activity 2 problem solving in genetics using dihybrid cross

How to Solve a Dihybrid Cross

Step 1: identify parental genotypes.

Determine the genotype of the parental organisms for the two traits you're studying. For example, if you're examining seed color (Y for yellow, y for green) and seed shape (R for round, r for wrinkled) in pea plants, you might start with parents of genotypes YYRR (pure-breeding for yellow round seeds) and yyrr (pure-breeding for green wrinkled seeds).

Create a "key" for the possible phenotypes

YY = Yellow Yy = Yellow yy = green RR = round Rr = round rr = wrinkled

Step 2: Determine Possible Gametes

Identify the possible gametes each parent can produce through the process of Mendelian segregation. Each parent will produce haploid gametes containing one allele for each trait.

For the parents YyRr and YyRr :

• Parent 1 gametes: YR, Yr, yR, yr
• Parent 2 gametes: YR, Yr, yR, yr

Use the "FOIL" method to determine the gametes - first, outer, inner, last

Step 3: Set Up a Punnett Square

Construct a Punnett square to visualize the potential genotypes of offspring resulting from the cross between the parental organisms. Label the rows and columns with the possible gametes from each parent.

Step 4: Analyze Offspring Genotypes

Fill in the Punnett square by combining the gametes from each parent. This will give you the genotype combinations of the offspring. There are 16 possible combinations of genotypes.

Step 5: Determine Phenotypes

Once you have the genotypes of the offspring, determine the phenotypes associated with each genotype based on the expression of the traits being studied. Refer to your "key" to help your determine the phenotypes. It may also be helpful to colorcode the squares to help you count.

Step 6: Calculate Phenotypic and Genotypic Ratios

Count the number of each phenotype and genotype in the offspring to determine the phenotypic and genotypic ratios.

Yellow, Round = 9/16 Yellow, Wrinkled= 3/16 Green, Round = 3/16 Green, Wrinkled = 1/6 9:3:3:1

Practice Sets on Genetic Crosses

Advanced Bunny Genetic Crosses with two traits - also includes fruit fly genetics

Bunny Genetics 2 - basic crosses, rabbits have gray or white hair, black or red eyes

Explore the Genetics of Corn Snakes - dihybrid crosses with corn snakes, color is polygenic

Multiple Allele Traits in Chickens - shows how combs are inherited (rrpp x RRpp)

AP® Biology

The dihybrid cross problem: ap® biology crash course.

• The Albert Team
• Last Updated On: March 1, 2022

The dihybrid cross problem can be a very challenging topic for AP® Biology students to master. Luckily, with practice and an organized method for completing the problem, the dihybrid cross problem can be solved! In this AP® Biology Crash Course Review , we will work together to solve a dihybrid cross problem. We will begin by reviewing the monohybrid cross. We will then discuss why a dihybrid cross would be done by scientists. Finally, we will walk through a dihybrid cross with a problem that you could see on your AP® Biology exam.

Monohybrid Cross

In order to understand a dihybrid cross, we must first review a monohybrid cross. A monohybrid cross is when mating occurs between two individuals with different alleles at a single locus of interest. When we consider these problems, plants will often be the focus because their mating can be most easily controlled by scientists.

We will review a case of tall-stemmed flowers and short-stemmed flowers. A tall stem is a dominant allele, and a short stem is a recessive allele (for the purpose of this example). Let’s review several different mating situations and discuss their outcomes.

In this case, we are crossing a tall-stemmed flower and a short-stemmed flower. As you can see, this cross will result in 100% heterozygous offspring. Heterozygous means that the individual has both a dominant and a recessive allele. Because the tall stem is a dominant allele, all of these plants will have tall stems.

In this case, we are crossing a tall-stemmed flower with a tall-stemmed flower. We will produce 100% homozygous dominant plants. Homozygous means that the individual has two of the same alleles. Each of the offspring will be tall stemmed.

Lastly, this case shows the cross of two heterozygous plants. In this example, there are a variety of outcomes that may occur.

Phenotypic and Genotypic Ratios

If the AP® Biology exam were to ask you about monohybrid crosses, the question will likely focus on the ratios that the cross produces. It will be necessary for you to first be able to fill out the Punnett square. After you have filled the Punnett square with the offspring genotypes, you can easily figure out the offspring ratios.

Let’s look at the Punnett squares that we completed above. The phenotypic ratio will be the ratio of phenotypes that you will see in the offspring organisms. For organisms with Tt or TT, the plant will have a tall stem. Organisms with tt will have a short stem. The genotypic ratio will be the ratio of genotypes in the offspring. In the genotypic ratio, even though Tt and TT both produce tall stem offspring, they are different.

In the final example, the genotypic ratio is 1:2:1. The order that we write the genotypic ratio is the dominant homozygous, the heterozygous, and the recessive homozygous. The phenotypic ratio is 3:1 because the heterozygous offspring will still be tall-stemmed.

Theory of Dihybrid Cross

Monohybrid crosses do occur in science; however, organisms are much more complicated than one single allele. Many times during cross-breeding, scientists will have to deal with organisms that have two different alleles at two different loci. For example, a dihybrid cross would be between a tall (TT) and blue plant (BB) with a short (tt) and red plant (bb).

It is important also to understand Mendel’s second law of independent assortment. Mendel’s second law states that alleles of one gene sort independently of alleles of another gene. Basically, when performing a dihybrid cross, you can think of it as two separate monohybrid crosses.

Now that we have reviewed the monohybrid cross and the theory of a dihybrid cross, we will work through an example of a dihybrid cross.

Dihybrid Cross Problem

We are going to work through a dihybrid cross problem step-by-step. These problems can be complicated to learn, but when we have walked through one problem, you should be able to use the same method for all of the dihybrid problems.

In guinea pigs, the allele for black fur (B) is dominant over that for brown fur (b). Similarly, the allele for short fur (S) is dominant over that for long fur (s).

We will do an example of a cross between BbSs and BBss.

The very first step that you should complete when doing a dihybrid cross is to figure out the possible gametes of the parents. We must figure out all of the ways possible for the alleles to sort themselves based on Mendel’s second law of independent assortment. We will fill this part of the table out first.

If you’re having trouble with this step, start with just one allele first and fill in the second allele after.

In this step, we will fill in the offspring by combining the gametes. This section can be complicated, but we will use the same method for filling in a dihybrid cross, as we did for a monohybrid cross.

In this final step, we will figure out the phenotypic ratios. The genotypic ratios can get very complicated in dihybrid cross problems. For that reason, the AP® Biology exam will not require you to understand those ratios.

In order to create the phenotypic ratios, we will have to go back to the problem to see which traits are dominant; black fur (B) is dominant over brown fur (b), and short fur (S) is dominant over long fur (s). Now, we can look at our cells and determine what the offspring will display as a phenotype based on their genotype.

Now we can create a ratio. The phenotypic ratio for this dihybrid cross will be 8 (black fur, short fur): 8(black fur, long fur): 0 (brown fur, short fur): 0 (brown fur, long fur). The ratios that you will see will always be in this convention: dominant trait 1, dominant trait 2 : dominant trait 1, recessive trait 2 : recessive trait 1, dominant trait 2 : recessive trait 1, recessive trait 2.

AP® Biology Exam Problem

Because these problems are a major sticking point for many students, and practice only makes these problems easier, we are going to do one more problem in this Dihybrid Cross: AP® Biology Crash Course Review. I won’t walk through all of the steps, but I will give you the problem and solution to see if you understand the dihybrid cross problems.

Consider pea plants, in which green color is dominant to yellow color, and two leaves are dominant to three. A cross between two plants that are heterozygous green and heterozygous tall is completed. What is the phenotypic ratio?

The phenotypic ratio for the problem above is 9:3:3:1. Note that when crossing two heterozygous (for both alleles) parents this will always be the ratio.

In this article Dihybrid Cross: AP® Biology Crash Course Review, we began by reviewing the monohybrid cross and the possibilities that could result from crossing parents with different alleles. We worked through several monohybrid crosses and learned that we can find the genotypic and phenotypic ratios by using Punnett squares and creating fractions of every genotype and phenotype. We then talked about the importance of dihybrid cross problems when dealing with very complex organisms. Finally, we walked through the steps to completing a dihybrid cross problem with guinea pigs, and then you completed a dihybrid cross independently to practice for the AP® Biology exam.

Thank you for reading this article! If you are interested in reading more, please check out our article Chi Square Test: AP® Biology Crash Course .

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Topics 5.3 – 5.5, Part 4: Dihybrid Crosses

1. introduction.

In the previous tutorials, we’ve looked at variations on crosses that involved a single gene, with two alleles (or three, in the case of blood type). But what happens when two (or more) genes are being transmitted at the same time?

To see the answer, we’re going to look at the genetics experiments performed by Gregor Mendel in the 1850s and 1860s.

While Mendel’s life and career are outside the scope of AP Biology, you can learn more about his story on Wikipedia .

2. Mendel’s 1 st Experiments: Monohybrid Crosses and Segregation of Alleles

The table below shows seven of the traits in garden peas that Mendel experimented with as he was figuring out the principles of genetics.

Why did Mendel work with peas? For one thing, they were easily available in true-breeding varieties. True breeding is roughly synonymous with homozygous. What it meant for Mendel is that a plant with yellow seeds could be crossed with another plant with yellow seeds and all the offspring would have yellow seeds. It’s similar to breeding a poodle with another poodle: the offspring will be poodles.

Mendel’s first series of experiments involved monohybrid crosses. 

From our previous study of genetics, we can figure out what happened. The green-seed trait disappears in the first generation, so we can assume that green is recessive and yellow is dominant. As a result, all of the F 1 s are yellow heterozygotes.

The Punnett square (a mathematical tool that Mendel didn’t have at his disposal), looks like this.

100% of the offspring are Yy: heterozygotes with yellow seeds.

Mendel’s next move was a monohybrid cross.  In a monohybrid cross, you cross two hybrids.

As we’ve previously seen when we were learning about sickle cell anemia, a monohybrid cross produces offspring in a 3 to 1 ratio, with 3 offspring showing the dominant trait for each one showing the recessive trait.

Here’s the Punnett square. We know that each of the F 1 s is a heterozygote.

75% of the offspring have genotypes YY or yy , giving them yellow seeds. 25% have green seeds (genotype yy ).

Mendel’s monohybrid crosses revealed some of the key principles of genetics. These include the idea of dominant and recessive traits, and what Mendel called the principle of segregation . Why segregation? The name reflects Mendel’s insight that parents possess two alleles for a trait. When that parent creates offspring, he/she/it only passes on one of its two alleles. The alleles separate (or segregate ) during gamete formation.

Note that we saw this in our introduction to genetics when we saw how sickle cell anemia can be inherited. Each of Hunter Haymore’s parents was a heterozygote carrier of the sickle cell allele. During meiosis, these alleles separated, with one of the two alleles sent to each parent’s gametes. That’s segregation of alleles.

3. Dihybrid Crosses and Independent Assortment

Mendel’s next series of experiments involved dihybrid crosses : crosses in which individuals were hybrid for two traits, such as stem height and flower color.

Follow what Mendel did through the interactive reading below.

[qwiz style=”min-height: 450px !important; width: 600px !important;” qrecord_id=”sciencemusicvideosMeister1961-Dihybrid Crosses, Interactive Reading (v2.0)”]

[h] Mendel’s Dihybrid Crosses Interactive Reading.

[q] We’ll start with seeing how Mendel produced his dihybrids. He started, as always, with true-breeding varieties. But these varieties were true-breeding for two traits. In this case, a pea plant that was true-breeding for tall stems and purple flowers was bred with a plant that was true-breeding for short stems and white flowers.

Predict what the offspring will be. When you’ve written your prediction in your student learning guide, click “continue.”

[q] All of the offspring were tall, with purple flowers. Note, however, that the offspring are dihybrid (hybrid tall stem AND hybrid purple-flowered)

[q] Mendel’s next move was to carry out a dihybrid cross : He crossed his F 1 hybrids, producing the F 2 s. Predict what happened.

Click after you’ve written down your prediction.

[q labels = “right”]Here’s what happened. There were four phenotypes. And, once Mendel analyzed his data, he found that they occurred in predictable ratios. Try to predict which ratio goes with which phenotype. Don’t worry if you don’t know the answer. Just guess and drag these fractions until they’re in the right place.

[f*] Correct!

[fx] No. Please try again.

[f*] Excellent!

[q] So, here are the F 2 results that you get from an F 1 dihybrid cross.

• 9/16 have both dominant traits (in this case, tall stems AND purple flowers)
• 3/16 are dominant for the first trait and recessive for the second (tall stems, white flowers)
• 3/16 are recessive for the first trait and dominant for the second (short stems, purple flowers), and
• 1/16 are recessive for both traits (short stems, white flowers)

Explaining this 9:3:3:1 ratio got Mendel to his second great insight, his Principle of Independent Assortment . Let’s reconstruct Mendel’s thinking.

[q labels = “right”] Dihybrid cross results:

• Tall stem, purple flowers: 9/16
• Tall stem, white flowers: 3/16
• Short stem, purple flowers: 3/16
• Short stem, white flowers: 1/16

Let’s look at each characteristic separately.

If there were 16 offspring, you could expect ______ individuals to have tall stems; and  ______ individuals to have short stems

Reducing that to a ratio you’d expect  ______   with tall stems for every  ______   with short stems.

Now, do the same for flower color. Out of 16, you’d expect ______ individuals with purple flowers; and ______ individuals with white flowers.

That’s a ratio of ______   with purple flowers to ______   white flowers.

[f*] Great!

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

[q] In the previous card, you determined that each trait separately has a 3:1 ratio of dominant phenotype to recessive phenotype (3 tall to 1 short; and 3 purple-flowered to 1 white flowered). That’s the ratio we’d expect for a monohybrid cross. But how do we explain the 9:3:3:1 ratio that we see when we combine these phenotypes? It’s the result that we’d get if we did two independent Punnett squares and then combined the results, using what’s called the rule of multiplication. Start by completing the Punnett squares below in your student learning guide.

Click to continue when you’re done.

[q] Here’s the solution:

Click when you’re ready.

[q] Now let’s talk about the rule of multiplication: when two events are independent, then the probability of them occurring together is the product of their independent probabilities. If I’m tossing a coin, the probability that four tosses will result in 4 heads is 1/2 x 1/2 x 1/2 x 1/2, or 1/16. Why? Because with each flip, the probability is 1/2, and these four events are independent.

Try this. Both the father and the mother in a family are heterozygous for cystic fibrosis (a recessive condition). They have two children. What’s the probability that both children will have cystic fibrosis? As you fill in the following blanks, you have to write out the numbers.

For the first child, the probability of being born with cystic fibrosis is one out of [hangman]. For the second child, it’s one out of [hangman]. That makes the probability for both children one over [hangman].

[c]IGZvdXI=[Qq]

[f]IEdyZWF0IQ==[Qq]

[f]IENvcnJlY3Qh[Qq]

[c]IHNpeHRlZW4=[Qq]

[f]IEV4Y2VsbGVudCE=[Qq]

[q labels = “top”]Likewise, if we conclude (as Mendel did) that the trait for stem length operates independently from the trait for flower color, then the probability of a plant being tall AND purple-flowered is the product of a plant being tall (3/4), multiplied by the probability of it being purple-flowered (3/4). Use that logic to complete the table below.

[q] So that’s Mendel’s principle of independent assortment, derived from the laws of probability.

However, we can also explain what happens in a dihybrid cross by thinking about meiosis,  where there’s also a process called independent assortment. Look at the diagram below, and write down a quick explanation of meiotic independent assortment. When you’re done, check it against the next slide.

[q] Here’s an explanation of meiotic independent assortment: w hen homologous pairs are lined up on the cell equator during metaphase 1, the way each pair lines up is independent of every other pair. For example, in the diagram below, the chromosome that originated from the mother is pink; the one from the father is blue (sorry for the stereotypes). In cell A, the maternal chromosome of one pair is facing up, as is the paternal chromosome of the second pair. In cell B, the maternal chromosome for the first pair is facing up, and the maternal chromosome for the second pair is facing up. After anaphase 1 and cytokinesis, this will result in different arrangements of chromosomes in the haploid gametes. Note that there are no other arrangements of cells A and B that produce different results from what’s shown in cells 1 through 4.

With two homologous pairs, you get four possible arrangements of chromosomes in the gametes, as shown below.

[q] Genes are located on chromosomes. So if chromosomes are independently assorting, then so are the genes that are on these chromosomes (as long as these genes are on different chromosomes: we’ll look at what happens when they’re not in the next tutorial). To make this as simple as possible, look at the diagram below.

In this germ cell (a diploid cell that’s going to make haploid gametes), there are two homologous pairs of chromosomes. Because this was a dihybrid cross, we’re going to put a dominant allele ( T or P ) on one member of each homolog, and a recessive allele ( t or p ) on the other homolog.

Each gamete has to have an allele for stem length, and an allele for flower color. So, what alleles can be in each of the four gametes?

Figure out the four different combinations, write them down in your student learning guide, then check your answer on the next card.

[q] Here’s the germ cell, and the four gametes that it could produce by independent assortment. Note you can use the FOIL algorithm from algebra (First, Outside, Inside, Last) to get these results.

In the next slide, you can use these gametes to create a Punnett square. Click when you’re ready.

[q labels = “right”]In the dihybrid cross we’ve been examining, each of the parents can produce four types of gametes: TP, Tp, tP, and tP. That means that to predict the genotypes and phenotypes of the offspring, you need a 4 x 4 square to accommodate all the possible zygotes that can be produced.

I’ve set up the square below. Your job is to drag the right genotypes into the right spot.

[q labels = “right”] Here’s the Punnett square you completed on the previous slide. Let’s analyze the results to make sure we’re still seeing that 9:3:3:1 ratio.

• Drag a ♠ if the phenotype is Tall stem, Purple Flowers
• Drag a ♣ for Tall, white-flowered
• Drag a ♥ for short, purple-flowered
• Drag a ♦ for short, white-flowered

[q] Here’s the table you produced on the previous slide.

If you count up all the ♠ (tall, purple) you get [hangman] (please write out the number).

All the ♣ (tall, white) come to [hangman]

All the ♥ (short, purple) come to [hangman], and

All the ♦  (short, white) come to [hangman].

[c]IG5pbmU=[Qq]

[f]IEdvb2Qh[Qq]

[c]IHRocmVl[Qq]

[c]IG9uZQ==[Qq]

4. Solving Dihybrid Crosses and Related Problems

Mendel’s principle of independent assortment  is important to know about in its own right. It’s also the basis of many dihybrid cross genetics problems that you’ll be asked to solve in an AP biology course.

In solving dihybrid cross problems, the toughest part, for most students, is figuring out the gametes that the parents can produce. Here’s an example, working with Mendel’s peas.

Try to solve these problems on your own before looking at the answer.

SAMPLE PROBLEM 1

In peas, round seed texture (R) is dominant over wrinkled seeds (r). Yellow seed color (Y) dominates over green (y). What is the genotype of an organism that is heterozygous round and heterozygous yellow? What gametes could this organism form?

Step 1: Figure out the genotype of the parents

The parent is described as heterozygous round and heterozygous yellow. That means it’ll be RrYy

Step 2:   Use the FOIL algorithm to determine what gametes this parent can make.

FOIL (which I mentioned in the interactive reading above) stands for F irst, O utside, I nside, L ast.

• “First” means the first allele in each pair. “R” is the first allele in “Rr,” and “Y” is the first allele in “Yy.” So, the first alleles in the genotype RrYy are RY . That’s the first possible gamete.
• Outside means the two outermost alleles. In RrYy that’s Ry. That’s the second possible gamete.
• Inside means the two innermost alleles. In RrYy that’s rY. 
• Last means the two last alleles in each pair. In RrYy that’s ry

So, the four possible gametes are RY, Ry, rY, and ry.

SAMPLE PROBLEM 2

The setup is the same as in problem 1: In peas, round seed texture (R) is dominant over wrinkled seeds (r). Yellow seed color (Y) dominates over green (y). What is the genotype of an organism that is homozygous round and heterozygous yellow? What gametes could this organism form?

The parent is described as homozygous round and heterozygous yellow. That means it’ll be RRYy.

• “First” means the first allele in each pair. So, the first alleles in the genotype RRYy are RY . That’s the first possible gamete.
• Outside means the two outermost alleles. In RRYy that’s Ry.  That’s the second possible gamete.
• Inside means the two innermost alleles. In RRYy that’s RY. 
• Last means the two last alleles in each pair. In RRYy that’s Ry

If you list the four alleles derived using FOIL, you’ll see RY, Ry, RY, and Ry. The last two are repeats, which means that this parent can produce only two types of gametes: RY and Ry

SAMPLE PROBLEM 3

Let’s put these two problems together.

In peas, the allele for round seeds (R) is dominant over the allele for wrinkled seeds (r). Yellow seed color (Y) dominates over green (y). What is the result of a cross between a dihybrid Round Yellow parent and a parent that is homozygous round and heterozygous yellow?

STEP 1: Figure out the genotype of the parents. Look at problems 1 and 2 above. The genotypes are RrYy x RRYy.

STEP 2: Figure out the genotype of the gametes each parent can make.

Look at problems 1 and 2 above for the solution.

RrYy: RY, Ry, rY, ry

RRYy: RY and Ry.

STEP 3:  Set up a Punnett Square.

In introducing dihybrid crosses, we used a Punnett square which was 4 squares x 4 squares, to accommodate all the possible gametes. But in the cross above, one parent can only form two types of gametes. So you can set up your Punnett square in a 4 x 2 grid, like this:

STEP 4: Complete your Punnett square by combining the alleles in the gametes to see the possible offspring.

STEP 5: Analyze the offspring.

When we were doing monohybrid crosses, I had you list both the genotypes and phenotypes of the offspring. With dihybrid crosses, you mostly have to report phenotypes and their frequency.

In this case, the analysis is simplified because none of the offspring are wrinkled, so there are only two phenotype categories. I’m going to put a ♥ in all the squares with offspring that are round and yellow (R_Y_) and a ♦ i n all the squares that have offspring that are round and green (R_yy).

75% of the offspring are round and yellow

25% of the offspring are round and green.

5. Practice Problems

Solve the following problems in your student learning guide. Then flip the card to check your answer.

[qdeck bold_text=”false” style=”width: 600px !important; min-height: 400px !important;” qrecord_id=”sciencemusicvideosMeister1961-Dihybrid Crosses Practice Problems (v2.0)”]

[h] Dihybrid Cross Problems (Problems 1, 2, and 3 were the sample problems above).

[i] Biohaiku

Dihybrid Crosses

Independent Assortment

Genes flowing through time

[q] Problem 4: In peas, the allele for round seeds (R) dominates over wrinkled seeds (r). Yellow seed color (Y) dominates over green (y). What is the genotype of an organism that is homozygous wrinkled and homozygous yellow? What would be the possible genotypes of its gametes?

[a] The genotype of the homozygous wrinkled and homozygous yellow parent is rrYY. It can only produce one type of gamete: rY.

[q] Problem 5: In peas, the allele for round seeds (R) dominates over wrinkled seeds (r). Yellow seed color (Y) dominates over green (y). Cross a homozygous round, heterozygous yellow plant with one that is heterozygous round and homozygous green. What are the phenotypes of the offspring?

[a] Step 1: Genotypes of the parents: RRYy x Rryy

Step 2: Genotypes of the parents’ gametes:  RRYy can create gametes RY and Ry. Rryy can create gametes Ry and ry.

Steps 3 and 4: Punnett square

STEP 5: Results

1 round and yellow: 1 round and green

[q] Problem 6: In watermelon, green (G) is dominant over striped (g). Short (S) is dominant to long (s).What is the genotype of an organism that is heterozygous green and heterozygous short?

[a] Heterozygous green and heterozygous short is genotype GgSs

[q] Problem 7: In watermelon, green (G) is dominant over striped (g). Short (S) is dominant over long (s).Create a Punnett square showing a dihybrid cross, and list the frequency of each phenotype.

[a] STEP 1 : Parental genotypes are GgSs x GgSs

STEP 2: Both parents will produce gametes GS, Gs, gS, and gs.

STEPS 3 and 4: Punnett Square

STEP 5: 9 Green short, 3 green long, 3 striped short, 1 striped long

[q] Problem 8: In humans, the gene for normal skin color (A) is dominant to the gene for albino skin (a). The gene for achondroplasic dwarfism (D) is dominant to the gene for normal height (d).Both genes are autosomal. A heterozygous normal skin and heterozygous achondroplasic man has children with a heterozygous normal skin woman of normal height. What are the resulting phenotypes?

[a] STEP 1: Parental Genotypes: AaDd (father) x Aadd (mother)

STEP 2: Genotypes of gametes: AD, Ad, aD, ad for the father; Ad and ad for the mother.

• ♠ if the phenotype is normal skin, normal height (A_dd)
• ♣ for albino, achondroplasic (aaD_)
• ♥ for normal skin, achondroplasic (A_D_)
• ♦ for albino, normal height (aadd)

STEP 5:  3 Normal skin, achondroplasic ♥ ; 3 normal skin normal height ♠ ; 1 albino, normal height ♦, 1 albino achondroplasic ♣ .

[q] Problem 9: In humans, the gene for normal skin color (A) is dominant to the gene for albino skin (a). The gene for Achondroplasic dwarfism (D) is dominant to the gene for normal height (d).A homozygous normal-skin homozygous achondroplasic man has children with an albino normal height woman. What are the resulting phenotypes?

[a] STEP 1: Parental Genotypes: AADD x aadd

STEP 2: Genotypes of gametes: AD for parent 1 (father); ad for parent 2 (mother)

STEP 5:  100% of offspring are AaDd

STEP 6: 100% of offspring are normal skin achondroplasics.

6. What’s next?

Continue to Topic 5.3 – 5.5 part 5: Linkage and Recombination (the next tutorial in AP Bio Unit 5)