Question

guinea pig coat color is determined by a single gene. the allele for black coat color is dominant to brown. in a cross between two black - haired guinea pigs, 20 offspring are born. if both parents were heterozygous, probability would predict that approximately how many of the 20 offspring would have brown hair?
a. 10
b. 15
c. 0
d. 5

Explanation:

Step1: Determine the genetic cross

Let the dominant allele for black coat be \( B \) and the recessive allele for brown coat be \( b \). Both parents are heterozygous, so their genotypes are \( Bb \). The cross between two \( Bb \) individuals is \( Bb\times Bb \).

Step2: Use Punnett square to find ratios

The Punnett square for \( Bb\times Bb \) gives the genotypes: \( BB \), \( Bb \), \( Bb \), \( bb \). The phenotypic ratio for black ( \( BB \) or \( Bb \)) to brown ( \( bb \)) is \( 3:1 \). So the probability of an offspring having brown hair ( \( bb \)) is \( \frac{1}{4} \).

Step3: Calculate the number of brown - haired offspring

We have a total of 20 offspring. To find the number of brown - haired offspring, we multiply the total number of offspring by the probability of having brown hair. So the number of brown - haired offspring \( = 20\times\frac{1}{4}=5 \).

Answer:

D. 5