal comes from the cross between II-5 and II-6.

If the condition was sex-linked recessive, then the Punnett square showing the cross between II-5 and II-6 would be as follows. II-5 (a male with the condition) would have to be XaY. II-6 (a female with the condition) would have to be XaXa.  As you can see in the Punnett square below, a girl without the condition would be impossible. Therefore the condition must be autosomal.

Xa Xa
Xa XaXa XaXa
Y XaY XaY

 

[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative FRQs|1a62562f0bd8d5″ question_number=”5″ topic=”5.3.Mendelian_Genetics”]

Note: the genes described in the following problem are fictional (for instructional purposes). In a certain breed of dog, eyes can be blue (B) or grey (b), the tail can be rigid (R) or stumpy (r), and hair can be long (Q) or short (q). B, R, and Q are the dominant alleles and the three genes are on different chromosomes.

PART 1: Betty is a dog that shows the three dominant traits. Indicate the genotype of the dog you should cross Betty with in order to determine her genotype. Explain your reasoning.

PART 2: All of the progeny of the cross described above have blue eyes and long hair. However, half have a stumpy tail, and half have a rigid tail. Identify Betty’s genotype. Explain your reasoning.

Part 3: Determine what fraction of the offspring from the cross BBRrQq X BbrrQq would be blue-eyed with a stumpy tail and short hair.

PART 4: The cross above produced 64 dogs from several matings. Predict how many dogs have the genotype BBrrqq.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]

Cg==

UEFSVCAxOiA=VGhlIGdlbm90eXBlIGlzIA==[Qq]bbrrqq. Explanation: This kind of cross is what is known as a test cross, and here’s how it works. To determine the genotype of an organism whose genotype is unknown, you cross it with an individual who is recessive (and therefore homozygous) for all the traits. If many offspring are produced and all of the offspring have the dominant trait, then the test-crossed parent has to be homozygous dominant. If there’s a mix of the dominant and recessive traits in the offspring, then you know that the test-crossed parent was a heterozygote for that trait.

PART 2. Betty’s genotype is BBRrQQ. Explanation: If all of the progeny of the cross described above have blue eyes and long hair, then Betty must have been homozygous dominant for those two traits, and her genotype must be BB and QQ. If you’re not clear on this, just draw a Punnett square that crosses BB x bb. All of the offspring will be Bb (with blue eyes). The same works for QQ x qq. On the other hand, if half of Betty’s offspring have a stumpy tail and half have a rigid tail, then Betty must have been heterozygous for tail type, or Rr. Again, imagine (or draw) a Punnett square that’s Rr x rr. Half the offspring will be Rr, and half will be rr.

PART 3: 1/8. Explanation: Use the rule of multiplication to solve this problem. In a cross between BB x Bb, all of the offspring will have blue eyes (BB or BB). Crossing Rr with rr, half of the offspring will have stumpy tails (rr). Crossing Qq with Qq, 1/4 will have short hair (qq). 1 x 1/2 x 1/4 = 1/8.

PART 4: Four dogs will have genotype BBrrqq. Explanation. Again, use the rule of multiplication to solve this problem. In a cross between BB x Bb, 1/2 of the genotype will have genotype BB. Crossing Rr with rr, 1/2 will have genotype rr. Crossing Qq with Qq, 1/4 will have genotype qq. 1/2 x 1/2 x 1/4 = 1/16.
64 times 1/16= 4.

[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative FRQs|1a623ee694f0d5″ question_number=”6″ topic=”5.3.Mendelian_Genetics”] A team of geneticists is investigating an inherited condition found in four generations within a family. They create the pedigree chart below.

PART 1: Propose the most likely mode of inheritance for this condition.

PART 2: Justify your proposal with at least two pieces of evidence from the pedigree chart

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IFBBUlQgMTogVGhlIG1vc3QgbGlrZWx5IG1vZGUgb2YgaW5oZXJpdGFuY2UgaXMgYXV0b3NvbWFsIHJlY2Vzc2l2ZS7CoA==

Cg==

UEFSVCAyOiBXZSBrbm93IHRoZSBjb25kaXRpb24gaXMgcmVjZXNzaXZlIGJlY2F1c2UgdHdvIHBhcmVudHMgd2hvIGFyZSB1bmFmZmVjdGVkIChzdWNoIGFzIEktMSBhbmQgSS0yIG9yIElJSS0zIGFuZCBJSUktNCBwcm9kdWNlIG9uZSBvciBtb3JlIG9mZnNwcmluZyB3aXRoIHRoZSBjb25kaXRpb24uIEZvciBleGFtcGxlLCBpZiB3ZSBhc3N1bWUgdGhhdCB0aGUgZ2Vub3R5cGUgb2YgYm90aCBJLTEgYW5kIEktMiBpcyAmIzgyMjA7QWEsJiM4MjIxOyB0aGVuIHRoZXkgaGF2ZSBhIDEvNCBjaGFuY2Ugb2YgYW55IG9uZSBvZiB0aGVpciBvZmZzcHJpbmcgKHN1Y2ggYXMgSUktMikgaW5oZXJpdGluZyBnZW5vdHlwZSAmIzgyMjA7YWEuJiM4MjIxOyA=[Qq]

We know that the condition is autosomal and not sex-linked because the condition shows up equally in males and females. Additional evidence against sex linkage (and for autosomal inheritance) is the mating between III-3 and III-4, which produce a girl (IV-3) with the condition. As the Punnett square below shows, that can only happen when a father with the condition has a child with a mother who is a carrier (as shown below). 

XA Xa
Xa XAXa XaXa
Y XAY XaY

III-3 does not have the condition. Therefore his genotype has to be XAY. A union between XAY and a carrier XAXa could not produce a girl with the condition, as shown below. 

XA Xa
XA XAXA XAXa
Y XAY XaY

[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative FRQs|1a6220a1fa5cd5″ question_number=”7″ topic=”5.4.Non-Mendelian_Genetics”] In a newly discovered species of fruit fly, the allele for black bristles (B) is dominant to green bristles (b) and red eyes (R) is dominant to white eyes (r).

Several BbRr flies are test crossed with bbrr flies. The observed distribution of offspring is black-red, 1070; black-white,177; green-red, 180; green-white 1072.

PART 1: Using a Punnett square, provide evidence that the alleles for bristle color and eye color do not independently assort.

PART 2: Calculate the recombination frequency between the bristle color and eye color alleles.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IFBBUlQgMTogSWYgdGhlIGFsbGVsZXMgZm9yIGJyaXN0bGUgY29sb3IgYW5kIGV5ZSBjb2xvciBpbmRlcGVuZGVudGx5IGFzc29ydGVkLCB0aGVuIHlvdSB3b3VsZCBleHBlY3QgYSAxOjE6MToxIHJhdGlvIG9mIHBoZW5vdHlwZXMgaW4gdGhlIG9mZnNwcmluZy4gVGhlIA==QmJScg==IHBhcmVudCBjb3VsZCBwcm9kdWNlIGZvdXIgdHlwZXMgb2YgZ2FtZXRlcywgYW5kIHRoZSA=YmJycg==IHBhcmVudCBjb3VsZCBwcm9kdWNlIG9uZSB0eXBlIG9mIGdhbWV0ZSwgYXMgc2hvd24gaW4gdGhlIG1vZGlmaWVkIFB1bm5ldHQgc3F1YXJlIGJlbG93Og==

[Qq]
BR Br bR br
br BbRr Bbrr bbRr bbrr
expected phenotype black-red black-white green-red green-white
expected proportion 25% 25% 25% 25%
actual results 1070 177 180 1072

 

PART 2: Instead of 1:1:1:1, 2142 offspring have parental phenotypes (black-red or green-white), while 357 have recombinant phenotypes (black-white or green-red). This can be explained by interpreting the results as follows: the bristle color and eye color alleles are linked on the same chromosome. As a result, they usually stay together (producing parental phenotypes), but because of crossing over, they’ll also recombine (producing recombinant phenotypes) with a frequency of 357/2499, or 14.2%. Another way to say this is that the bristle color and eye color alleles are on the same chromosome, 14.2 cM apart.

[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative FRQs|1a62000953e4d5″ question_number=”8″ topic=”5.4.Non-Mendelian_Genetics”] You are studying the inheritance of flower color in poppies. Suppose that flower color is determined by a single gene with 3 alleles:

H = purple allele

I = orange allele

J = red allele

The phenotypes produced by the different genotypes are:

HH = purple flowers

HI = purple flowers

II = orange flowers

HJ = white flowers

JJ = red flowers

IJ = lethal (seed does not germinate and no flowers are produced)

Suppose you cross a plant producing purple flowers (genotype HI) with a
plant producing white flowers (genotype HJ).

PART 1: Predict the frequencies of the genotypes and the phenotypes in the surviving offspring.

PART 2: Justify your prediction.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IA==UEFSVCAxOiBUaGUgZ2Vub3R5cGVzIG9mIHRoZSBzdXJ2aXZpbmcgb2Zmc3ByaW5nIGFyZSAxLzMgSEgsIDEvMyBISSwgYW5kIDEvMyBISi4gVGhlIHBoZW5vdHlwZXMgYXJlIDIvMyBwdXJwbGUgZmxvd2VycyBhbmQgMS8zIHdoaXRlIGZsb3dlcnMu

Cg==

UEFSVCAyOsKgIFdlJiM4MjE3O3JlIHRvbGQgdGhhdCB0aGUgcGFyZW50cyBhcmUgSEkgYW5kIEhKLg==

Cg==

[Qq]Here’s the Punnett square.

H I
H HH HI
J HJ IJ

 

Three genotypes (HH, HI, and HJ) appear in equal proportion. The IJ genotype is lethal and isn’t counted in the survivors. To determine the phenotypes look up the phenotype associated with each genotype. HH is purple, as is HI. HJ is white. IJ is lethal. So, 2/3 of the offspring are purple, and 1/3 are white.

[q json=”true” xx=”1″ multiple_choice=”false” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative FRQs|1a61e1c4b950d5″ question_number=”9″ topic=”5.4.Non-Mendelian_Genetics”] Cynthia and Neil have just had twins, a boy named Jadin and a girl named Kacee. Both of the babies’ grandfathers are red-green color blind, while their grandmothers have normal vision. Neil, who is an only child, has normal vision. Cynthia and her brother also have normal vision.

PART A: Construct a family pedigree to show the inheritance of this recessive sex-linked genetic disorder. In your pedigree, use circles for females and squares for males; shade those affected by color blindness; use partial shading to indicate carriers.

PART B: Explain what can be deduced about the genotypes of both Cynthia and Neil with respect to red-green color blindness from the given information.

PART C: Use your answers to part (b) to predict the probabilities that Jadin and Kacee will be color blind. Justify your response.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]IFBBUlQgQS4gSGVyZSYjODIxNztzIHRoZSBwZWRpZ3JlZS4gTm90ZSB0aGF0IGluIHlvdXIgcGVkaWdyZWUgdGhlIHBvc2l0aW9ucyBvZiB0aGUgaW5kaXZpZHVhbHMgbWlnaHQgYmUgZGlmZmVyZW50LCBidXQgdGhlIG92ZXJhbGwgaW5mb3JtYXRpb24gbXVzdCBiZSB0aGUgc2FtZS4=

Cg==

Cg==

[Qq]PART B: XC denotes the allele for normal vision. Xc denotes the allele for normal vision. Y denotes the Y chromosome. Neil must be XCY because he has normal vision (his only X chromosome must carry the gene for normal vision). Cynthia (because her dad was colorblind and therefore had to pass on his Xc allele to his daughters), must be a carrier, making her genotype XCXc.

PART C: Jadin’s genotype, as shown in the Punnett Square below, could be XCY or XcY). Therefore, he has a 50% chance of being colorblind. Kacee’s genotype could be XCXC or XCXc, which means that she has a 0% chance of being colorblind and a 50% chance of being a carrier.

[q json=”true” multiple_choice=”false” unit=”5.Heredity” topic=”5.6.Chromosomal_Inheritance” dataset_id=”Unit 5 Cumulative FRQs|1a615169d818d5″ question_number=”10″] Alport syndrome is a genetic condition that is characterized by hearing loss and kidney disease. Often, the disease is associated with abnormalities in the COL4A5 gene.

To study mechanisms of inheritance in Alport syndrome, scientists genetically modified rats so that they lacked the COL4A5 gene. Then, they measured the survival of the offspring of the rats with the mutated version and the non-mutated (wild-type) version. Their results are shown below.

PART 1: Identify two controlled variables (constants) that the scientists should have implemented in this study.

PART 2: Make a claim about the type of inheritance pattern that Alport Syndrome shows in mice. Justify your claim with evidence or reasoning.

[c]IFNob3cgdGhl IGFuc3dlcg==[Qq]

[f]

Cg==

UEFSVCAxOiBBbnkgdHdvIG9mIHRoZSBmb2xsb3dpbmcgYXJlIGFjY2VwdGFibGUgKGFuZCBvdGhlciBjb3JyZWN0IHJlc3BvbnNlcyBhcmUgcG9zc2libGUp

Cg==
    Cg==
  • [Qq]The rats needed to be fed the same diet
  • The rats needed to be given the same space/enclosure
  • The rats needed to have been screened for other genetic disorders causing premature death
  • The rats needed to have been from the same line/pedigree.

PART 2: This pattern of inheritance is most likely sex-linked. The evidence is that the male mutant rats have a lower survival rate than the female mutant rats. This would happen if the disorder is X-linked. In that case, males with the mutation would completely lack the gene (since males only have one X chromosome) and produce none of the gene’s product (resulting in the symptoms of the disease). Females with one mutation would still have one normal copy of the gene on their second X chromosome, and would still produce some of the gene product associated with the COL4A5 gene.

[/qwiz]


6. Meiosis Click-On Challenge

[qwiz random=”true” style=”width: 600px !important;” quiz_timer=”true” use_dataset=”Meiosis Click-on dataset” dataset_intro=”true” spaced_repetition=”false” qrecord_id=”sciencemusicvideosMeister1961-Unit 5 Meiosis Click On Challenge”]

[h] Meiosis Click-on Challenge

[i] Note the timer in the top right. Your goal is accuracy and speed. A good strategy: once through slowly, then additional trials for improvement.
[/qwiz]