1.  Introducing Genetics

[qwiz style=”width: 550px !important; min-height: 400px !important;” qrecord_id=”sciencemusicvideosmeister1961-Introducing Genetics (v2.0)”]

[h] Introducing Genetics: The story of Hunter Haymore

[i]

Hunter Haymore (center) with her parents. Hunter was born with sickle cell anemia.

[q]Genetics is the study of how genetic information gets passed from parents to offspring. The basic principles of genetics were figured out by an Austrian monk named Gregor Mendel, who carried out a series of genetic experiments in the 1860s. In his honor, these principles are called Mendelian Genetics.

Gregor Mendel

 

[q]We’ll look at Mendel’s experiments later in this unit. In this tutorial, as a way of introducing genetics concepts, we’re going to look at the inheritance of a disease called sickle cell anemia, which we discussed in the context of protein structure in Unit 1.

As we do, we’re going to focus on Hunter Haymore, a person who was born with sickle cell anemia.

 

[q]In sickle cell anemia, an inherited mutation changes a single amino acid in hemoglobin, the protein that transports oxygen in red blood cells. As a result, hemoglobin molecules cluster together (see the lower part of the diagram), causing red blood cells to deform into a sickled shape. These sickled blood cells get stuck in blood vessels, causing tissue damage and debilitating pain.

Over the past decades, treatment options have vastly increased the lifespan and quality of life of people born with sickle cell anemia. In addition, recent advances involving gene therapy and stem cells have put a cure for the disease on the horizon.

[q]Sickle cell anemia is the most common inherited genetic disease among Americans of African descent, occurring in 1 in 365 African-American births. About 1 in 13 African Americans (7%) carry the gene.

In some parts of Africa, the prevalence of people who carry the gene is much higher (above 14%).

Note that later in this tutorial we’ll address

  1. The difference between being born with the condition and carrying the gene, and
  2. Why this gene is in such high frequency in Africa.

[q multiple_choice=”true”] Let’s return to Hunter. The image on the right is a pedigree chart. It shows the inheritance of a trait. Here’s how to read it.

  • Squares represent males.
  • Circles represent females.
  • An empty square or circle means the trait is absent.
  • A filled-in square or circle means the trait is present.

True or false: Hunter’s mother and father suffer from sickle cell anemia.

[c]IFRydWU=[Qq]

[f]IE5vLiBUaGUgY2lyY2xlIGZvciBIdW50ZXImIzgyMTc7cyBtb3RoZXIgYW5kIHRoZSBzcXVhcmUgZm9yIEh1bnRlciYjODIxNztzIGZhdGhlciBhcmUgYm90aCBlbXB0eS4gVGhhdCBtZWFucyB0aGF0IHRoZXkgZG9uJiM4MjE3O3Q=IGhhdmUgdGhlIHRyYWl0LiBJZiBuZWl0aGVyIG9mIHRoZW0gaGFzIHRoZSBkaXNlYXNlLCB0aGVuIGhvdyBkaWQgSHVudGVyIGluaGVyaXQgaXQ/[Qq]

[c]IEZh bHNl[Qq]

[f]IENvcnJlY3QuIFRoZSBjaXJjbGUgZm9yIEh1bnRlciYjODIxNztzIG1vdGhlciBhbmQgdGhlIHNxdWFyZSBmb3IgSHVudGVyJiM4MjE3O3MgZmF0aGVyIGFyZSBib3RoIGVtcHR5LiBUaGF0IG1lYW5zIHRoYXQgdGhleSA=ZG9uJiM4MjE3O3Q=IGhhdmUgdGhlIHRyYWl0LiBJZiBuZWl0aGVyIG9mIHRoZW0gaGFzIHRoZSBkaXNlYXNlLCB0aGVuIGhvdyBkaWQgSHVudGVyIGluaGVyaXQgaXQ/[Qq]

[q] The hemoglobin gene is found on human chromosome 11. A gene is a unit of heredity that is transferred from parent to offspring and determines some trait.

[q] The hemoglobin gene comes in two forms. As we discussed in relation to meiosis, an alternative form of a gene is known as an allele.

In the case of sickle cell anemia, the two alleles are HBA, which codes for normal hemoglobin, and HBS, which codes for the mutated form of hemoglobin that causes sickle cell anemia.

[q] Let’s make sure the distinction between the two terms we just discussed is clear. A [hangman] is a unit of heredity that determines some trait.

[c]IGdlbmU=[Qq]

[f]IENvcnJlY3Qh[Qq]

[q] Alternative versions of genes are called [hangman].

[c]IGFsbGVsZXM=[Qq]

[f]IEdyZWF0IQ==[Qq]

[q] Hunter inherited two copies of the mutant HBS allele. That makes Hunter, for this particular allele, a homozygote: an organism that has two identical alleles for the same trait.

[q] Hunter’s parents are heterozygotes. That means that they have two different alleles. One is the normal version, HBA, and the second is the mutated version, HBS.

 

 

 

[q]Based on what you know about Hunter and her parents, designate each member of the family as a “homozygote” or “heterozygote” by typing the correct term into the boxes below.

[hangman]
[hangman]
[hangman]

[c]IGhldGVyb3p5Z290ZQ==[Qq]

[f]IEV4Y2VsbGVudCE=[Qq]

[c]IGhvbW96eWdvdGU=[Qq]

[f]IENvcnJlY3Qh[Qq]

[c]IGhldGVyb3p5Z290ZQ==[Qq]

[f]IENvcnJlY3Qh[Qq]

[q] But if Hunter’s parents both have the HBS allele, why don’t they have sickle cell anemia?

[q] HBS, the sickle cell allele, is recessive. It only has an effect on an individual who inherits two copies. In other words, a recessive allele like HBS will only cause sickle cell anemia in an individual who (like Hunter) is a [hangman].

[c]IGhvbW96eWdvdGU=[Qq]

[f]IEdyZWF0IQ==[Qq]

[q]

HBA, the normal allele, is dominant. If you inherit a dominant allele, you’ll express the dominant trait (even if you also inherit the recessive allele). In the case of sickle cell anemia, the normal allele is normal hemoglobin and normal blood cells.

[q]Based on what you’ve learned about sickle cell anemia, label the diagram below with the terms “recessive” or “dominant.”

[hangman]
[hangman]

[c]IGRvbWluYW50[Qq]

[f]IEdvb2Qh[Qq]

[c]IHJlY2Vzc2l2ZQ==[Qq]

[f]IEV4Y2VsbGVudCE=[Qq]

[q] A few additional terms are going to be useful in our study of genetics:

  • Phenotype: an organism’s appearance.
  • Genotype: an organism’s underlying genes.

For example, Mr. W’s eye-color phenotype is brown-eyed. And while the genetic basis of eye color inheritance is complex (involving up to three genes), Mr. W knows (based on his children’s and father’s blue-eyed phenotype) that he has to be a heterozygote.

[q]

Neither of Hunter’s parents has sickle cell anemia. That’s a description of their [hangman]. At the same time, we know that they’re heterozygotes, and have the sickle cell gene. That’s a description of their [hangman]

[c]cGhlbm90eXBl[Qq]

[c]Z2Vub3R5cGU=[Qq]

[q]

And here’s another term. Heterozygotes like Hunter’s parents possess the recessive HBS allele but don’t have sickle cell disease. That makes them carriers. Note that this term is only used to refer to heterozygotes, who are carrying a recessive allele (in each of their cells) but not expressing it. Homozygous recessive individuals like Hunter, even though they obviously have the HBS allele (two copies, in each of their cells) are referred to as homozygotes or homozygous recessive.

[q] When Hunter’s parents created the gametes that led to Hunter, each had a 1 in 2 chance of creating a gamete with a copy of chromosome 11 that carried the HBS allele. So what was the chance that Hunter would inherit two mutant alleles, leading her to have sickle cell anemia?

Fill in your answer below: it’s a 1 in ____ chance?

[textentry single_char=”true”]

[c]ID Q=[Qq]

[f]IEV4Y2VsbGVudC4gVGhlIGNoYW5jZSBvZiBIdW50ZXIgaW5oZXJpdGluZyB0d28gSEI=Uw==IGFsbGVsZXMgd2FzIGEgMSBpbiAyIGNoYW5jZSBvZiByZWNlaXZpbmcgSEI=Uw==IGluIGhlciBtb3RoZXImIzgyMTc7cyBlZ2csIG11bHRpcGxpZWQgYnkgYSAxIGluIDIgY2hhbmNlIG9mIHJlY2VpdmluZyBIQg==[Qq]S in her father’s sperm. That makes for a 1/4 chance.

[c]Kg==[Qq]

[f]Tm86IGhlcmUmIzgyMTc7cyBob3cgdG8gdGhpbmsgYWJvdXQgaXQuIFRoZSBjaGFuY2Ugb2YgSHVudGVyIGluaGVyaXRpbmcgdHdvIEhCUw==IGFsbGVsZXMgd2FzIGEgMSBpbiAyIGNoYW5jZSBvZiByZWNlaXZpbmcgSEI=Uw==IGluIGhlciBtb3RoZXImIzgyMTc7cyBlZ2csIG11bHRpcGxpZWQgYnkgYSAxIGluIDIgY2hhbmNlIG9mIHJlY2VpdmluZyBIQg==[Qq]S in her father’s sperm. That makes for a 1/4 chance.

[q] Here’s a more elaborate representation that might make this 1 in 4 chance clearer. This diagram uses the capital letter S to represent the dominant allele (HBA) for normal hemoglobin and the lowercase letter s to represent the recessive (HBS) allele for sickle cell hemoglobin. Study the diagram for a moment, and then go on to the next card.

[q labels=”top”]Complete the table below.

Genotype Homozygous or heterozygous? Phenotype
SS ______________________ ______________________
Ss ______________________ ______________________
ss ______________________ ______________________

[l]homozygous

[fx] No, that’s not correct. Please try again.

[f*] Good!

[l]heterozygous

[fx] No. Please try again.

[f*] Excellent!

[l]dominant

[fx] No, that’s not correct. Please try again.

[f*] Excellent!

[l]recessive

[fx] No, that’s not correct. Please try again.

[f*] Great!

[q]

When a heterozygote (such as Hunter’s mother and father) creates gametes, there’s a 1:2 chance that they’ll pass on their recessive allele. To figure out the chance that two heterozygotes will have a child with two recessive alleles, you multiply 1/2 X 1/2 = 1/4. Hunter had the misfortune to be on the receiving end of that 1 in [hangman] chance.

[c]Zm91cg==[Qq]

[q]There’s the happy ending to this story. Hunter had a bone marrow transplant that cured her of sickle cell anemia.

And there’s more good news on the horizon: diseases like sickle cell anemia are current targets for genetic therapies that might provide other ways to cure people of the disease (at lower cost and less risk than a bone marrow transplant).

[q]Here’s a final point. Why is the sickle cell allele so frequent among African Americans (and Africans)? It’s for an evolutionary reason. When looking from the outside, there’s no phenotypic difference between people who are homozygous dominant (SS) and heterozygotes (Ss). But carrying a single copy of the allele causes a difference in the protein chemistry of their red blood cells. This difference provides heterozygotes with protection against a mosquito-borne disease called malaria. As a result, in areas with a lot of malaria, there’s been natural selection to increase the number of heterozygotes, despite the harm caused to individuals who are homozygous recessive.

We’ll talk more about this phenomenon, called heterozygote advantage, in AP Bio Unit 7 (evolution).

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[/qwiz]

2. Genetics Vocabulary Check

[qwiz qrecord_id=”sciencemusicvideosmeister1961-Genetics Vocabulary (v2.0)”]

[h]Genetics Vocabulary Check

[q]The founder of genetics was an Austrian monk named Gregor [hangman]. In his honor, the basic principles of genetics are referred to as [hangman] genetics.

[c]TWVuZGVs[Qq]

[c]TWVuZGVsaWFu[Qq]

[q] An allele that always shows up in the offspring is a(n) [hangman] allele.

[c]ZG9taW5hbnQ=[Qq]

[q] An alternative form of a gene is called a(n) [hangman].

[c]YWxsZWxl[Qq]

[q]A(n) [hangman] allele will only show up in the offspring if the offspring inherits two copies of that allele.

[c]cmVjZXNzaXZl[Qq]

[q] An organism that has two identical alleles for a particular gene is known as a(n) [hangman].

[c]aG9tb3p5Z290ZQ==[Qq]

[q] An organism that has two different alleles for a particular gene is known as a(n) [hangman].

[c]aGV0ZXJvenlnb3Rl[Qq]

[q]Another term for a heterozygote refers to the fact that they possess the allele, don’t display the trait, and can pass it on to the next generation. That term is [hangman].

[c]Y2Fycmllcg==[Qq]

[q] A unit of heredity that is passed from parent to offspring is known as a(n) [hangman].

[c]Z2VuZQ==[Qq]

[q] The chance that a heterozygote parent will pass on a recessive or a dominant allele is one in [hangman] (write out the number).

[c]dHdv[Qq]

[q] The chance that the offspring of two heterozygotes will inherit two recessive alleles is one in [hangman](write out the number).

[c]Zm91cg==[Qq]

[q] Your underlying genes make up your [hangman]. Your appearance is your [hangman].

[c]Z2Vub3R5cGU=[Qq]

[c]cGhlbm90eXBl[Qq]

[q]A chart (like the one below) used to show the inheritance of a trait is a(n) [hangman] chart.

[c]cGVkaWdyZWU=[Qq]

[q]A person with a genotype that would be symbolized as Aa or Cc is

[c]aG9tb3p5Z291cyBkb21pbmFudA==[Qq]

[f]Tm8uIEhvbW96eWdvdXMgbWVhbnMgdGhlIHR3byBhbGxlbGVzIGFyZSB0aGUgc2FtZS4gSW4gdGhpcyBjYXNlLCB0aGUgdHdvIGFsbGVsZXMgYXJlIGRpZmZlcmVudC4gQSBob21venlnb3VzIGRvbWluYW50IGdlbm90eXBlIGxvb2tzIGxpa2UgdGhpczogU1MsIEFBLCBHRywgZXRjLiBBbHdheXMgdHdvIHVwcGVyY2FzZSBsZXR0ZXJzLg==[Qq]

[c]aGV0ZXJvenlnb3VzIC hhbmQgZG9taW5hbnQp[Qq]

[f]TmljZS4gQ2Mgb3IgQWEgZ2Vub3R5cGVzIGFyZSBoZXRlcm96eWdvdXMgKGFuZCBoZXRlcm96eWdvdGVzIGFsd2F5cyBkaXNwbGF5IHRoZSBkb21pbmFudCB0cmFpdCwgd2hpY2ggaXMgd2h5ICYjODIyMDtkb21pbmFudCYjODIyMTsgaXMgaW4gcGFyZW50aGVzaXMu[Qq]

[c]aGV0ZXJvenlnb3VzIChhbmQgcmVjZXNzaXZlKQ==[Qq]

[f]Tm8uIElmIHlvdSYjODIxNztyZSBoZXRlcm96eWdvdXMsIHlvdSYjODIxNztsbCBkaXNwbGF5IHRoZSBkb21pbmFudCB0cmFpdC4gWW91IGNhbiYjODIxNzt0IGJlICYjODIyMDtoZXRlcm96eWdvdXMgcmVjZXNzaXZlLiYjODIyMTs=[Qq]

[c]aG9tb3p5Z291cyByZWNlc3NpdmU=[Qq]

[f]Tm8uIEhvbW96eWdvdXMgcmVjZXNzaXZlIHBoZW5vdHlwZXMgbG9vayBsaWtlIHRoaXM6IGNjLCBzcywgYWEsIGV0Yy4gQWx3YXlzIHR3byBsb3dlcmNhc2UgbGV0dGVycy4=[Qq]

[q]A person with a genotype that would be symbolized as aa or cc is

[c]aG9tb3p5Z291cyBkb21pbmFudA==[Qq]

[f]Tm8uIEEgZG9taW5hbnQgZ2Vub3R5cGUgYWx3YXlzIGhhcyBhbiB1cHBlcmNhc2UgbGV0dGVyLiBBIGhvbW96eWdvdXMgZG9taW5hbnQgZ2Vub3R5cGUgbG9va3MgbGlrZSB0aGlzOiBTUywgQUEsIEdHLCBldGMu[Qq]

[c]aGV0ZXJvenlnb3VzIChhbmQgZG9taW5hbnQp[Qq]

[f]Tm8uwqAgSGV0ZXJvenlnb3VzIG1lYW5zIHRoYXQgdGhlIHR3byBhbGxlbGVzIGFyZSBkaWZmZXJlbnQuIEZvciBleGFtcGxlLCAmIzgyMjA7QWEmIzgyMjE7IGlzIGEgaGV0ZXJvenlnb3VzIGdlbm90eXBlLg==[Qq]

[c]aGV0ZXJvenlnb3VzIChhbmQgcmVjZXNzaXZlKQ==[Qq]

[f]Tm8uIEhldGVyb3p5Z291cyBtZWFucyB0aGF0IHRoZSB0d28gYWxsZWxlcyBhcmUgZGlmZmVyZW50LiBGb3IgZXhhbXBsZSwgJiM4MjIwO0FhJiM4MjIxOyBpcyBhIGhldGVyb3p5Z291cyBnZW5vdHlwZS4gQWxzbywgeW91IGNhbiYjODIxNzt0IGJlIGhldGVyb3p5Z291cyBhbmQgcmVjZXNzaXZlLiBIZXRlcm96eWdvdGVzIGFsd2F5cyBkaXNwbGF5IHRoZSBkb21pbmFudCB0cmFpdC4=[Qq]

[c]aG9tb3p5Z291cy ByZWNlc3NpdmU=[Qq]

[f]V2F5IHRvIGdvLiBIb21venlnb3VzIHJlY2Vzc2l2ZSBwaGVub3R5cGVzIGFsd2F5cyBoYXZlIHR3byBsb3dlcmNhc2UgbGV0dGVycy4=[Qq]

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3. Understanding Cystic Fibrosis

To consolidate what you learned above, let’s look at the most common genetic disease to affect North Americans of European descent: cystic fibrosis.

Cystic fibrosis is primarily a lung disease. The key symptom is the buildup of sticky mucus that leads to chronic infections in the lungs. In addition, this same mucus causes problems throughout the body. Even with the best medical care, people with cystic fibrosis have much shorter lifespans than people without the disease. As you read about this disease, fill in the blanks as needed.

[qwiz qrecord_id=”sciencemusicvideosmeister1961-Cystic Fibrosis Interactive Reading (v2.0)”]

[h]Cystic Fibrosis, Interactive Reading

[i]

Symptoms of cystic fibrosis

[q labels=”top”]

Cystic fibrosis is caused by a defect in a protein found in cell membranes. The _______ for this protein is found on chromosome 7. That makes the gene an ____________ gene (because it’s not on the X or Y chromosome).

If you inherit a working ________ (version) of this gene, you’ll be fine, even if you inherit one defective allele. That’s because the disease-causing allele is ___________. If you inherit the working, ____________ allele, then this allele will be translated into the working membrane protein. Your cells will function as they should, and there’ll be no build-up of the sticky mucus that underlies the disease.

[l]allele

[fx] No, that’s not correct. Please try again.

[f*] Good!

[l]autosomal

[fx] No, that’s not correct. Please try again.

[f*] Good!

[l]dominant

[fx] No, that’s not correct. Please try again.

[f*] Great!

[l]gene

[fx] No. Please try again.

[f*] Great!

[l]genotypes

[fx] No. Please try again.

[f*] Great!

[l]homozygous

[fx] No, that’s not correct. Please try again.

[f*] Great!

[l]heterozygous

[fx] No. Please try again.

[f*] Good!

[l]phenotype

[fx] No. Please try again.

[f*] Excellent!

[l]recessive

[fx] No. Please try again.

[f*] Excellent!

[q labels=”top”]

You can only have the disease if you inherit two defective alleles. Another way to say this is that to have cystic fibrosis, you have to be ______________ for the cystic fibrosis allele. Being homozygous means that you’ll have two copies of the allele. People who are _______________, with one dominant allele and one recessive allele, don’t get the disease. They have a normal ____________ (no sticky mucus).

[l]allele

[fx] No, that’s not correct. Please try again.

[f*] Good!

[l]autosomal

[fx] No, that’s not correct. Please try again.

[f*] Good!

[l]dominant

[fx] No, that’s not correct. Please try again.

[f*] Great!

[l]gene

[fx] No. Please try again.

[f*] Great!

[l]genotypes

[fx] No. Please try again.

[f*] Great!

[l]homozygous

[fx] No, that’s not correct. Please try again.

[f*] Great!

[l]heterozygous

[fx] No. Please try again.

[f*] Good!

[l]phenotype

[fx] No. Please try again.

[f*] Excellent!

[l]recessive

[fx] No. Please try again.

[f*] Excellent!

[q labels=”top”]

If we use C to represent the normal allele and c to represent the cystic fibrosis allele, then we have three ____________:

  • _______________ dominant: CC
  • ________________: Cc
  • homozygous ___________: cc

[l]allele

[fx] No, that’s not correct. Please try again.

[f*] Good!

[l]autosomal

[fx] No, that’s not correct. Please try again.

[f*] Good!

[l]dominant

[fx] No, that’s not correct. Please try again.

[f*] Great!

[l]gene

[fx] No. Please try again.

[f*] Great!

[l]genotypes

[fx] No. Please try again.

[f*] Great!

[l]homozygous

[fx] No, that’s not correct. Please try again.

[f*] Great!

[l]heterozygous

[fx] No. Please try again.

[f*] Good!

[l]phenotype

[fx] No. Please try again.

[f*] Excellent!

[l]recessive

[fx] No. Please try again.

[f*] Excellent!

[q]Just to check your understanding. Cystic fibrosis is an autosomal [hangman] condition. That means you’ll only have the phenotype if you inherit [hangman] copies of the allele.

[c]cmVjZXNzaXZl[Qq]

[c]dHdv[Qq]

[/qwiz]

4. Cystic Fibrosis: Extending your learning

Follow this link (it’ll open in a new tab) to the Genetics Home Reference about Cystic Fibrosis. Spend about five minutes reading the first three sections. Record what you’ve learned on your student learning guide.

5. Guided Punnett Square 1: Autosomal Inheritance

Gregor Mendel, By Iltis, Hugo via Wikimedia Commons

In AP Bio, you’ll be tested on your ability to understand genetics through problem-solving. While that might feel daunting, you’ll find that solving genetics problems will lead you to deep and substantial understanding (which always feels, great!).

Many of the problems that you’ll encounter in a typical AP Bio textbook involve peas. Why peas? It’s for historical reasons. Much of the foundational work in genetics was discovered by Gregor Mendel, an Austrian monk who lived from 1822 to 1884. Mendel discovered the key principles of genetics by breeding different varieties of garden pea plants and analyzing the results.

Table displaying illustrations of the different characteristics Mendel observed in garden peas. Seed traits include form (round or wrinkled) and cotyledons (yellow or green). Flower color (white or purple) was an observed trait. Pod traits included form (full or constricted) and color (yellow or green). Stem traits included place (axial or terminal flowers) and size (tall or short).
Mendel experimented with various genetic traits in garden peas

The most basic type of genetics problem involves dominant and recessive autosomal alleles.

I’ll use a sample problem to demonstrate how you can go about finding a solution.

Sample Problem: In one of the varieties that Mendel experimented with, the allele for purple flower color (P) is dominant to the allele for white flower color (p). What genotypes and phenotypes will be produced if a heterozygous purple-flowered pea plant is crossed with a white-flowered pea plant?

STEP 1: Figure out the alleles of the parents
Heterozygous means that the alleles are different. Therefore the first parent has to be Pp. The second parent is showing the recessive phenotype and has to be homozygous recessive, or pp.
STEP 2: Draw a Punnett square.

 A  A
 A  AA AA
 A  AA  AA

STEP 3: Place the alleles of the parents on the left and to the top of the Punnett square. Remember that these alleles represent the possible alleles in the gametes (sex cells) that this parent could create.

P p
p
p

STEP 4: Bring the alleles in the parents’ gametes over and down to fill in the four inner squares. Each of those squares represents a potential zygote.

P p
p Pp pp
p Pp pp

STEP 5: Represent the possible genotypes as a ratio.
50% of the offspring will have the genotype Pp. 50% will have the genotype pp. You can also state this as
1 Pp: 1pp. Or 1/2 Pp and 1/2 pp.
STEP 6: Represent the possible phenotypes as a ratio
50% purple flowers: 50% white flowers (or 1 purple-flowered: 1 white flowered)

Got it? Try the problems in the quiz below. On some, you’ll be guided, and on some, you’ll need paper and a pencil to find the solution.

[qwiz qrecord_id=”sciencemusicvideosmeister1961-Guided Punnett Square, Autosomal (v2.0)”]

[h]Guided Punnett Square 1: Autosomal inheritance

[q] In peas, the tall allele (T) is dominant to the short allele (t). Cross a plant that is homozygous tall with a plant that is heterozygous tall.

Step 1: Determine the genotypes of the parents

Parent 1) Homozygous tall: ___ ___

Parent 2) Heterozygous tall: ___ ___

Steps 2 through 4: Complete your Punnett square. Note that for this quiz, you have to place the alleles of the gametes of the 1st parent to the left of the Punnett square, and put the alleles of the second parent’s gametes on top. On paper, you could do it in either order.

 

Parent 2
       __    __
 Parent 1 __ ___ ___ ___ ___
__ ___ ___ ___ ___

 

Step 5: Genotype ratio: _____ TT: ____Tt: ____tt

Step 6: Phenotype ratio: _____ Tall: ______Short

 

[l]T

[fx] No, that’s not correct. Please try again.

[f*] Excellent!

[l]t

[fx] No. Please try again.

[f*] Great!

[l]0%

[fx] No, that’s not correct. Please try again.

[f*] Good!

[l]25%

[fx] No, that’s not correct. Please try again.

[f*] Correct!

[l]50%

[fx] No, that’s not correct. Please try again.

[f*] Good!

[l]75%

[fx] No, that’s not correct. Please try again.

[f*] Correct!

[l]100%

[fx] No. Please try again.

[f*] Great!

 

[q]In peas, the gene for purple flowers (P) is dominant to the gene for white flowers (p). What will happen if a plant homozygous for purple flowers is crossed with a plant homozygous for white flowers?

Step 1: Determine the genotypes of the parents

Parent 1) Homozygous purple flowers: ___ ___
Parent 2) Homozygous white flowers: ___ ___

Step 2: Draw the Punnett square (already done below)

Step 3: Place the alleles of the gametes that the 1st parent can make to the left of the Punnett square. Put the alleles that the 2nd parent can make on top. (note: these required positions (left and top) are solely so that this quiz program will work. On paper, you can put them in either position.

Step 4: Move the gametes to the right or downward into the square, simulating fertilization and zygote formation.

Parent 2
 __  __
 Parent 1   __ ___ ___ ___ ___
__ ___ ___ ___ ___

Step 5: Genotype ratio: _____ PP: ____Pp: ____pp

Step 6: Phenotype ratio: _____ Purple-flowered, ____white-flowered

[l]P

[fx] No. Please try again.

[f*] Excellent!

[l]p

[fx] No, that’s not correct. Please try again.

[f*] Correct!

[l]0%

[fx] No. Please try again.

[f*] Correct!

[l]25%

[fx] No. Please try again.

[f*] Good!

[l]75%

[fx] No. Please try again.

[f*] Great!

[l]100%

[fx] No, that’s not correct. Please try again.

[f*] Correct!

[q]In peas, the gene for tall (T) is dominant to the gene for short (t). What genotypes and phenotypes will result in the offspring if two heterozygous plants are crossed with each other?

Step 1: Determine the genotypes of the parents

Parent 1) Heterozygous tall: ___ ___
Parent 2) Heterozygous tall: ___ ___

Step 2: Draw the Punnett square (I did that one for you)

Step 3: Place the alleles of the gametes that the 1st parent can make to the left of the Punnett square. Put the alleles that the 2nd parent can make on top. (note: these required positions (left and top) are solely so that this quiz program will work. On paper, you can put them in either position).

Step 4: Move the gametes to the right or downward into the square, simulating fertilization and zygote formation.

Parent 2
     __  __
 Parent 1  __ ___ ___ ___ ___
__ ___ ___ ___ ___

Step 5: Genotype ratio: _____ TT: ____Tt: ____tt

Step 6: Phenotype ratio:

_____ Tall, _____medium, _____ short

[l]T

[fx] No. Please try again.

[f*] Great!

[l]t

[fx] No, that’s not correct. Please try again.

[f*] Great!

[l]0%

[fx] No, that’s not correct. Please try again.

[f*] Excellent!

[l]25%

[fx] No. Please try again.

[f*] Excellent!

[l]50%

[fx] No, that’s not correct. Please try again.

[f*] Excellent!

[l]75%

[fx] No. Please try again.

[f*] Correct!

[l]100%

[fx] No. Please try again.

[f*] Great!

[/qwiz]

What’s next?

Proceed to the next tutorial in AP Bio Unit 5: Topics 5.3-5.5 Part 2: Solving ABO Blood Type Inheritance Problems