AP Biology : Understanding Punnett Squares and Test Crosses

Study concepts, example questions & explanations for AP Biology

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Example Questions

Example Question #11 : Understanding Punnett Squares And Test Crosses

A breed of dog can be either of two colors. In this breed, brown is dominant to yellow. A brown dog mates with a yellow dog and produces a litter of six brown dogs. What is the genotype of the parental brown dog?

Possible Answers:

Unable to determine from the given information

Homozygous dominant

Heterozygous

Homozygous recessive

Correct answer:

Unable to determine from the given information

Explanation:

We know that the yellow dog must be homozygous recessive and that the brown dog must be either heterozygous or homozygous dominant. If B is used to represent the dominant brown allele and b is used to represent the recessive yellow allele, this means that the yellow parent must be bb and the brown parent could be either BB or Bb.

We know that all of the puppies must carry a dominant allele, since they all express the dominant phenotype. The yellow parent only carries the recessive allele. This indicates that every puppy must have inherited a dominant allele from the brown parent. The most likely genotype of the brown parent would be homozygous dominant, which would lead to all puppies being heterozygous and brown.

BB x bb

All offspring will be Bb and express the dominant brown phenotype.

Note, however, that we cannot determine the genotype of the brown parent for certain. Due to independent assortment, it is possible that a heterozygous brown parent gave the recessive allele to all offspring by chance.

Bb x bb

Half offspring will be Bb and half will be bb.

The probability of getting six brown puppies from this cross would be equal to one-half to the sixth power, or 1.56%. While this is a very small chance, it is not impossible and we cannot rule out a heterozygous genotype for the brown dog.

Example Question #12 : Understanding Punnett Squares And Test Crosses

In a species of lizard, green coloration is dominant to blue coloration. A breeder wants to produce as many blue lizards as possible from one cross. Which of the following crosses will produce the highest percentage of blue lizards?

Possible Answers:

Cc x Cc

cc x cc

Cc x cc

CC x cc

Correct answer:

cc x cc

Explanation:

The question tells us that green is dominant to blue. This means that the C allele must correspond to green and the c allele must correspond to blue. We are looking for the cross that will produce the most blue offspring. Since blue is recessive, all blue offspring will be homozygous recessive; thus, we are essentially looking for the cross that carries the most recessive alleles.

Of the given crosses, cc x cc contains the greatest number of recessive alleles. Every offspring from this cross will be homozygous recessive and display the blue phenotype. This cross thus gives the highest percentage of blue offspring, with that percentage being 100%.

Example Question #13 : Understanding Punnett Squares And Test Crosses

If a homozygous brown rat mates with a heterozygous brown rat, what percent of the offspring will have white fur?

Possible Answers:

Correct answer:

Explanation:

Heterozygous organisms carry one dominant allele and one recessive allele. The dominant allele is expressed over the recessive allele, giving the organism the dominant phenotype. If the heterozygous rat in the question is brown, then we can conclude that brown is dominant to white.

The cross of these two rats would be:

Parents: BB (brown) x Bb (brown)

Offspring: half BB (brown), half Bb (brown)

Though half of the offspring will be homozygous and half will be heterozygous, all offspring will be brown. None of the offspring from this cross will show the white phenotype.

Example Question #211 : Evolution And Genetics

If a male is heterozygous for the dimples allele and marries a female who does not possess the allele needed for dimples, what is the chance their child will have dimples? Dimples are an autosomal dominant trait.

Possible Answers:

0%

50%

100%

25%

Correct answer:

50%

Explanation:

By creating a Punnett square in which the male carries the label Dd and the female has the label dd, one can see the possible combinations of alleles that the child may have. D will symbolize the allele for dimples, and d will symbolize the allele for no dimples. The cross will be Dd x dd.

There is a 50% chance that the child does not obtain the allele needed for dimples (dd), and a 50% chance that the child is heterozygous (Dd). Because dimples is an autosomal dominant trait, heterozygosity will express dimples, leading to a 50% chance that the child will have dimples. 

Example Question #15 : Understanding Punnett Squares And Test Crosses

A new species of insect was recently discovered. Scientists have found that orange coloration is recessive to yellow coloration. If a heterozygous yellow insect is mated to an orange insect, what are the expected phenotypic percentages of the offspring?

Possible Answers:

100% yellow

100% orange

75% yellow, 25% orange

50% yellow, 50% orange

25% yellow, 75% orange

Correct answer:

50% yellow, 50% orange

Explanation:

We are told that the yellow parent is heterozygous, so we can set up a simple cross. We know that the orange insect must be homozygous recessive due to the fact that it displays the orange phenotype.

In our cross, we will use A as the dominant allele and a as the recessive allele. This gives the yellow parent a genotype of Aa and the orange parent a genotype of aa.

Aa x aa

Possible offspring: Aa, Aa, aaaa

We can see that there are only two possible genotypic results. Half of the offspring will be Aa and half will be aa. The Aa offspring will be yellow and the aa offspring will be orange, giving a 1:1 phenotypic ratio.

Example Question #16 : Understanding Punnett Squares And Test Crosses

In pea plants, tall is dominant for height, green is dominant for color and round is dominant for pea shape. 

A tall green plant is crossed with a short yellow plant. 50% of the offspring are tall and green, and 50% are tall and yellow.

What are the genotypes of the parent plants?

Possible Answers:

Tall-green: TtGg
Short-yellow: ttgg

Tall-green: TtGG
Short-yellow: ttgg

Tall-green: TTGG
Short-yellow: ttgg

Tall-green: ttgg
Short-yellow: TTGg

Tall-green: TTGg
Short-yellow: ttgg

Correct answer:

Tall-green: TTGg
Short-yellow: ttgg

Explanation:

Let's look at each trait individually. We can see that all of the offspring from this cross will be tall; the tall allele must be dominant over the recessive allele. We also know that one parent must be homozygous dominant for the tall allele, since it is passed to every single offspring. The parent genotypes for height are TT (tall) and tt (short). The short plant must be homozygous recessive to show the recessive phenotype.

Now, let's look at color. Half of the offspring are green and half are yellow. This tells us that whichever parent displays the dominant phenotype is also heterozygous, allowing the recessive phenotype to appear in the offspring. Since we know that green is dominant, we know that the green plant must be heterozygous. In order to display the recessive yellow phenotype, the yellow plant must be homozygous recessive. The parent genotypes for color must be Gg (green) and gg (yellow).

Together, we can see that the tall-green plant will be TTGg and the short-yellow plant will be ttgg.

Example Question #17 : Understanding Punnett Squares And Test Crosses

A biologist genetically crosses two cats, both with black fur, which is dominant over white fur. The pair have 12 kittens, 9 with black fur and 3 with white fur. What is the parental cross?

Possible Answers:

Bb X Bb

bb X BB

bb X bb

BB X BB

BB X Bb

Correct answer:

Bb X Bb

Explanation:

The ratio of kittens with black fur to kittens to white fur is 3:1. Since the recessive phenotype is shown in the offspring, we know that both of the parents are carriers of the recessive genotype and are thus both heterozygous, Bb.

Example Question #18 : Understanding Punnett Squares And Test Crosses

Assume that in the organism in question, the allele for brown fur (B) is dominant to the allele for white fur (b). Also assume that the allele for curly fur (C) is dominant to the allele for straight fur (c). Finally, assume these genes are independent.

Given parents with the genotypes BBCc and Bbcc, what fraction of offspring will display straight, brown fur?

Possible Answers:

Correct answer:

Explanation:

Each parent will contribute one allele at random from each of the two genes. The odds of having brown fur, given the parental genotypes, is 100% because all children will receive a dominant allele for brown fur (B) from the first parent. This means that the odds of having brown straight fur will depend only on the odds of having straight fur. Having straight fur requires one recessive allele (c) from each parent. The second parent will always contribute a recessive allele while the first will contribute a recessive allele half the time. Thus, the odds of straight fur is 50%, as are the odds of straight, brown fur. Alternatively, a Punnett square may be used. The square below shows the 50% offspring combinations with straight, brown fur highlighted in red.

Square

Example Question #19 : Understanding Punnett Squares And Test Crosses

If two parents are heterozygous for a trait and they have children, what is the percentage of the children that are heterozygous for the trait?

Possible Answers:

25%

100%

50%

75%

Correct answer:

50%

Explanation:

For simplicity, we will assign the letter "a" for the gene of interest. Thus, the heterozygous genotype is Aa. You may sketch a punnet square of the cross: Aa x Aa to help illustrate that the combinations result in 50% chance of heterozygous offspring.

Example Question #20 : Understanding Punnett Squares And Test Crosses

In a cross in which both parents are heterozygous, what would be the percentage of offspring that are homozygous recessive for the trait?

Possible Answers:

40%

25%

100%

50%

Correct answer:

25%

Explanation:

Arbitrarily, we may assign the letter "b" for the gene of interest. The cross then is as follows: Bb x Bb. Each parent has a 50% chance of donating a recessive (b) allele to the offspring. We must multiply these probabilities to get the chance of a homozygous recessive offspring. 

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