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

Explain how meiosis transmits genetic material from one generation to the next.
Compare and contrast diploid and haploid cells, and explain how these terms connect to somatic cells and germ cells.
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

Explain how meiosis generates genetic diversity.
Define “homologous” chromosomes (their origin, their relationship in terms of genetic information) and explain what happens to homologous pairs during meiosis
Explain what crossing over is, and how it generates genetic diversity.
Explain fertilization (in terms of haploid and diploid chromosome numbers), as well as fertilization’s contribution to genetic diversity.
Compare meiosis 1 and meiosis 2, and explain what happens during each process.
Connect the events of meiosis and fertilization to how sexual reproduction creates variation.
Connect the events of meiosis to Mendel’s laws of segregation and independent assortment; and recombination of linked alleles.
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

Explain Mendel’s laws of segregation and independent assortment, and connect them to what happens during meiosis.
Explain relevant rules of probability that apply to genetics.
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

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.
Explain the inheritance patterns of sex-linked genes, and be able to solve genetics problems involving sex linkage.
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.
Define polygenic traits, and describe why these usually have a bell-curve-shaped distribution pattern.
Explain non-nuclear inheritance

Topic 5.5. Environmental Effects on Phenotype

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 2⁴chromosomally 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 Questions

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[h] Unit 5 Cumulative Multiple Choice Questions

[i]

Heredity

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e38515643062c” question_number=”1″ 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]SSA=[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]

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMuIFlvdSYjODIxNztyZSBpbnN0cnVjdGVkIHRvIGxvb2sgZm9yIHRoZSBhbW91bnQgb2YgRE5BIGluIGFuIA==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[Qq]

[f]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[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e37eae437d62c” question_number=”2″ 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==[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e37891a446e2c” question_number=”3″ 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 a cell during prophase 1 of meiosis I?

[c]IEkg[Qq][c]IElJ SSA=[Qq][c]IElWIA==[Qq][c]IFZJ

Cg==[Qq]

[f]IE5vLiBZb3UmIzgyMTc7cmUgdG9vIGVhcmx5IGluIHRoZSBwcm9jZXNzLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMuIEZyb20gdGhlIHF1ZXN0aW9uLCB5b3Uga25vdyB0aGF0IHRoaXMgaXMgbWVpb3Npcy4gTWVpb3NpcyBiZWdpbnMgYnkgZG91YmxpbmcgdGhlIEROQSAoc2hvd24gYXQgJiM4MjIwO0lJJiM4MjIxOyksIHRoZW4gZGl2aWRlcyB0aGUgRE5BIHR3aWNlLiBQcm9waGFzZSAxIHdvdWxkIGJlIGFmdGVyIHRoZSBETkEgZG91YmxpbmcgYnV0IGJlZm9yZSB0aGUgZmlyc3QgZGl2aXNpb24uIEZpZ3VyZSBpdCBvdXQgZnJvbSB0aGVyZS4=[Qq]

[f]IEZhYnVsb3VzLiBQcm9waGFzZSAxIHdvdWxkIGJlIGFmdGVyIHRoZSBkb3VibGluZyBvZiBETkEgaW4gaW50ZXJwaGFzZSBJIChzaG93biBhdCBJSSkgYnV0IGJlZm9yZSB0aGUgY2VsbCBkaXZpc2lvbiBvZiBtZWlvc2lzIEkgKHNob3duIGF0IElWKS4=[Qq]

[f]IE5vLiBZb3UmIzgyMTc7cmUgdG9vIGxhdGUgaW4gdGhlIHByb2Nlc3MuIEhlcmUmIzgyMTc7cyBob3cgdG8gdGhpbmsgYWJvdXQgdGhpcy4gRnJvbSB0aGUgcXVlc3Rpb24sIHlvdSBrbm93IHRoYXQgdGhpcyBpcyBtZWlvc2lzLiBNZWlvc2lzIHN0YXJ0cyB3aXRoIGEgZG91Ymxpbmcgb2YgRE5BIChzaG93biBhdCAmIzgyMjA7SUkmIzgyMjE7KSwgdGhlbiBkaXZpZGVzIHRoZSBETkEgdHdpY2UuIFByb3BoYXNlIDEgd291bGQgYmUgYWZ0ZXIgdGhlIEROQSBkb3VibGluZyBidXQgYmVmb3JlIHRoZSBmaXJzdCBkaXZpc2lvbi4gRmlndXJlIGl0IG91dCBmcm9tIHRoZXJlLg==[Qq]

[f]IE5vLiBZb3UmIzgyMTc7cmUgdG9vIGxhdGUgaW4gdGhlIHByb2Nlc3MuIEhlcmUmIzgyMTc7cyBob3cgdG8gdGhpbmsgYWJvdXQgdGhpcy4gRnJvbSB0aGUgcXVlc3Rpb24sIHlvdSBrbm93IHRoYXQgdGhpcyBpcyBtZWlvc2lzLiBNZWlvc2lzIHN0YXJ0cyB3aXRoIGEgZG91Ymxpbmcgb2YgRE5BLCB0aGVuIGRpdmlkZXMgdGhlIEROQSB0d2ljZS4gUHJvcGhhc2UgMSB3b3VsZCBiZSBhZnRlciB0aGUgRE5BIGRvdWJsaW5nIGJ1dCBiZWZvcmUgdGhlIGZpcnN0IGRpdmlzaW9uLiBGaWd1cmUgaXQgb3V0IGZyb20gdGhlcmUu[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e371e0021762c” question_number=”4″ topic=”5.1.Meiosis”] In the diagram below, maternal chromosomes are colored white, and paternal chromosomes are colored black.

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” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e36a995ceee2c” 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 could develop into an individual with Turner syndrome?

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

Cg==[Qq]

[f]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[Qq]

[f]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[Qq]

[f]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[Qq]

[f]IEV4Y2VsbGVudC4gVGhlIHp5Z290ZSBhdCAmIzgyMjA7ZCYjODIyMTsgaXMgbWlzc2luZyBhIGNocm9tb3NvbWUuIFNpbmNlIHBlb3BsZSB3aXRoIFR1cm5lciBzeW5kcm9tZSBhcmUgZmVtYWxlcyB3aG8gYXJlIG1pc3NpbmcgYW4gWCBjaHJvbW9zb21lLCB0aGVuIHRoYXQgY2VsbCBmaXRzIHRoZSBkZXNjcmlwdGlvbiBvZiBhIHp5Z290ZSB0aGF0IGNvdWxkIGxlYWQgdG8gYSBwZXJzb24gd2l0aCBUdXJuZXIgc3luZHJvbWUu[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e3632d770822c” 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==[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e35b9c506322c” question_number=”7″ 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]

[f]IFllcy4gRE5BIHJlcGFpciB3aWxsIHRha2UgcG90ZW50aWFsIG11dGF0aW9ucyBhbmQgZWxpbWluYXRlIHRoZW0sIHdoaWNoIHJlZHVjZXMgZ2VuZXRpYyBkaXZlcnNpdHku[Qq]

[c]IHJhbmRvbSBhc3NvcnRtZW50IG9mIG1hdGVybmFsIGFuZCBwYXRlcm5hbCBjaHJvbW9zb21lcyBhc3NvY2lhdGVkIHdpdGggbWV0YXBoYXNlIDEgYW5kIGFuYXBoYXNlIDEu[Qq]

[f]IE5vLiBSYW5kb20gYXNzb3J0bWVudCBvZiBtYXRlcm5hbCBhbmQgcGF0ZXJuYWwgY2hyb21vc29tZXMgKGluZGVwZW5kZW50IGFzc29ydG1lbnQpIHByb2R1Y2VzIGNocm9tb3NvbWFsIGNvbWJpbmF0aW9ucyB0aGF0IGFyZSBicmFuZCBuZXcmIzgyMzA7VGhpcyBpcyBhIG1ham9yIHNvdXJjZSBvZiB2YXJpYXRpb24uIFdoYXQmIzgyMTc7cyBvbiB0aGUgbGlzdCB0aGF0IHdvdWxkIA==ZGVjcmVhc2U=IHZhcmlhdGlvbj8=[Qq]

[c]IHJhbmRvbSBmZXJ0aWxpemF0aW9uIG9mIGdhbWV0ZXMgdGhhdCBmb2xsb3dzIG1laW9zaXMu[Qq]

[f]IE5vLiBSYW5kb20gZmVydGlsaXphdGlvbiBvZiBnYW1ldGVzIGluY3JlYXNlcyB2YXJpYXRpb24sIGJ5IGNvbWJpbmluZyB0aGUgZ2VuZXMgb2YgdW5yZWxhdGVkIG1hdGVzLiBXaGF0JiM4MjE3O3Mgb24gdGhlIGxpc3QgdGhhdCB3b3VsZCA=ZGVjcmVhc2U=IHZhcmlhdGlvbj8=

Cg==

[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e355ca32a922c” question_number=”8″ 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]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcHJvYmxlbS4gWW91IGtub3cgdGhhdCB0aGUgbWFja2VyZWwgdGFiYnkgdHJhaXQgaXMgZG9taW5hbnQgKGJlY2F1c2UgYWxsIG9mIHRoZSBG_MQ==IG9mZnNwcmluZyBhcmUgbWFja2VyZWwpLiBBbmQgeW91IGFsc28ga25vdyB0aGF0IHRoZSBG_MQ==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.

	M	m
M
m

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

	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” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e34ea8ce3ee2c” question_number=”9″ 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]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcHJvYmxlbS4gWW91IGtub3cgdGhhdCB0aGUgbWFja2VyZWwgdGFiYnkgdHJhaXQgaXMgZG9taW5hbnQgKGJlY2F1c2UgYWxsIG9mIHRoZSBG_MQ==IG9mZnNwcmluZyBhcmUgbWFja2VyZWwpLiBBbmQgeW91IGFsc28ga25vdyB0aGF0IHRoZSBG_MQ==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]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcHJvYmxlbS4gWW91IGtub3cgdGhhdCB0aGUgbWFja2VyZWwgdGFiYnkgdHJhaXQgaXMgZG9taW5hbnQgKGJlY2F1c2UgYWxsIG9mIHRoZSBG_MQ==IG9mZnNwcmluZyBhcmUgbWFja2VyZWwpLiBBbmQgeW91IGFsc28ga25vdyB0aGF0IHRoZSBG_MQ==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]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcHJvYmxlbS4gWW91IGtub3cgdGhhdCB0aGUgbWFja2VyZWwgdGFiYnkgdHJhaXQgaXMgZG9taW5hbnQgKGJlY2F1c2UgYWxsIG9mIHRoZSBG_MQ==IG9mZnNwcmluZyBhcmUgbWFja2VyZWwpLiBBbmQgeW91IGFsc28ga25vdyB0aGF0IHRoZSBG_MQ==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]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcHJvYmxlbS4gWW91IGtub3cgdGhhdCB0aGUgbWFja2VyZWwgdGFiYnkgdHJhaXQgaXMgZG9taW5hbnQgKGJlY2F1c2UgYWxsIG9mIHRoZSBG_MQ==IG9mZnNwcmluZyBhcmUgbWFja2VyZWwpLiBBbmQgeW91IGFsc28ga25vdyB0aGF0IHRoZSBG_MQ==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” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e344e8dc7422c” question_number=”10″ 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]IE5vLiBUaGUga2V5IGlkZWEgaXMgdGhhdCB0aGVzZSBnZW5lcyAoYW5kIHRoZWlyIGFsbGVsZXMpIGFyZSBub3QgbGlua2VkLiBBcyBhIHJlc3VsdCwgdGhlcmUgd29uJiM4MjE3O3QgYmUgcGFydGljdWxhciBjb21iaW5hdGlvbnMgb2YgYWxsZWxlcyB0aGF0IGFyZSBhbHdheXMgaW5oZXJpdGVkIHRvZ2V0aGVyLiBJbnN0ZWFkLCB0aGV5JiM4MjE3O2xsIGJlIGluaGVyaXRlZCBpbmRlcGVuZGVudGx5LiBSZW1lbWJlciB0aGF0IHRoZSBuZXh0IHRpbWUgeW91JiM4MjE3O2xsIHNlZSB0aGlzIHF1ZXN0aW9uLCBhbmQgY2hvb3NlIGFub3RoZXIgYW5zd2VyLg==[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e33d57b5cf22c” question_number=”11″ 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]

[f]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[Qq]

[f]IEV4Y2VsbGVudCEgVGhlIHByb2JhYmlsaXR5IGlzIDEvMiB4IDEvMiB4IDEvMiwgb3IgMS84Lg==[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e33556ccef62c” question_number=”12″ 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?

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e32f3a2db8e2c” question_number=”13″ 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]IE5vLiBUaGluayBhYm91dCBpdCB0aGlzIHdheS4gSW4gYSBtb25vaHlicmlkIGNyb3NzIGJldHdlZW4gQWE=IGFuZCA=QWE=LCB0aGUgb2Zmc3ByaW5nIGNhbiBiZSBob21venlnb3VzIGluIHR3byB3YXlzOiB0aGV5IGNhbiBiZSA=[Qq]aa or AA. If you imagine the Punnett square for Aa x Aa, you’ll see that 1/2 of the squares are filled with aa or AA. The same is true for Bb x Bb, Cc x Cc, and Dd x Dd. Now use the rule of multiplication, and you’ll have the answer.

[f]IEV4Y2VsbGVudCEgSW4gYSB0ZXRyYWh5YnJpZCBjcm9zcyAoQWFCYkNjRGQgeCBBYUJiQ2NEZCksIHRoZSBudW1iZXIgb2Ygb2Zmc3ByaW5nIHRoYXQgd2lsbCBiZSBob21venlnb3VzIGZvciBhbGwgZm91ciBnZW5lcyBpcyAxLzIgeCAxLzIgeCAxLzIgeCAxLzIsIG9yIDEvMTYu[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e32659c063a2c” question_number=”14″ 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 show a recessive phenotype for all four genes? (Helpful note from Mr. W: Do not solve this by creating some huge Punnett square. Use the rules of probability.)

[c]IDEvMj U2IA==[Qq][c]IDEvMzIg[Qq][c]IDEvMTYg[Qq][c]IDEvOCA=[Qq][c]IDEvNA==

Cg==[Qq]

[f]IEV4Y2VsbGVudCEgSW4gYSB0ZXRyYWh5YnJpZCBjcm9zcyAoQWFCYkNjRGQgeCBBYUJiQ2NEZCkgdGhlcmUmIzgyMTc7cyBhIDEvMjU2IGNoYW5jZSBvZiB0aGUgb2Zmc3ByaW5nIHNob3dpbmcgdGhlIHJlY2Vzc2l2ZSBwaGVub3R5cGUgZm9yIGFsbCBmb3VyIGdlbmVzICgxLzQgeCAxLzQgeCAxLzQgeCAxLzQp[Qq]

[f]IE5vLiBUaGVyZSBhcmUgdHdvIHN0ZXBzIHRvIGdldHRpbmcgdGhpcyBwcm9ibGVtIHJpZ2h0LiBGaXJzdCwgbm90ZSB0aGF0IGluIGEgbW9ub2h5YnJpZCBjcm9zcyAoQWEgeCBBYSkgdGhlcmUmIzgyMTc7cyBhIDEvNCBjaGFuY2Ugb2YgZ2V0dGluZyBhbiBvZmZzcHJpbmcgdGhhdCB3aWxsIHNob3cgdGhlIHJlY2Vzc2l2ZSBwaGVub3R5cGUuIFNlY29uZCwga25vd2luZyB0aGF0LCB5b3UgdXNlIHRoZSBydWxlIG9mIG11bHRpcGxpY2F0aW9uIHRvIGNhbGN1bGF0ZSB0aGUgcHJvYmFiaWxpdHkgb2YgdGhpcyBoYXBwZW5pbmcgZm9yIGFsbCBmb3VyIGFsbGVsZSBwYWlycy4gV2hhdCYjODIxNztzIDEvNCB4IDEvNCB4IDEvNCB4IDEvND8=[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e31f131b3b22c” question_number=”15″ 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]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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]IDEvMg==[Qq]

Cg==Cg==Cg==Cg==

[Qq]	T	t
T	TT	Tt
t	Tt	tt

	X	Y
X	XX	XY
X	XX	XY

[c]IDEvNA==[Qq]

[f]IE5vLiBUYXkgU2FjaHMgaXMgY2F1c2VkIGJ5IGFuIGF1dG9zb21hbCByZWNlc3NpdmUgYWxsZWxlLCBzbyB3ZSBjYW4gcmVwcmVzZW50IHRoYXQgYWxsZWxlIHdpdGggYSBsb3dlciBjYXNlICYjODIyMDt0LiYjODIyMTsgVGhlIG5vcm1hbCBhbGxlbGUgd2UgY2FuIHJlcHJlc2VudCBieSAmIzgyMjA7VC4mIzgyMjE7IElJSS0xIGFuZCBJSUktMiBib3RoIGhhdmUgdG8gYmUgaGV0ZXJvenlnb3RlcyAob3RoZXJ3aXNlLCB0aGVyZSYjODIxNztzIG5vIHdheSB0aGF0IHRoZXNlIHR3byB1bmFmZmVjdGVkIHBhcmVudHMgd291bGQgcGFzcyBvbiB0aGUgYWxsZWxlIHRvIG9uZSBvZiB0aGVpciBjaGlsZHJlbiwgc3VjaCBhcyBJVi0xKS4gU28gd2UgY2FuIHJlcHJlc2VudCBib3RoIHBhcmVudHMmIzgyMTc7IGdlbm90eXBlcyBhcyBUdC4gSWYgeW91IHNldCB1cCBhIFB1bm5ldHQgc3F1YXJlLCBpdCYjODIxNztsbCBsb29rIGxpa2UgdGhpczo=

Cg==Cg==Cg==Cg==

[Qq]	T	t
T	TT	Tt
t	Tt	tt

	X	Y
X	XX	XY
X	XX	XY

[c]IDEv OA==[Qq]

[f]IEV4Y2VsbGVudCEgVGhlcmUmIzgyMTc7cyBhIDEvNCBjaGFuY2UgdGhhdCB0aGVpciBjaGlsZCB3aWxsIGhhdmUgdGhlIGNvbmRpdGlvbiwgYW5kIGEgMS8yIGNoYW5jZSB0aGF0IHRoZWlyIGNoaWxkIHdpbGwgYmUgYSBib3kuIE11bHRpcGx5IHRoZSB0d28gdG9nZXRoZXIsIGFuZCB5b3UgZ2V0IDEvOC4=[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e317f1b6d0e2c” question_number=”16″ 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[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e31244d9d522c” question_number=”17″ 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
X^D	X^DX^D	XDY
X^d	X^DX^d	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]IE5vLiBYLWxpbmtlZCByZWNlc3NpdmUgKGEgY29uZGl0aW9uIGxpa2UgaGVtb3BoaWxpYSkgY2FuJiM4MjE3O3QgZXhwbGFpbiB0aGlzIHBlZGlncmVlLiBXZSYjODIxNztsbCB1c2UgJiM4MjIwO2gmIzgyMjE7IHRvIHJlcHJlc2VudCB0aGUgcmVjZXNzaXZlIGFsbGVsZSwgYW5kICYjODIyMDtIJiM4MjIxOyB0byByZXByZXNlbnQgaXRzIGRvbWluYW50IGNvdW50ZXJwYXJ0LiBMb29rIGF0IHRoZSBjcm9zcyBiZXR3ZWVuIElJLTUgYW5kIElJLTYuIElmIHRoaXMgY29uZGl0aW9uIHdlcmUgWC1saW5rZWQgcmVjZXNzaXZlLCB0aGVuIHRoZSBmYXRoZXIgd291bGQgaGF2ZSB0byBiZSBY^aA==WSBhbmQgdGhlIG1vdGhlciBY^RA==WA==^[Qq]d. The Punnett square would be

	Xh	Y
X^h	X^hX^h	XhY
X^h	X^hX^h	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” xx=”1″ multiple_choice=”true” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e30c02f9e062c” question_number=”18″ 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.

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.

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==[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e304e1957622c” question_number=”19″ topic=”5.3.Mendelian_Genetics”] In Drosophila, allele W codes for normal wings, while allele w codes for vestigial wings. Allele G codes for the normal body color, while allele g codes for ebony. A heterozygous normal winged, ebony male is crossed with a vestigial winged, heterozygous normal body color female.

What is the genotype of the male fly?

[c]IFd3R2c=[Qq]

[f]IE5vLiBUaGUgbWFsZSBpcyBkZXNjcmliZWQgYXMgaGV0ZXJvenlnb3VzIG5vcm1hbCB3aW5nZWQsIGVib255LiA=V3dHZw==IHdvdWxkIGJlIGRlc2NyaWJlZCBhcyBoZXRlcm96eWdvdXMgbm9ybWFsIHdpbmdlZCwgaGV0ZXJvenlnb3VzIG5vcm1hbCBib2R5IGNvbG9yLiBXaGF0IGdlbm90eXBlIHdvdWxkIGEgZmx5IGhhdmUgdG8gaGF2ZSB0byBiZSBlYm9ueT8=[Qq]

[c]IFdXR0c=[Qq]

[f]IE5vLiBUaGUgbWFsZSBpcyBkZXNjcmliZWQgYXMgaGV0ZXJvenlnb3VzIG5vcm1hbCB3aW5nZWQsIGVib255LiA=V1dHRw==IHdvdWxkIGJlIGRlc2NyaWJlZCBhcyBob21venlnb3VzIG5vcm1hbCB3aW5nZWQsIGhvbW96eWdvdXMgbm9ybWFsIGJvZHkgY29sb3IuIFdoYXQgZ2Vub3R5cGUgd291bGQgYSBmbHkgaGF2ZSB0byBoYXZlIHRvIGJlIGhldGVyb3p5Z291cyBmb3Igbm9ybWFsIHdpbmdzLCBhbmQgZWJvbnkgYm9kaWVkPw==[Qq]

[c]IFd3 Z2c=[Qq]

[f]IEF3ZXNvbWUhIEEgZ2Vub3R5cGUgb2YgV3dnZw==IG1lYW5zIHRoYXQgdGhlIGZseSBpcyBoZXRlcm96eWdvdXMgZm9yIG5vcm1hbCB3aW5ncyBhbmQgZWJvbnkgY29sb3Iu[Qq]

[c]IHd3Z2c=[Qq]

[f]IE5vLiBUaGUgbWFsZSBpcyBkZXNjcmliZWQgYXMgaGV0ZXJvenlnb3VzIG5vcm1hbCB3aW5nZWQsIGVib255LiBUaGUgZ2Vub3R5cGUgd3dnZw==IHdvdWxkIGJlIGRlc2NyaWJlZCBhcyBob21venlnb3VzIHZlc3RpZ2lhbCB3aW5nZWQsIGhvbW96eWdvdXMgZWJvbnkgYm9keSBjb2xvciAoc28geW91IGdvdCB0aGUgc2Vjb25kIHBhcnQgcmlnaHQpLiBXaGF0IGdlbm90eXBlIHdvdWxkIGEgZmx5IGhhdmUgdG8gaGF2ZSB0byBiZSBoZXRlcm96eWdvdXMgZm9yIG5vcm1hbCB3aW5ncz8=[Qq]

[c]IHd3R0c=[Qq]

[f]IE5vLiBUaGUgbWFsZSBpcyBkZXNjcmliZWQgYXMgaGV0ZXJvenlnb3VzIG5vcm1hbCB3aW5nZWQsIGVib255LiBUaGUgZ2Vub3R5cGUgd3dHRw==IHdvdWxkIGJlIGRlc2NyaWJlZCBhcyBob21venlnb3VzIHZlc3RpZ2lhbCB3aW5nZWQsIGhvbW96eWdvdXMgbm9ybWFsIGJvZHkgY29sb3IuIFdoYXQgZ2Vub3R5cGUgd291bGQgYSBmbHkgaGF2ZSB0byBoYXZlIHRvIGJlIGhldGVyb3p5Z291cyBmb3Igbm9ybWFsIHdpbmdzLCBhbmQgZWJvbnktYm9kaWVkIChhbmQgcmVtZW1iZXIgdGhhdCBlYm9ueS1ib2RpZWQgaXMgYSByZWNlc3NpdmUgdHJhaXQpLg==[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e2fd75af8f62c” question_number=”20″ topic=”5.3.Mendelian_Genetics”] In Drosophila, allele W codes for normal wings, while allele w codes for vestigial wings. Allele G codes for normal body color, while allele g codes for ebony coloring. A heterozygous normal winged, ebony male is crossed with a vestigial winged, heterozygous normal body color female. What ratio of of phenotypes is expected in the offspring?

[c]IDEgbm9ybWFsIHdpbmdzLCBub3JtYWwgYm9keTogMSBub3JtYWwgd2luZ3MsIGVib255IGJvZHk6IDEg dmVzdGlnaWFsIHdpbmdzLCBub3JtYWwgYm9keTogMSB2ZXN0aWdpYWwgd2luZ3MsIGVib255IGJvZHk=[Qq]

[f]IEV4Y2VsbGVudC4gQXMgeW91IGNhbiBzZWUgYmVsb3csIGEgY3Jvc3MgYmV0d2VlbiA=V3dnZw==IGFuZCA=d3dHZw==IHdpbGwgZ2l2ZSB5b3UgYSAxOjE6MToxIHJhdGlvIG9mIHBoZW5vdHlwZXMgaW4gdGhlIG9mZnNwcmluZy4=

[Qq]

	wG	wg
Wg	WgGg	Wwgg
wg	wwGg	wwgg

[c]IDMgbm9ybWFsIHdpbmdzLCBub3JtYWwgYm9keSA6IDMgbm9ybWFsIHdpbmdzLCBlYm9ueSBib2R5IDogMyB2ZXN0aWdpYWwgd2luZ3MsIG5vcm1hbCBib2R5IDogMyB2ZXN0aWdpYWwgd2luZ3MsIGVib255IGJvZHk=[Qq]

[f]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

Cg==Cg==Cg==Cg==

[Qq]	wG	wg
Wg	WgGg
wg

[c]IDMgbm9ybWFsIHdpbmdzLCBub3JtYWwgYm9keTogMSBub3JtYWwgd2luZ3MsIGVib255IGJvZHkgOiAzIHZlc3RpZ2lhbCB3aW5ncywgbm9ybWFsIGJvZHkgOiAxIHZlc3RpZ2lhbCB3aW5ncywgZWJvbnkgYm9keQ==[Qq]

[f]IE5vLiBTdGFydCBieSByZWFkaW5nIHRoZSBkZXNjcmlwdGlvbiBvZiB0aGUgcGFyZW50cywgYW5kIGNvbnZlcnRpbmcgdGhlbSBpbnRvIGdlbm90eXBlcy4gVGhlIG1hbGUgaXMgaGV0ZXJvenlnb3VzIG5vcm1hbCB3aW5nZWQgYW5kIGVib255LWJvZGllZCwgc28gaGlzIGdlbm90eXBlIGlzIA==V3dnZw==LiBUaGUgZmVtYWxlIGlzIHZlc3RpZ2lhbCB3aW5nZWQsIGhldGVyb3p5Z291cyBub3JtYWwgYm9keSBjb2xvciwgc28gaGVyIGdlbm90eXBlIGlzIA==d3dHZw==LiBJbiB0aGUgUHVubmV0dCBzcXVhcmUgYmVsb3csIEkgcHV0IHRoZSBtYWxlJiM4MjE3O3MgZ2FtZXRlcyBvbiB0aGUgbGVmdCwgYW5kIHRoZSBtb3RoZXImIzgyMTc7cyBnYW1ldGVzIG9uIHRvcCwgYW5kIEkgZXZlbiBmaWxsZWQgaW4gdGhlIGZpcnN0IHNxdWFyZSBmb3IgeW91LiBZb3UgZG8gdGhlIHJlc3QgYW5kIHRoZW4gY29udmVydCB0aGUgZ2Vub3R5cGVzIGJhY2sgaW50byBwaGVub3R5cGVzLiBUaGF0IHNob3VsZCBnaXZlIHlvdSB0aGUgYW5zd2VyIHdoZW4geW91IG5leHQgc2VlIHRoaXMgcXVlc3Rpb24u

[Qq]

	wG	wg
Wg	WgGg
wg

[c]IDkgbm9ybWFsIHdpbmdzLCBub3JtYWwgYm9keSA6IDMgbm9ybWFsIHdpbmdzLCBlYm9ueSBib2R5IDogMyB2ZXN0aWdpYWwgd2luZ3MsIG5vcm1hbCBib2R5OiAxIHZlc3RpZ2FsIHdpbmdzLCBlYm9ueSBib2R5[Qq]

Cg==Cg==Cg==Cg==

[Qq]	wG	wg
Wg	WgGg
wg

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e2f62f0a66e2c” question_number=”21″ topic=”5.3.Mendelian_Genetics”] The pedigree below shows inheritance of an X-linked recessive trait.

If the child indicated by ? is a girl, what is the probability she will be affected by the trait?

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

Cg==[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcXVlc3Rpb24uIEltYWdpbmUgdGhhdCB0aGUgdHJhaXQgdW5kZXIgZGlzY3Vzc2lvbiBpcyBoZW1vcGhpbGlhLiBGb3IgY2xhcml0eSwgbGV0JiM4MjE3O3MgaGF2ZSAmIzgyMjA7SCYjODIyMTsgc3RhbmQgZm9yIHRoZSBub3JtYWwgYWxsZWxlIGFuZCAmIzgyMjA7aCYjODIyMTsgZm9yIHRoZSBhbGxlbGUgdGhhdCBjYXVzZXMgdGhlIHJlY2Vzc2l2ZSBjb25kaXRpb24uIFRoZSBjaGlsZCBhdCA=Pw==IGhhcyBhIGZhdGhlciB3aG8gaGFzIHRoZSB0cmFpdDogdGhlcmVmb3JlIHRoZSBmYXRoZXImIzgyMTc7cyBnZW5vdHlwZSBpcyBY^aA==WS4gVGhlIGNoaWxkJiM4MjE3O3MgbW90aGVyIGhhZCBhIGJyb3RoZXIgd2hvIHdhcyBoYXMgdGhlIHJlY2Vzc2l2ZSB0cmFpdC4gVGhlcmVmb3JlID8mIzgyMTc7cyBncmFuZG1vdGhlciBoYWQgdG8gYmUgYSBjYXJyaWVyLCBhbmQgPyYjODIxNztzIG1vdGhlciBoYXMgYSA1MCUgY2hhbmNlIG9mIGJlaW5nIGEgY2Fycmllci4=

[Qq]

Now use the rule of multiplication. It’s already established in the question that the child is a girl. There’s a 100% chance that she’ll inherit the recessive allele from her father (that’s how she wound up being a girl in the first place, right?) There’s a 1/2 chance that the mother is a carrier. And, if mom’s a carrier, there’s a 1/2 chance that she’ll pass on her X^h allele, producing a daughter that’s X^hX^h. Multiply these probabilities out, and you’ll have your answer.

[f]IEV4Y2VsbGVudCEgSXQmIzgyMTc7cyBhbHJlYWR5IGVzdGFibGlzaGVkIGluIHRoZSBxdWVzdGlvbiB0aGF0IHRoZSBjaGlsZCBpcyBhIGdpcmwuIFRoZXJlJiM4MjE3O3MgYSAxMDAlIGNoYW5jZSB0aGF0IHNoZSYjODIxNztsbCBpbmhlcml0IHRoZSByZWNlc3NpdmUgYWxsZWxlIGZyb20gaGVyIGZhdGhlciAodGhhdCYjODIxNztzIGhvdyBzaGUgd291bmQgdXAgYmVpbmcgYSBnaXJsIGluIHRoZSBmaXJzdCBwbGFjZSwgcmlnaHQ/KSBUaGVyZSYjODIxNztzIGEgMS8yIGNoYW5jZSB0aGF0IHRoZSBtb3RoZXIgaXMgYSBjYXJyaWVyLiBBbmQsIGlmIG1vbSYjODIxNztzIGEgY2FycmllciwgdGhlcmUmIzgyMTc7cyBhIDEvMiBjaGFuY2UgdGhhdCBzaGUmIzgyMTc7bGwgcGFzcyBvbiBoZXIgWA==^aA==IGFsbGVsZSwgcHJvZHVjaW5nIGEgZGF1Z2h0ZXIgdGhhdCYjODIxNztzIFg=^aA==WA==^[Qq]h. 1/2 x 1/2 = 1/4.

[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcXVlc3Rpb24uIEltYWdpbmUgdGhhdCB0aGUgdHJhaXQgdW5kZXIgZGlzY3Vzc2lvbiBpcyBoZW1vcGhpbGlhLiBGb3IgY2xhcml0eSwgbGV0JiM4MjE3O3MgaGF2ZSAmIzgyMjA7SCYjODIyMTsgc3RhbmQgZm9yIHRoZSBub3JtYWwgYWxsZWxlIGFuZCAmIzgyMjA7aCYjODIyMTsgZm9yIHRoZSBhbGxlbGUgdGhhdCBjYXVzZXMgdGhlIHJlY2Vzc2l2ZSBjb25kaXRpb24uIFRoZSBjaGlsZCBhdCA=Pw==IGhhcyBhIGZhdGhlciB3aG8gaGFzIHRoZSB0cmFpdDogdGhlcmVmb3JlIHRoZSBmYXRoZXImIzgyMTc7cyBnZW5vdHlwZSBpcyBY^aA==WS4gVGhlIGNoaWxkJiM4MjE3O3MgbW90aGVyIGhhZCBhIGJyb3RoZXIgd2hvIGhhZCB0aGUgcmVjZXNzaXZlIHRyYWl0LiBUaGVyZWZvcmUgPyYjODIxNztzIGdyYW5kbW90aGVyIGhhZCB0byBiZSBhIGNhcnJpZXIsIGFuZCA/JiM4MjE3O3MgbW90aGVyIGhhcyBhIDUwJSBjaGFuY2Ugb2YgYmVpbmcgYSBjYXJyaWVyLg==

[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcXVlc3Rpb24uIEltYWdpbmUgdGhhdCB0aGUgdHJhaXQgdW5kZXIgZGlzY3Vzc2lvbiBpcyBoZW1vcGhpbGlhLiBGb3IgY2xhcml0eSwgbGV0JiM4MjE3O3MgaGF2ZSAmIzgyMjA7SCYjODIyMTsgc3RhbmQgZm9yIHRoZSBub3JtYWwgYWxsZWxlIGFuZCAmIzgyMjA7aCYjODIyMTsgZm9yIHRoZSBhbGxlbGUgdGhhdCBjYXVzZXMgdGhlIHJlY2Vzc2l2ZSBjb25kaXRpb24uIFRoZSBjaGlsZCBhdCA=Pw==IGhhcyBhIGZhdGhlciB3aG8gaGFzIHRoZSB0cmFpdDogdGhlcmVmb3JlIHRoZSBmYXRoZXImIzgyMTc7cyBnZW5vdHlwZSBpcyBY^aA==WS4gVGhlIGNoaWxkJiM4MjE3O3MgbW90aGVyIGhhZCBhIGJyb3RoZXIgd2hvIGhhcyB0aGUgcmVjZXNzaXZlIHRyYWl0LiBUaGVyZWZvcmUgPyYjODIxNztzIGdyYW5kbW90aGVyIGhhZCB0byBiZSBhIGNhcnJpZXIsIGFuZCA/JiM4MjE3O3MgbW90aGVyIGhhcyBhIDUwJSBjaGFuY2Ugb2YgYmVpbmcgYSBjYXJyaWVyLg==

[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e2ef58277922c” question_number=”22″ 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]

[c]IHNleC1saW5rZWQgcmVjZXNzaXZl[Qq]

[f]IE5vLCBidXQgeW91IGNob3NlIHRoZSBzZWNvbmQtYmVzdCBhbnN3ZXIuIEp1c3QgcmUtcmVhZCB0aGUgcXVlc3Rpb24gY2FyZWZ1bGx5LCBsb29raW5nIHNwZWNpZmljYWxseSBmb3IgYSBkZXNjcmlwdGlvbiBvZiB0aGUgYmVoYXZpb3Igb2YgdGhlIEJhcnJpbmcgYWxsZWxlLiBSZW1lbWJlciB3aGF0IHlvdSYjODIxNzt2ZSBsZWFybmVkLCBhbmQgdGhlIGFuc3dlciB3aWxsIGJlIG9idmlvdXMgdGhlIG5leHQgdGltZSB5b3Ugc2VlIHRoaXMgcXVlc3Rpb24u[Qq]

[c]IHNleC1saW5rZW QgZG9taW5hbnQ=[Qq]

[f]IFdheSB0byBnbyEgQmFycmluZyBpcyBzZXgtbGlua2VkIChpdCYjODIxNztzIGNhcnJpZWQgb24gdGhlIFogY2hyb21vc29tZSkgYW5kIGRvbWluYW50IChpdCYjODIxNztzIHJlcHJlc2VudGVkIGJ5IGEgY2FwaXRhbCBsZXR0ZXIpLg==[Qq]

[c]IG1pdG9jaG9uZHJpYWw=[Qq]

[f]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[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e2e8f106c622c” question_number=”23″ 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=[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	X^HX^H	X^HX^h
Y	X^HY	X^hY

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	X^HX^H	X^HX^h
Y	X^HY	X^hY

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	X^HX^H	X^HX^h
Y	X^HY	X^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	X^HX^H	X^HX^h
Y	X^HY	X^hY

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.

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	X^HX^H	X^HX^h
Y	X^HY	X^hY

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” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e2e3b3ec0522c” question_number=”24″ 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

c^R: brown coat

c: white coat

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

Which of the following predictions is correct?

[c]IENj^Ug==IHggQ2M=^Ug==OiBhbGwgYmxhY2sgbWljZS4=[Qq]

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

	C	cR
C	CC	CcR
c^R	Cc^R	cRcR

CC is black; Cc^Ris black. But c^Rc^Ris brown.

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

[f]IE5vLiBIZXJlJiM4MjE3O3MgdGhlIFB1bm5ldHQgc3F1YXJlIGZvciA=Yw==^Ug==Yw==^Ug==[Qq] x c^Rc.

	c^R	c^R
c^R	c^Rc^R	cRcR
c	c^Rc	cRc

c^Rc^R is brown; c^Rc is fawn.

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

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

	c^R	c^R
c	c^Rc	cRc
c	c^Rc	cRc

All the offspring have the c^Rc 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” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e2dd4ccb5222c” question_number=”25″ 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.

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 Z^BZ^b 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]

[c]IE5vbi1iYXJyZWQgbWFsZQ==[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e2d834f14f62c” question_number=”26″ 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).

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 Z^BW will, as an adult, have which phenotype?

[c]IEJhcnJlZCBtYWxl[Qq]

[f]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[Qq]

[c]IEJhcnJlZC BmZW1hbGU=[Qq]

[f]IFRlcnJpZmljLiBZb3UmIzgyMTc7cmUgY2xlYXJseSB1bmRlcnN0YW5kaW5nIGhvdyBzZXggZGV0ZXJtaW5hdGlvbiBhbmQgc2V4IGxpbmtlZCBhbGxlbGVzIHdvcmsgaW4gYmlyZHMh[Qq]

[c]IE5vbi1iYXJyZWQgZmVtYWxl[Qq]

[c]IE5vbi1iYXJyZWQgbWFsZQ==[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e2d1834f1fe2c” question_number=”27″ 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).

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

In a cross between a barred female and a non-barred male, what percentage of the offspring will be non-barred females?

[c]IDAl[Qq]

[f]IE5vLiBZb3UgaGF2ZSB0byB0cmVhdCB0aGlzIGxpa2UgYW55IGdlbmV0aWNzIHByb2JsZW0gdGhhdCBpbnZvbHZlcyBzZXggbGlua2FnZS4gSGVyZSBhcmUgdGhlIGdlbm90eXBlczogdGhlIGJhcnJlZCBmZW1hbGUgaXMgWg==^Qg==[Qq]W. The non-barred male is Z^bZ^b. Get a pencil, set up a Punnett square like the one below, and solve the problem (determining the proportion of each phenotype).

	Z^B	W
Z^b
Z^b

Hold on to the answer until the next time you see the question.

[c]IDI1JQ==[Qq]

	Z^B	W
Z^b
Z^b

Hold on to the answer until the next time you see the question.

[c]IDUw JQ==[Qq]

[f]IFlvdSYjODIxNztyZSBleGhpYml0aW5nIG1hcnZlbG91cyBtYXN0ZXJ5IG9mIChwb3N0KSBNZW5kZWxpYW4gZ2VuZXRpY3MhIDUwJSBvZiB0aGUgb2Zmc3ByaW5nIGFyZSBub24tYmFycmVkIGZlbWFsZXMuIElmIHlvdSYjODIxNztyZSBsaWtlIG1lLCB5b3Ugc2V0IHVwIHRoaXMgY3Jvc3MuIEFsbCB0aGUgZmVtYWxlcyBhcmUgbm9uLWJhcnJlZCAoWg==^Yg==[Qq]W), and that’s 50% of the offspring.

	Z^B	W
Z^b	Z^BZ^b	Z^bW
Z^b	Z^BZ^b	Z^bW

[c]IDc1JQ==[Qq]

	Z^B	W
Z^b
Z^b

Hold on to the answer until the next time you see the question.

[c]IDEwMCU=[Qq]

	Z^B	W
Z^b
Z^b

Hold on to the answer until the next time you see the question.

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e2cc20f3a0a2c” question_number=”28″ topic=”5.4.Non-Mendelian_Genetics”] Thryothorus ludovicianus, also known as the Carolina wren, has a white streak called a “supercilium” above its eye. The supercilium may extend well behind the eye, as shown below.

A family of wrens is found with a short supercilium that barely extends past the eye. To determine the type of inheritance that underlies supercilium length, a group of Carolina wrens was bred in captivity, with the following results.

Experiment 1: individuals with a long supercilium were bred together. All of the offspring had long supercilia.
Experiment 2: individuals with a short supercilium were bred together. All of the offspring had short supercilia.
Experiment 3: individuals with a short supercilium were bred with individuals with a long supercilium. All of the offspring had short supercilia.
Experiment 4: the short supercilia offspring from experiment three were test crossed with the true-breeding long supercilia offspring of experiment 1. Approximately 50% of the offspring had long supercilia, and 50% had short supercilia.

The data above are consistent with the theory that … If this theory is correct, then when the short supercilia offspring from experiment 3 are mated with each other the offspring of that cross will be ….

[c]IHN1cGVyY2lsaXVtIGxlbmd0aCBpcyBzZXgtbGlua2VkJiM4MjMwOy4gYWxsIGxvbmcu[Qq]

[f]IE5vLiBTZXggbGlua2FnZSB3b3VsZCBsZWFkIHRvIGNlcnRhaW4gcGhlbm90eXBlcyBiZWluZyBhc3NvY2lhdGVkIHdpdGggb25lIHNleCBvciB0aGUgb3RoZXIuIFRoZXJlJiM4MjE3O3Mgbm8gZXZpZGVuY2Ugb2Ygc2V4IGxpbmthZ2UgaGVyZS4gSWYgYSB0cmFpdCBpc24mIzgyMTc7dCBzZXgtbGlua2VkLCB0aGVuIGl0JiM4MjE3O3MgbW9zdCBsaWtlbHkgX19fX19fLiBGaWd1cmUgb3V0IHRoZSBibGFuaywgYW5kIHlvdSYjODIxNztsbCBiZSBvbiB5b3VyIHdheSB0byB0aGUgYW5zd2VyLg==[Qq]

[c]IHN1cGVyY2lsaXVtIGxlbmd0aCBpcyBhdXRvc29tYWwsIHdpdGggdGhlIGxvbmcgYWxsZWxlIGJlaW5nIHRoZSBkb21pbmFudCBvbmUmIzgyMzA7LjMvNCBsb25nIGFuZCAxLzQgc2hvcnQu[Qq]

[f]IE5vLiBZb3UmIzgyMTc7cmUgY29ycmVjdCBpbiB0aGF0IGl0JiM4MjE3O3MgYXV0b3NvbWFsLCBidXQgbG9vayBhdCB0aGUgb2Zmc3ByaW5nIG9mIHRoZSBmaXJzdCBjcm9zcywgYW5kIHJlYXNzZXNzIHdoaWNoIGFsbGVsZSBpcyBkb21pbmFudC4=[Qq]

[c]IHN1cGVyY2lsaXVtIGxlbmd0aCBpcyBhdXRvc29tYWwsIHdpdGggdGhlIHNob3J0IGFsbGVs ZSBiZWluZyB0aGUgZG9taW5hbnQgb25lJiM4MjMwOzMvNCBzaG9ydCBhbmQgMS80IGxvbmcu[Qq]

[f]IEZhYnVsb3VzLiBCYXNlZCBvbiB0aGUgZGF0YSBhYm92ZSwgdGhlIGFsbGVsZSBpcyBhdXRvc29tYWwsIHdpdGggc2hvcnQgYmVpbmcgZG9taW5hbnQuIFRoZSBG_Mg==IGdlbmVyYXRpb24gd2lsbCBoYXZlIGEgcmF0aW8gb2YgdGhyZWUgc2hvcnQgdG8gb25lIGxvbmcu[Qq]

[c]IHN1cGVyY2lsaXVtIGxlbmd0aCBpcyBhdXRvc29tYWwsIHdpdGggdGhlIHNob3J0IGFsbGVsZSBiZWluZyB0aGUgZG9taW5hbnQgb25lICYjODIzMDsgYWxsIHNob3J0Lg==[Qq]

[f]IE5vLCBidXQgeW91JiM4MjE3O3JlIHZlcnkgY2xvc2UuIElmIHNob3J0IGlzIGRvbWluYW50IChhbmQgcmVwcmVzZW50ZWQgYnkgUyksIHRoZW4gd2hhdCB3b3VsZCBiZSB0aGUgcmVzdWx0IG9mIGNyb3NzaW5nIFNzIHdpdGggU3M/[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e2c56f517122c” question_number=”29″ 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==IGhhcyBhIGZhdGhlciB3aG8gaGFzIHRoZSB0cmFpdDogdGhlcmVmb3JlIHRoZSBmYXRoZXImIzgyMTc7cyBnZW5vdHlwZSBpcyBY^aA==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 X^h allele. Multiply all of these probabilities together, and you’ll have your answer.

[f]IEV4Y2VsbGVudC4gVGhlcmUmIzgyMTc7cyBhIDEvMiBjaGFuY2UgdGhhdCB0aGUgY2hpbGQgd2lsbCBiZSBhIGJveS4gVGhlcmUmIzgyMTc7cyBhIDEvMiBjaGFuY2UgdGhhdCB0aGUgbW9tIChpbmRpdmlkdWFsIDEpIGlzIGEgY2Fycmllci4gQW5kIHRoZXJlJiM4MjE3O3MgYSAxLzIgY2hhbmNlIHRoYXQgc2hlIChpbmRpdmlkdWFsIDEpIHdpbGwgcGFzcyBvbiBoZXIgcmVjZXNzaXZlIGFsbGVsZS4gMS8yIHggMS8yIHggMS8yID0gMS84LCBvciAxMi41JQ==[Qq]

[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 X^h allele. Multiply all of these probabilities together, and you’ll have your answer.

[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 herX^h allele. Multiply all of these probabilities together, and you’ll have your answer.

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e2c057776e62c” question_number=”30″ 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]

Cg==

SWYgSmFzbWluZSYjODIxNztzIG1vdGhlciB3YXMgYSBjYXJyaWVyIGFuZCBoZXIgZmF0aGVyIHdhcyBub3QgYSBoZW1vcGhpbGlhYyAoYWxzbyBrbm93biBmcm9tIHRoZSBwcm9ibGVtKSwgdGhlbiBKYXNtaW5lIGhlcnNlbGYgaGFzIGEgMS8yIGNoYW5jZSBvZiBiZWluZyBhIGNhcnJpZXIuIEFuZCBpZiBzaGUgd2VyZSBhIGNhcnJpZXIsIHRoZW4gc2hlJiM4MjE3O2QgaGF2ZSBhIDEvMiBjaGFuY2Ugb2YgcGFzc2luZyBvbiB0aGUgYWxsZWxlIHRvIGhlciBtYWxlIGNoaWxkLiBXaGF0JiM4MjE3O3MgMS8yIHRpbWVzIDEvMj8=[Qq]

Cg==

SWYgSmFzbWluZSYjODIxNztzIG1vdGhlciB3YXMgYSBjYXJyaWVyIGFuZCBoZXIgZmF0aGVyIHdhcyBub3QgYSBoZW1vcGhpbGlhYyAoYWxzbyBrbm93biBmcm9tIHRoZSBwcm9ibGVtKSwgdGhlbiBKYXNtaW5lIGhlcnNlbGYgaGFzIGEgMS8yIGNoYW5jZSBvZiBiZWluZyBhIGNhcnJpZXIuIEFuZCBpZiBzaGUgd2VyZSBhIGNhcnJpZXIsIHRoZW4gc2hlJiM4MjE3O2QgaGF2ZSBhIDEvMiBjaGFuY2Ugb2YgcGFzc2luZyBvbiB0aGUgYWxsZWxlIHRvIGhlciBtYWxlIGNoaWxkLiBXaGF0JiM4MjE3O3MgMS8yIHRpbWVzIDEvMj8=[Qq]

Cg==

SWYgSmFzbWluZSYjODIxNztzIG1vdGhlciB3YXMgYSBjYXJyaWVyIGFuZCBoZXIgZmF0aGVyIHdhcyBub3QgYSBoZW1vcGhpbGlhYyAoYWxzbyBrbm93biBmcm9tIHRoZSBwcm9ibGVtKSwgdGhlbiBKYXNtaW5lIGhlcnNlbGYgaGFzIGEgMS8yIGNoYW5jZSBvZiBiZWluZyBhIGNhcnJpZXIuIEFuZCBpZiBzaGUgd2VyZSBhIGNhcnJpZXIsIHRoZW4gc2hlJiM4MjE3O2QgaGF2ZSBhIDEvMiBjaGFuY2Ugb2YgcGFzc2luZyBvbiB0aGUgYWxsZWxlIHRvIGhlciBtYWxlIGNoaWxkLiBXaGF0JiM4MjE3O3MgMS8yIHRpbWVzIDEvMj8=[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e2ba8559b462c” question_number=”31″ 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]

Cg==

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[Qq]

[f]IEV4Y2VsbGVudCEgVGhlcmUmIzgyMTc7cyBhIDUwJSBwcm9iYWJpbGl0eSB0aGF0IEphc21pbmUgaXMgYSBjYXJyaWVyLCBhbmQgdGhlcmUmIzgyMTc7cyBhIDUwJSBwcm9iYWJpbGl0eSB0aGF0IGEgY2FycmllciB3aWxsIHBhc3Mgb24gYW4gWCBjaHJvbW9zb21lIHdpdGggdGhlIGhlbW9waGlsaWEgYWxsZWxlIG9uIHRvIGhlciBzb24uIEFzIHlvdSBtdXN0IGhhdmUgdW5kZXJzdG9vZCwgdGhlIGZhdGhlciYjODIxNztzIHN0YXR1cyBhcyBhIGhlbW9waGlsaWFjIGlzIGlycmVsZXZhbnQgc2luY2UgaGUmIzgyMTc7bGwgb25seSBwYXNzIG9uIGhpcyBYIGNocm9tb3NvbWUgdG8gaGlzIGRhdWdodGVycy4gNTAlIG9mIDUwJSBpcyAyNSUu[Qq]

Cg==

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e2b1352a2462c” question_number=”32″ topic=”5.4.Non-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-pigmented, or red-pigmented. If there’s no fruit pigment, then the fruits are 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

The fruit color of plants with the genotype IiYy would be

[c]IHllbGxvdy4=[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcXVlc3Rpb24uIExvb2sgY2FyZWZ1bGx5IGF0IHRoZSB0YWJsZSBzaG93aW5nIHRoZSBlZmZlY3RzIG9mIHRoZSB0d28gYWxsZWxlcyBmb3IgZ2VuZSA=MQ==LiBJZiBhIHBsYW50IGhhcyB0aGUgZG9taW5hbnQgYWxsZWxlIEk=LCB3aGF0IGhhcHBlbnMgdG8gZnJ1aXQgY29sb3I/IE9uY2UgeW91IGhhdmUgdGhhdCBmaWd1cmVkIG91dCwgeW91IHNob3VsZCBoYXZlIG5vIHRyb3VibGUgd2l0aCB0aGlzIHF1ZXN0aW9uIHRoZSBuZXh0IHRpbWUgeW91IHNlZSBpdC4=[Qq]

[c]IHJlZC4=[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcXVlc3Rpb24uIExvb2sgY2FyZWZ1bGx5IGF0IHRoZSB0YWJsZSBzaG93aW5nIHRoZSBlZmZlY3RzIG9mIHRoZSB0d28gYWxsZWxlcyBmb3IgZ2VuZQ==IDE=LiBJZiBhIHBsYW50IGhhcyB0aGUgZG9taW5hbnQgYWxsZWxlIEk=LCB3aGF0IGhhcHBlbnMgdG8gZnJ1aXQgY29sb3I/IE9uY2UgeW91IGhhdmUgdGhhdCBmaWd1cmVkIG91dCwgeW91IHNob3VsZCBoYXZlIG5vIHRyb3VibGUgd2l0aCB0aGlzIHF1ZXN0aW9uIHRoZSBuZXh0IHRpbWUgeW91IHNlZSBpdC4=[Qq]

[c]IGdy ZWVu[Qq]

[f]IEV4Y2VsbGVudCEgaWYgdGhlIHBsYW50IGhhcyB0aGUgYWxsZWxlIEk=LCB0aGVuIGl0IHdvbiYjODIxNzt0IGV4cHJlc3MgYW55IGZydWl0IHBpZ21lbnQsIGFuZCB0aGUgZnJ1aXQgd2lsbCBiZSBncmVlbi4=[Qq]

[c]IGdyZWVuIGFuZCB5ZWxsb3cu[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e2a6707325e2c” question_number=”33″ 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==[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e29f29cdfd62c” question_number=”34″ topic=”5.4.Non-Mendelian_Genetics”] In shorthorn cattle, coat color can be solid red, solid white, or red-roan (a mixture of white and pigmented hairs). When a true breeding solid red male is crossed with a white female, the offspring are roan. This demonstrates that

[c]IGNvYXQgY29sb3IgaXMgY29udHJvbGxlZCBieSBhdCBsZWFzdCB0aHJlZSBhbGxlbGVzLg==[Qq]

[f]IE5vLiBZb3UgbWlnaHQgYmUgdGhpbmtpbmcgb2Ygc3lzdGVtcyBsaWtlIGluaGVyaXRhbmNlIG9mIGJsb29kIHR5cGUsIHdoZXJlIHRoZSBBIGFsbGVsZSwgdGhlIEIgYWxsZWxlLCBhbmQgdGhlIG8gYWxsZWxlIHByb2R1Y2UgYmxvb2QgdHlwZXMgdHlwZSBBLCBCLCBBQiwgYW5kIE8gKHRoZXNlIHRocmVlIGFsbGVsZXMgYXJlIGFsc28gcmVwcmVzZW50ZWQgYXMgSQ==^QQ==LCA=[Qq]I^B, and i). But there are other systems of inheritance that allow you to produce an intermediate phenotype with just two alleles (as opposed to three). Keep that in mind when you see this question again.

[c]IHJlZCBhbmQgd2hpdGUgc2hvdyBp bmNvbXBsZXRlIGRvbWluYW5jZS4=[Qq]

[f]IFllcy4gSWYgcg==IGlzIHRoZSBhbGxlbGUgZm9yIHJlZCBhbmQgdw==IGlzIHRoZSBhbGxlbGUgZm9yIHdoaXRlLCB0aGVuIHRoZSBjcm9zcyBhYm92ZSBjb3VsZCBiZSByZXByZXNlbnRlZCBhcyA=[Qq]rr x ww, with all the offspring having the genotype rw, and a roan phenotype.

[c]IHRoZSByZWQgYWxsZWxlIGlzIGRvbWluYW50IG92ZXIgdGhlIHdoaXRlIGFsbGVsZS4=[Qq]

[f]IE5vLiBJZiA=Ug==IHJlcHJlc2VudGVkIHRoZSBkb21pbmFudCBhbGxlbGUgZm9yIHJlZCwgYW5kIA==cg==IHJlcHJlc2VudGVkIHRoZSByZWNlc3NpdmUgYWxsZWxlIGZvciB3aGl0ZSwgdGhlbiBhIGNyb3NzIGJldHdlZW4gdHJ1ZSBicmVlZGluZyByZWQgYW5kIHdoaXRlIHdvdWxkIGJlIFJSIHggcnIuIEFsbCB0aGUgb2Zmc3ByaW5nIHdvdWxkIGJlIA==[Qq]Rr, and we’d expect them to be red, not roan. What kind of inheritance system would result in pure bred red crossed with purebred white resulting in an intermediate phenotype?

[c]IENyb3NzaW5nIG92ZXIgYW5kIHJlY29tYmluYXRpb24gb2NjdXJyaW5nIGR1cmluZyBtZWlvc2lzLg==[Qq]

[f]IE5vLiBDcm9zc2luZyBvdmVyIGFuZCByZWNvbWJpbmF0aW9uIGFsbW9zdCBjZXJ0YWlubHkgZGlkIG9jY3VyIGR1cmluZyBtZWlvc2lzLCBidXQgaXQgd291bGRuJiM4MjE3O3QgYWNoaWV2ZSB0aGUgcmVzdWx0IGRlc2NyaWJlZCBhYm92ZS4gV2hhdCBraW5kIG9mIGluaGVyaXRhbmNlIHN5c3RlbSB3b3VsZCByZXN1bHQgaW4gcHVyZSBicmVkIHJlZCBjcm9zc2VkIHdpdGggcHVyZWJyZWQgd2hpdGUgcmVzdWx0aW5nIGluIGFuIGludGVybWVkaWF0ZSBwaGVub3R5cGU/[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e2982daa5162c” question_number=”35″ 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 I^A and I^B 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 I^Bi). But what genotype is impossible for either of the parents to have?

[c]IFR5cG UgQUI=[Qq]

[f]IFdheSB0byBnby7CoEEgY2hpbGQgd2l0aCB0eXBlIE8gYmxvb2QgbmVlZHMgZWFjaCBvZiB0aGVpciBwYXJlbnRzIHRvIGdpdmUgdGhlbSBhbg==IGk=IGFsbGVsZS4gR2l2ZW4gdGhhdCwgdHlwZSBBQiBpcyBpbXBvc3NpYmxlIChiZWNhdXNlIGEgdHlwZSBBQiBwYXJlbnQgY291bGQgZ2l2ZSB0aGVpciBjaGlsZCBlaXRoZXIgSQ==^QSA=[Qq]or I^B, but not ai.

[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 I^Ai). But what genotype is impossible for either of the parents to have?

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e28e92f944e2c” question_number=”36″ 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 I^A and I^B as codominant to one another, with each of these dominant over i. If a man with type B blood has children with a woman with type O blood, which of the following blood types would be impossible in their children.

[c]IFR5cGUgTw==[Qq]

[f]IE5vLiBUaGUgbWFuIHdpdGggdHlwZSBCIGJsb29kIGNvdWxkIGhhdmUgb25lIG9mIHR3byBnZW5vdHlwZXM6IA==SQ==^Qg==aSA=KGlmIGhlIHdlcmUgaGV0ZXJvenlnb3VzKSw=[Qq] or I^BI^B, (if he were homozygous type B). The woman is type O, so her genotype has to be ii. If you can set up two Punnett squares on your own, then do so, and you’ll see the two genotypes that are impossible. If not, then continue reading below.

Here’s the Punnett square that shows the father as a heterozygote:

	I^B	i
i
i		ii

Here’s the Punnett square that shows the father as a homozygote:

	I^B	I^B
i
i		ii

Complete both Punnett squares by bringing the alleles down and over. As you can see by the first Punnett square, a type O child is possible. But which blood type is impossible?

[c]IFR5cGUgQg==[Qq]

Here’s the Punnett square that shows the father as a heterozygote:

	I^B	i
i
i

Here’s the Punnett square that shows the father as a homozygote:

	I^B	I^B
i
i

Complete both Punnett squares by bringing the alleles down and over. As you can see, in either Punnett square, a type B child is possible (any child with genotype I^Bi), But which blood type is impossible?

[c]IFR5cG UgQUI=[Qq]

[f]IENvcnJlY3QhIFRoZSBtYW4gd2l0aCB0eXBlIEIgYmxvb2QgY291bGQgaGF2ZSBvbmUgb2YgdHdvIGdlbm90eXBlczogSQ==^Qg==aSA=KGlmIGhlIHdlcmUgaGV0ZXJvenlnb3VzKSw=[Qq] or I^BI^B, (if he were homozygous type B). The woman is type O, so her genotype has to be ii. The offspring will all be type B or type O. Type AB is impossible (as is type A).

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e288e61c4922c” question_number=”37″ topic=”5.4.Non-Mendelian_Genetics”] Conditions like hemophilia, Red-green colorblindness, and Duchenne’s muscular dystrophy are inherited as sex-linked recessive disorders. The term “sex-linked” means

[c]IHRoZSBhbGxlbGUgaXMgY2FycmllZCBvbiBhbiBhdXRvc29tZS4=[Qq]

[f]IE5vLiBUaGUgcmVhc29uIHRoZXNlIGNvbmRpdGlvbnMgYXJlICYjODIyMDtzZXggbGlua2VkJiM4MjIxOyBpcyBiZWNhdXNlIHRoZXkmIzgyMTc7cmUgbm90IGNhcnJpZWQgb24gYXV0b3NvbWVzLiBJZiB0aGV5IHdlcmUgY2FycmllZCBvbiBhdXRvc29tZXMsIHRoZWlyIGluaGVyaXRhbmNlIHdvdWxkIG5vdCBiZSBsaW5rZWQgdG8gc2V4LiBXaGF0IGNocm9tb3NvbWVzIHdvdWxkIGEgY29uZGl0aW9uIGhhdmUgdG8gYmUgbGlua2VkIHRvIGluIG9yZGVyIHRvIGJlIGxpbmtlZCB0byBzZXg/[Qq]

[c]IHRoZSBhbGxlbGUgaXMgY2FycmllZCBvbiB0aGUgWSBjaHJvbW9zb21lLg==[Qq]

[f]IE5vLCBidXQgeW91JiM4MjE3O3JlIHRoaW5raW5nIGFib3V0IHRoaXMgdGhlIHJpZ2h0IHdheS4gSGVtb3BoaWxpYSwgcmVkLWdyZWVuIGNvbG9yYmxpbmRuZXNzIGFuZCBEdWNoZW5uZSYjODIxNztzIG11c2N1bGFyIGR5c3Ryb3BoeSBzaG93IHVwIG1vcmUgb2Z0ZW4gaW4gbWFsZXMgdGhhbiBmZW1hbGVzLCBidXQgbm90IGJlY2F1c2UgdGhleSBhcmUgb24gdGhlIFkgY2hyb21vc29tZS4gV2hhdCBjaHJvbW9zb21lIHdvdWxkIGEgY29uZGl0aW9uIGhhdmUgdG8gYmUgbGlua2VkIHRvIGluIG9yZGVyIHRvIGJlIGxpbmtlZCB0byBzZXg/[Qq]

[c]IHRoZSBhbGxlbGUgaXMgY2FycmllZC BvbiB0aGUgWCBjaHJvbW9zb21lLg==[Qq]

[f]IEZhYnVsb3VzISBUaGUgYWxsZWxlcyBmb3IgaGVtb3BoaWxpYSwgcmVkLWdyZWVuIGNvbG9yYmxpbmRuZXNzLCBhbmQgRHVjaGVubmUmIzgyMTc7cyBtdXNjdWxhciBkeXN0cm9waHkgYXJlIGFsbCBmb3VuZCBvbiB0aGUgWCBjaHJvbW9zb21lLCBtYWtpbmcgdGhlbSBzZXgtbGlua2VkLg==[Qq]

[c]IHRoZSBhbGxlbGUgaXMgY2FycmllZCBpbiB0aGUgbWl0b2Nob25kcmlhbCBETkEu[Qq]

[f]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[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e281e9f89d22c” question_number=”38″ topic=”5.4.Non-Mendelian_Genetics”] A man and a woman, both with normal vision, have a son who is red-green colorblind. Which of the following is the most likely explanation?

[c]IEJvdGggbWF0ZXJuYWwgZ3JhbmRwYXJlbnRzIGFyZSBjb2xvcmJsaW5k[Qq]

[f]IE5vLiBUaGUga2V5IHRoaW5nIHRvIHJlbWVtYmVyIGhlcmUgaXMgdGhhdCByZWQtZ3JlZW4gY29sb3JibGluZG5lc3MgaXMgYSByZWNlc3NpdmUsIHNleC1saW5rZWQgY29uZGl0aW9uLCB3aXRoIHRoZSBhbGxlbGUgZm9yIHRoZSBjb25kaXRpb24gZm91bmQgb24gdGhlIFggY2hyb21vc29tZS4gVGhlIGdlbm90eXBlIG9mIGEgY29sb3JibGluZCBtYWxlIGlzIHJlcHJlc2VudGVkIGFzIFg=^Yw==WSwgd2hlcmVhcyBhIG1hbGUgd2l0aCBub3JtYWwgdmlzaW9uIHdvdWxkIGhhdmUgdGhlIGdlbm90eXBlIFg=^Qw==WS4gTm93LCB0aGluayBhYm91dCB3aGF0IHlvdSBrbm93IGFib3V0IGluaGVyaXRhbmNlLCBhbmQgYXNrIHlvdXJzZWxmIHdobyB0aGF0IGNvbG9yYmxpbmQgbWFsZSB3b3VsZCBoYXZlIGluaGVyaXRlZCBoaXMgWCBjaHJvbW9zb21lIGZyb20uIFJlbWVtYmVyIHRoYXQgdGhlIG5leHQgdGltZSB5b3Ugc2VlIHRoaXMgcXVlc3Rpb24u[Qq]

[c]IFRoZSBtb3RoZXIgY2Fycmllcy BhIHJlY2Vzc2l2ZSBhbGxlbGU=[Qq]

[f]IEV4Y2VsbGVudCEgQmVjYXVzZSByZWQtZ3JlZW4gY29sb3JibGluZG5lc3MgaXMgYSByZWNlc3NpdmUgc2V4LWxpbmtlZCBjb25kaXRpb24sIGl0JiM4MjE3O3MgYWx3YXlzIHRyYW5zbWl0dGVkIHRvIHRoZSBzb25zIGJ5IHRoZSBtb3RoZXIsIHdobyBwYXNzZXMgYSBjb3B5IG9mIHRoZSByZWNlc3NpdmUgYWxsZWxlIHRvIGhlciBzb25zIHRocm91Z2ggYW4gWCBjaHJvbW9zb21lIGluIGhlciBlZ2cu[Qq]

[c]IFRoZSBmYXRoZXIgcGFzc2VkIGhpcyBhbGxlbGUgZm9yIGNvbG9yYmxpbmRuZXNzIHRvIGhpcyBzb24u[Qq]

[f]IE5vLiBUaGVyZSBhcmUgdHdvIGtleSB0aGluZ3MgdG8gcmVtZW1iZXIgaGVyZS4gVGhlIGZpcnN0IGlzIHRoYXQgcmVkLWdyZWVuIGNvbG9yYmxpbmRuZXNzIGlzIGEgcmVjZXNzaXZlLCBzZXgtbGlua2VkIGNvbmRpdGlvbiwgd2l0aCB0aGUgYWxsZWxlIGZvciB0aGUgY29uZGl0aW9uIGZvdW5kIG9uIHRoZSBYIGNocm9tb3NvbWUuIFRoZSBzZWNvbmQgaXMgdGhhdCB0aGUgd2F5IHRoYXQgYSBtYWxlIGh1bWFuIHByb2R1Y2VzIG1hbGUgb2Zmc3ByaW5nIGlzIGJ5IHBhc3Npbmcgb24gaGlzIFkgY2hyb21vc29tZSAoYW5kIG5vdCBoaXMgWCBjaHJvbW9zb21lKS4=

Cg==

QmVjYXVzZSBpdCYjODIxNztzIHNleC1saW5rZWQsIHRoZSBnZW5vdHlwZSBvZiBhIGNvbG9yYmxpbmQgbWFsZSBpcyByZXByZXNlbnRlZCBhcyBY^Yw==WSwgd2hlcmVhcyBhIG1hbGUgd2l0aCBub3JtYWwgdmlzaW9uIHdvdWxkIGhhdmUgdGhlIGdlbm90eXBlIFg=^[Qq]CY. Now, think about what you know about inheritance, and ask yourself who that colorblind male would have inherited his X chromosome from. Remember that the next time you see this question.

[c]IEJvdGggdGhlIG1vdGhlciBhbmQgdGhlIGZhdGhlciB3ZXJlIGhldGVyb3p5Z291cyBmb3IgdGhlIGNvbG9yLWJsaW5kIGdlbmUu[Qq]

[f]IE5vLiBUaGUga2V5IHRoaW5nIHRvIHJlbWVtYmVyIGhlcmUgaXMgdGhhdCByZWQtZ3JlZW4gY29sb3JibGluZG5lc3MgaXMgYSByZWNlc3NpdmUsIHNleC1saW5rZWQgY29uZGl0aW9uLCB3aXRoIHRoZSBhbGxlbGUgZm9yIHRoZSBjb25kaXRpb24gZm91bmQgb24gdGhlIFggY2hyb21vc29tZS4gVGhlIGdlbm90eXBlIG9mIGEgY29sb3JibGluZCBtYWxlIGlzIHJlcHJlc2VudGVkIGFzIFg=^Yw==WSwgd2hlcmVhcyBhIG1hbGUgd2l0aCBub3JtYWwgdmlzaW9uIHdvdWxkIGhhdmUgdGhlIGdlbm90eXBlIFg=^Qw==WS4gV2hpbGUgYSBmZW1hbGUgY2FuIGJlIGhldGVyb3p5Z291cyBub3JtYWwsIHdpdGggdGhlIGdlbm90eXBlIFg=^[Qq]CX^c, a male can’t be heterozygous. He either inherits the normal allele, or the recessive one (with no countervailing allele on his Y chromosome). Remember that and make a different choice the next time you see this question.

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e27cacdddc22c” question_number=”39″ topic=”5.5.Environmental_Effects_on_Phenotype”] In a group of organisms, individuals are genetically identical at a particular single gene locus. However, these individuals show a variety of phenotypes for the trait. The most likely explanation for the variation in phenotypes for this trait is

[c]IGNvZG9taW5hbnQgYWxsZWxlcyB0aGF0IGFyZSBkb21pbmFudCB0byBhIHRoaXJkIGFsbGVsZSBhdCB0aGF0IGxvY3VzLg==[Qq]

[f]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[Qq]

[c]IHBvbHlnZW5pYyBpbmhlcml0YW5jZSBmb3IgdGhlIHRyYWl0IHVuZGVyIGludmVzdGlnYXRpb24u[Qq]

[f]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[Qq]

[c]IGVudmlyb25tZW50YWwgaW5mbHVlbmNlcyBjcmVh dGluZyBwaGVub3R5cGljIGRpZmZlcmVuY2VzLg==[Qq]

[f]IE5pY2UhIElmIGluZGl2aWR1YWxzIGFyZSBnZW5ldGljYWxseSBpZGVudGljYWwsIHRoZW4gdGhlIG1vc3Qgc3RyYWlnaHRmb3J3YXJkIGV4cGxhbmF0aW9uIGZvciB0aGUgZGlmZmVyZW5jZSB3b3VsZCBiZSBlbnZpcm9ubWVudGFsIGluZmx1ZW5jZXMuIFBsYW50cyB3aXRoIGlkZW50aWNhbCBzZWVkcyBwbGFudGVkIGluIGRpZmZlcmVudCB0eXBlcyBvZiBzb2lsIHdpbGwgaGF2ZSBkaWZmZXJlbnQgcGhlbm90eXBlcywgYW5kIHRoYXQmIzgyMTc7cyB3aGF0IHlvdSBzZWUgaGVyZS4=[Qq]

[c]IG11bHRpcGxlIGFsbGVsZXMgYXQgdGhlIGdlbmUgbG9jdXMgd2l0aGluIHRoZSBnZW5lIHBvb2wu[Qq]

[f]IE5vLiBJbiB0aGlzIHF1ZXN0aW9uLCBpbmRpdmlkdWFscyBhcmUgZGVzY3JpYmVkIGFzIGJlaW5nIGlkZW50aWNhbCBhIHBhcnRpY3VsYXIgZ2VuZSBsb2N1cy4gRXZlbiBpZiB0aGVyZSB3ZXJlIG11bHRpcGxlIGFsbGVsZXMgZm9yIHRoaXMgZ2VuZSwgaXQgd291bGRuJiM4MjE3O3QgZXhwbGFpbiB0aGVpciBwaGVub3R5cGljIHZhcmlhdGlvbiAoYmVjYXVzZSB0aGV5IGFsbCBoYXZlIHRoZSBzYW1lIGFsbGVsZSkuIE5leHQgdGltZSB5b3Ugc2VlIHRoaXMsIGFzayB5b3Vyc2VsZiB0aGlzIHF1ZXN0aW9uOiBXaGF0IGNvdWxkIHByb2R1Y2UgcGhlbm90eXBpYyB2YXJpYXRpb24gaW4gZ2VuZXRpY2FsbHkgaWRlbnRpY2FsIG9yZ2FuaXNtcz8=[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e274f67678e2c” question_number=”40″ topic=”5.5.Environmental_Effects_on_Phenotype”] The diagram below shows the results of an experiment in which swallowtail butterflies were collected from Alaska and Michigan. Each population was then bred in the laboratory at different temperatures (with all other environmental factors kept the same). At each temperature, the orange bar on the left represents the butterflies originating in Alaska, and the blue bar on the right indicates the butterflies originating in Michigan.

Which of the following is the best conclusion to draw from this experiment?

[c]IFRoZSBncm93dGggcmF0ZXMgYXJlIGRldGVybWluZWQgYnkgdGhlIGVudmlyb25tZW50Lg==[Qq]

[f]IE5vLiBOb3RlIHRoYXQgZm9yIGJvdGggdGhlIEFsYXNrYSBhbmQgTWljaGlnYW4gcG9wdWxhdGlvbnMsIHRoZSByZWxhdGl2ZSBncm93dGggcmF0ZSBpcyBzaWduaWZpY2FudGx5IGhpZ2hlciBhdCAyNMKwQy4gV2hpbGUgdGhhdCBpbmRpY2F0ZXMgdGhhdCwgYXQgbGVhc3QgdG8gc29tZSBleHRlbnQsIHRoZSBncm93dGggcmF0ZXMgYXJlIGluZmx1ZW5jZWQgYnkgdGhlIGVudmlyb25tZW50LCBpdCBkb2VzbiYjODIxNzt0IGV4cGxhaW4gdGhlIA==ZGlmZmVyZW5jZXM=IGJldHdlZW4gdGhlIEFsYXNrYSBhbmQgTWljaGlnYW4gcG9wdWxhdGlvbnMu[Qq]

[c]IFRoZSBncm93dGggcmF0ZXMgYXJlIGdlbmV0aWNhbGx5IGRldGVybWluZWQu[Qq]

[f]IE5vLiBUaGVyZSYjODIxNztzIGNsZWFybHkgYSBnZW5ldGljIGluZmx1ZW5jZS4gT3RoZXJ3aXNlLCBob3cgY291bGQgeW91IGV4cGxhaW4gdGhlIGRpZmZlcmVuY2UgYmV0d2VlbiB0aGUgQWxhc2thIGFuZCBNaWNoaWdhbiBwb3B1bGF0aW9ucyB3aGVuIGdyb3duIGF0IHRoZSBzYW1lIHRlbXBlcmF0dXJlPyBCdXQgZ2VuZXRpY3MgY2FuJiM4MjE3O3QgYmUgdGhlIGVudGlyZSBzdG9yeSwgYmVjYXVzZSB5b3UgYWxzbyBoYXZlIHRvIGV4cGxhaW4gd2h5IGJvdGggcG9wdWxhdGlvbnMgZ3JvdyBmYXN0ZXIgYXQgMjTCsEMgdGhhbiB0aGV5IGRvIGF0IDEywrBDLiBMb29rIG92ZXIgdGhlIGNob2ljZXMgYWdhaW4gYW5kIGZpbmQgYSBiZXR0ZXIgYW5zd2VyLg==[Qq]

[c]IFRoZSBncm93dGggcmF0ZXMgZXhoaWJpdCBhIGdlbm 90eXBlL2Vudmlyb25tZW50IGludGVyYWN0aW9uLg==[Qq]

[f]IEV4Y2VsbGVudC4gQXMgd2l0aCBtb3N0IGJpb2xvZ2ljYWwgcGhlbm9tZW5hLCBwaGVub3R5cGUgZW1lcmdlcyBhcyBhIHJlc3VsdCBvZiBpbnRlcmFjdGlvbnMgYmV0d2VlbiB0aGUgZ2Vub3R5cGUgYW5kIHRoZSBlbnZpcm9ubWVudC4=[Qq]

[c]IFRoZSBncm93dGggcmF0ZXMgYXJlIGNvbnRyb2xsZWQgYnkgdGVtcGVyYXR1cmUu[Qq]

[f]IE5vLiBXaGlsZSB0ZW1wZXJhdHVyZSBoYXMgYW4gZWZmZWN0IG9uIHRoZSBncm93dGggcmF0ZSwgaXQgY2FuJiM4MjE3O3QgYmUgdGhlIGVudGlyZSBzdG9yeS4gVGVtcGVyYXR1cmUgYWxvbmUgZG9lc24mIzgyMTc7dCBleHBsYWluIHdoeSBhdCB0aGUgc2FtZSB0ZW1wZXJhdHVyZSwgdGhlIEFsYXNrYSBzd2FsbG93dGFpbHMgZ3JvdyBhdCBhIGRpZmZlcmVudCByYXRlIGZyb20gdGhlIE1pY2hpZ2FuIHN3YWxsb3d0YWlscy4gTG9vayBvdmVyIHRoZSBjaG9pY2VzIGFnYWluIGFuZCBmaW5kIGEgYmV0dGVyIGFuc3dlci4=

Cg==

Cg==[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e26dd5120ea2c” question_number=”41″ 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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[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e253590acd22c” question_number=”42″ 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]IE5vLiBBIGNocm9tb3NvbWFsIGludmVyc2lvbiB3b3VsZG4mIzgyMTc7dCBiZSBwb3NzaWJsZSB0byBkaXNjZXJuIGluIHRoaXMgdHlwZSBvZiBibGFjayBhbmQgd2hpdGUga2FyeW90eXBlLiBBbiBpbnZlcnNpb24gcmVzdWx0cyB3aGVuIGNocm9tb3NvbWFsIHNlZ21lbnRzIGFyZSBmbGlwcGVkLCBhcyBpcyBzaG93biBpbiB0aGUgZGlhZ3JhbSBiZWxvdyAoZnJvbSBub2JlbHByaXplLm9yZyk=

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” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e24ac31ef462c” question_number=”43″ 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 that this is a very difficult question.

[c]IEE=[Qq]

[f]IE5vLiBIZXJlJiM4MjE3O3MgaG93IHRvIHRoaW5rIGFib3V0IHRoaXMgcXVlc3Rpb24uIEEgdGVzdCBjcm9zcw==IGlzIGFsd2F5cyB3aXRoIGFuIG9yZ2FuaXNtIHRoYXQgaXMgcmVjZXNzaXZlIGZvciBib3RoIGFsbGVsZXMuIFRoZXJlZm9yZSBvbmUgcGFyZW50IGlzIGdvaW5nIHRvIGJlIHRiLy90YiAod2hpY2ggc2hvdWxkIGxldCB5b3UgZWxpbWluYXRlIHR3byBvZiB0aGUgY2hvaWNlcyBhYm92ZSkuIEZvciB0aGUgb3RoZXIgcGFyZW50LCBjaG9vc2Ugb25lIHdob3NlIGFsbGVsZXMgYXJlIG9yZ2FuaXplZCBpbiBzdWNoIGEgd2F5IHRoYXQg[Qq]crossing over will produce recombinant gametes with alleles tB/ and /Tb. Then, after the test cross, the recombinant offspring will be tB//tb and Tb//tb.

[c]IEI=[Qq]

[c]IE M=[Qq]

[f]IENvcnJlY3QhIEEgY3Jvc3MgYmV0d2VlbiBUQi8vdGIgYW5kIHRiLy90YiB3aWxsIHJlc3VsdCBpbiB0aGUgdHdvIHJlY29tYmluYW50IG9mZnNwcmluZyBzaG93biBhYm92ZSwgYXMgd2VsbCBhcyB0d28gcGFyZW50YWwgdHlwZSBvZmZzcHJpbmcgKFRCLy90YiBhbmQgdGIvL3RiKS4=[Qq]

[c]IEQ=[Qq]

[f]IE5vLCBidXQgeW91JiM4MjE3O3JlIG9uIHRoZSByaWdodCB0cmFjay4gWW91IGNvcnJlY3RseSBpZGVudGlmaWVkIGEgY2hvaWNlIHdoZXJlIG9uZSBwYXJlbnQgaXMgc3VpdGFibGUgdG8gYmUgYSB0ZXN0IGNyb3NzIHBhcmVudCAocmVjZXNzaXZlIGZvciBib3RoIGFsbGVsZXM6IHRiLy90YikuIEZvciB0aGUgb3RoZXIgcGFyZW50LCBjaG9vc2UgdGhlIG9uZSB3aG9zZSBhbGxlbGVzIGFyZSBvcmdhbml6ZWQgaW4gc3VjaCBhIHdheSB0aGF0IA==Y3Jvc3Npbmcgb3Zlcg==IHdpbGwgcHJvZHVjZSByZWNvbWJpbmFudCBnYW1ldGVzIHdpdGggYWxsZWxlcyB0Qi8gYW5kIC9UYi4gVGhlbiwgYWZ0ZXIgdGhlIHRlc3QgY3Jvc3MsIHRoZSByZWNvbWJpbmFudCBvZmZzcHJpbmcgd2lsbCBiZSB0Qi8vdGIgYW5kIFRiLy90Yi4=[Qq]

[q json=”true” xx=”1″ multiple_choice=”true” unit=”5.Heredity” dataset_id=”Unit 5 Cumulative Multiple Choice Questions (v2.0)|e226d1a764a2c” question_number=”44″ 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]

[f]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[Qq]

[f]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[Qq]

[x][restart]

[/qwiz]

4. 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]