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 two heterozygous guinea pigs ( $BbSs$ ) are mated and produce a total of $400$ offspring, how many offspring are expected to display the black coat and long hair phenotype?

Answer: 75 offspring

Answer

75 offspring are expected to display the black coat and long hair phenotype.

In a dihybrid cross between two individuals heterozygous for both traits (

BbSs \times BbSs

), independent assortment produces an

F_2

phenotypic ratio of

9:3:3:1

. The phenotype with one dominant and one recessive trait (black coat and long hair,

B\_ss

) represents

\frac{3}{16}

of the total offspring population. Out of

400

offspring, the expected count is calculated as

\frac{3}{16} \times 400 = 75

Step-by-Step Solution

Determine the phenotypic ratio from the dihybrid cross

BbSs \times BbSs

The expected ratio is

9 : 3 : 3 : 1

for (black coat, short hair) : (black coat, long hair) : (white coat, short hair) : (white coat, long hair).

Mendel's Law of Independent Assortment dictates that the two gene pairs segregate independently during gamete formation.

Determine the proportion of offspring showing the black coat and long hair phenotype (

B\_ss

The proportion is

\frac{3}{16}

The probability of inheriting the dominant coat trait (

B\_

) is

\frac{3}{4}

, and the probability of inheriting the recessive hair length trait (

ss

) is

\frac{1}{4}

. Combined probability =

\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}

Multiply the proportion by the total population size to find the expected number of offspring.

\frac{3}{16} \times 400 = 75

Applying the theoretical phenotypic probability to the sample size of 400 yields the expected count.

Key Concept

Mendel's Law of Independent Assortment and Dihybrid Phenotypic Ratios

Estimated Time:1m 30s

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