# AP Biology : Understanding Punnett Squares and Test Crosses

## Example Questions

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### Example Question #1 : Understanding Punnett Squares And Test Crosses

Let us assume that flower color is either purple or white in a particular species. Color is determined by complete dominance, and purple is dominant to white.

A purple flower sprouts in the garden one day, and the gardener would like to know if it will only create purple flowers if it is pollinated. If she wants to use only one generation to determine its genotype, how should she pollinate the flower?

Self-pollinate the purple flower

Pollinate the purple flower with a homozygous purple flower

Pollinate the purple flower with any other purple flower

Pollinate the purple flower with a white flower

Pollinate the purple flower with a white flower

Explanation:

When attempting to determine if an organism is heterozygous or homozygous for a dominant trait, it is best to use a test cross. A test cross involves crossing the flower in question with a homozygous recessive flower. Since a white flower can only contribute a white allele, we can determine if the purple flower in question is heterozygous or homozygous. Any white flowers in the next generation will confirm that the purple flower is heterozygous. If they are all purple, we can confirm that the flower is homozygous for the trait.

### Example Question #2 : Understanding Punnett Squares And Test Crosses

Let us assume that purple flowers are dominant to white flowers. A pure breeding purple flower is crossed with a pure breeding white flower. The F1 generation is self pollinated.

What percentage of the F2 generation will be homozygous?

25%

Further information is needed to determine the answer

100%

50%

50%

Explanation:

The F1 generation will be entirely heterozygous, thus the Fgeneration is the result of a heterozygous self-cross. A Punnet square reveals that 75% of the generation will be purple (PP or Pp) and 25% will be white (pp). Of the three purple flowers in the punnett square, two of them are heterozygous for color (Pp). The other flower is homozygous for the purple allele (PP). In addition, the white flower is homozygous for the recessive white allele (pp).

There are two homozygous flowers, and two heterozygous flowers in the punnett square, so 50% is the correct answer.

### Example Question #3 : Understanding Punnett Squares And Test Crosses

A botanist finds a pea plant growing in his backyard and wants to know its genotype for color, knowing that the allele for green is dominant to the allele for yellow. He breeds the unknown plant with a known homozygous dominant pea plant. If the F2 generation is three-fourths green and one-fourth yellow, what was the genotype of the unknown plant?

Homozygous recessive

Heterozygous

Either homozygous dominant or heterozygous

Homozygous dominant

Impossible to tell

Homozygous recessive

Explanation:

The possible genotypes of the unknown plant are GG, Gg, or gg. To create the F1 generation, it is crossed with a plant with genotype GG.

Scenario 1: GG x GG, result is all GG in F1; F2 cannot possibly contain a yellow (gg) plant

Scenario 2: Gg x GG, result is half Gg and half GG; F2 will contain both green and yellow plants of all genotypes

Scenario 3: gg x GG, result is all Gg in F1; F2 will be 75% green and 25% yellow

In order to have the ratios described in the question, the unknown plant must be homozygous recessive.

### Example Question #1 : Mendelian And Population Genetics

In a certain species of bird, yellow beaks are dominant to orange beaks, and blue feathers are dominant to black feathers.

Two heterozygous birds are crossed. What fraction of the offspring would be expected to have yellow beaks and blue feathers?

Explanation:

This question requires us to do a dihybrid cross. We can represent the gene for beak color with the symbol "A" for dominant yellow and "a" for the recessive orange. Likewise, for feather color, we can use "B" for blue feathers and "b" for black.

The problem states that both birds are heterozygous for each trait, implying that our cross is between two birds with the genotype AaBb.

Now we look at the gametes that can be produced by these parents: AB, Ab, aB, and ab. These gametes can then be used to make a punnet square.

Offspring: 1 AABB, 3 Aabb, 8 AaBa, 3 aaBb, 1 aabb

There are 16 total offspring. 12 of them carry the dominant A allele, giving them the yellow beak phenotype.

Yellow beaks: 1 AABB, 3Aabb, 8 AaBb

Finally, of these 12, 9 carry the dominant B allele for blue feathers.

Yellow beaks and blue feathers: 1 AABB, 8 AaBb

This gives a total of 9 out of the 16 offspring that will express both the yellow beak and blue feather phenotypes.

You should be familiar with the 9:3:3:1 phenotypic ratio resulting from dihybrid crosses.

### Example Question #1 : Understanding Punnett Squares And Test Crosses

A new type of plant is shown to have two distinct traits for their seeds: seed color and seed shape. Green color is dominant to white, and a long shape is dominant to round. If two plants heterozygous for both traits were crossed, what is the probability that an offspring would show the dominant phenotype for both traits?

Explanation:

This question requires that we do a dihybrid cross. The cross in question is AaBb x AaBb, using A to represent dominant green color and B to represent dominant long shape. The parents are heterozygous for both traits, meaning they will carry one dominant color allele and one dominant shape allele.

The result of a punnet square for a dihybrid cross is: 1 AABB, 3 Aabb, 8 AaBb, 3 aaBb, 1 aabb.

This gives a total of sixteen different offspring. Two different genotypes carry dominant alleles for both traits: AABB and AaBb. There are a total of nine offspring between these two genotypes. The probability of an offspring being dominant for both traits is thus nine out of sixteen.

### Example Question #6 : Understanding Punnett Squares And Test Crosses

In a newly discovered species, albinism is recessive to natural coloring. If an albino parent were crossed with a naturally colored parent, what would be the expected phenotypic ratios of the offspring?

100% albino

50% albino and 50% natural

100% natural color, or 50% albino and 50% natural

100% natural color

100% natural color, or 50% albino and 50% natural

Explanation:

Because we are only told the phenotype of the naturally colored parent, we do not know if it is homozygous dominant or heterozygous. To account for this, we must anticipate the phenotypic ratios of both dominant genotypes. Two crosses must be performed: one between a homozygous dominant parent and a homozygous recessive parent, and one between a heterozygous parent and a homozygous recessive parent. The resulting ratios would be 100% natural and 50% albino/50% natural respectively.

AA x aa: all offspring Aa (natural)

Aa x aa: half offspring Aa (natural), half offspring aa (albino)

### Example Question #2 : Understanding Punnett Squares And Test Crosses

A new type of plant is shown to have two distinct traits for their seeds: color and shape. Green color is dominant to white, and a long shape is dominant to round. If two plants heterozygous for both traits are crossed, what would be the expected phenotypic ratios of the offspring?

9 green round: 3 green long: 3 white round: 1 white long

9 green long: 3 green round: 3 white long: 1 white round

3 green long: 1 white round

8 green long: 3 green round: 3 white long: 1 white round

9 green long: 3 green round: 3 white long: 1 white round

Explanation:

This question requires that we do a dihybrid cross. The cross in question is AaBb x AaBb, using A to represent dominant green color and B to represent dominant long shape. The parents are heterozygous for both traits, meaning they will carry one dominant color allele and one dominant shape allele.

The result of a punnet square for a dihybrid cross is: 1 AABB, 3 Aabb, 8 AaBb, 3 aaBb, 1 aabb.

This gives a total of sixteen different offspring. Two different genotypes carry dominant alleles for both traits: AABB and AaBb. There are a total of nine offspring between these two genotypes that will be green and long. The other genotypes all represent different phenotypes. Aabb will be green and round. aaBb will be white and long. aabb will be white and round. Together, this gives a final phenotypic ratio of 9:3:3:1.

### Example Question #261 : Evolution And Genetics

In a new species of beetle, black coloration is recessive to blue coloration. If a homozygous dominant blue beetle is mated to a black beetle, what are the expected phenotypic ratios?

100% black

50% blue, 50% black

75% blue, 25% black

100% blue

100% blue

Explanation:

Because we are told that the blue parent is homozygous dominant, we can set up a simple cross. We know that the black beetle must be homozygous recessive to present the black phenotype. Using B as the dominant blue allele and b as the recessive black allele, we can see that the parent beetles have genotypes of BB and bb.

BB x bb

Offspring: All offspring will be Bb and present the dominant (blue) phenotype.

All resulting offspring will have at least one dominant allele, giving us a 100% blue ratio.

### Example Question #3 : Understanding Punnett Squares And Test Crosses

A breeder wants to know her dog’s genotype. The breed she works with comes in two varieties: black and yellow. Black is dominant to yellow. She breeds her black dog with a yellow dog and gets a litter of three black dogs and three yellow dogs. What is the genotype of the parental black dog?

Homozygous dominant

Heterozygous

Homozygous recessive

Unable to determine from the given information

Heterozygous

Explanation:

We know that the yellow dog must be homozygous recessive and that the black dog must be either heterozygous or homozygous dominant. When crossed, their offspring can show either the dominant (black) phenotype or the recessive (yellow) phenotype. In order for offspring to show the recessive phenotype, they must inherit a recessive allele from each parent. To have this outcome, the black dog must carry a recessive allele even though it expresses the dominant trait; this makes the black dog heterozygous.

We can look at a punnett square to verify the result. We will use B as the dominant allele and b as the recessive allele.

Bb (black dog) x bb (yellow dog)

Offspring: Half Bb (black) and half bb (yellow)

### Example Question #2 : Understanding Punnett Squares And Test Crosses

Two individuals with the following genotypes are crossed:

AABbccDdEeFF x AaBBCCDdEeff

What is the probability that their offspring will have the genotype AaBbCcddEEFf?

Explanation:

To obtain the overall probability, multiply the individual probabilities for each locus.

Parent cross: AABbccDdEeFF x AaBBCCDdEeff

Offspring: AaBbCcddEEFf

A locus cross is AA x Aa. Half of the offspring will be AA and half will be Aa. The probability of Aa is one half.

B locus cross is Bb x BB. Half of the offspring will be BB and half will be Bb. The probability of Bb is one half.

C locus cross is cc x CC. All offspring will be Cc. The probability of Cc is one.

D locus cross is Dd x Dd. One-fourth of the offspring will be DD, half will be Dd, and one-fourth will be dd. The probability of dd is one-fourth.

E locus cross is Ee x Ee. One-fourth of the offspring will be EE, half will be Ee, and one-fourth will be ee. The probability of EE is one-fourth.

F locus cross is FF x ff. All offspring will be Ff. The probability of Ff is one.

Multiply all the probability values.

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