In guinea pigs (*Cavia porcellus*), black coat color (B) is dominant over white coat color… — JAMB UTME

In guinea pigs (*Cavia porcellus*), black coat color ( $B$ ) is dominant over white coat color ( $b$ ), and short hair ( $S$ ) is dominant over long hair ( $s$ ). If a heterozygous black, short-haired guinea pig ( $BbSs$ ) is mated with a white, long-haired guinea pig ( $bbss$ ) and they produce a total of 640 offspring, how many of the offspring are expected to display a black coat and long hair?

Answer: 160 offspring

Answer

160 offspring are expected to have a black coat and long hair.

In a dihybrid testcross involving a double heterozygote (

BbSs

) and a homozygous recessive individual (

bbss

), the offspring phenotypes appear in equal ratios of 1:1:1:1 (25% for each phenotypic class). The black coat, long hair phenotype (

Bbss

) corresponds to 1/4 of the total offspring. Multiplying 1/4 by 640 yields exactly 160 expected offspring.

Step-by-Step Solution

Determine the type of genetic cross and parental genotypes.

The cross is a dihybrid testcross between

BbSs

and

bbss

One parent is heterozygous for both independently assorting traits (

BbSs

), and the other parent is homozygous recessive (

bbss

Determine the proportion of offspring expected to have the phenotype black coat and long hair (

Bbss

The proportion of

Bbss

offspring is

\frac{1}{4}

(or

25\%

The

BbSs

parent produces four types of gametes (

BS

Bs

bS

bs

) in equal proportions (

\frac{1}{4}

each). Combining

Bs

with

bs

yields

Bbss

Calculate the expected count out of 640 total offspring.

\frac{1}{4} \times 640 = 160

Multiplying the expected phenotypic fraction by the total offspring count yields the absolute expected count.

Key Concept

Dihybrid testcross ratio and probability calculation

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic