In domestic fowl (*Gallus gallus*), rose comb (R) is dominant over single comb (r), and fe… — JAMB UTME

In domestic fowl (*Gallus gallus*), rose comb ( $R$ ) is dominant over single comb ( $r$ ), and feathered legs ( $F$ ) are dominant over clean legs ( $f$ ). A rooster heterozygous for both traits ( $RrFf$ ) is test-crossed with a hen possessing a single comb and clean legs ( $rrff$ ). If a total of $400$ chicks are hatched from this cross, how many are expected to exhibit a rose comb and clean legs?

A
$75$
$100$ Answer
C
$200$
D
$225$

Answer

The expected number of offspring displaying rose comb and clean legs is

100

A cross between a double heterozygote (

RrFf

) and a double recessive organism (

rrff

) is a dihybrid test cross. According to Mendel's Law of Independent Assortment, the heterozygous parent forms four types of gametes (

RF

Rf

rF

rf

) in equal

1:1:1:1

frequencies, while the recessive parent yields only

rf

gametes. This produces four phenotypic classes in equal proportions (

\frac{1}{4}

each). Out of

400

total offspring, the expected count for rose comb and clean legs (

Rrff

) is

\frac{1}{4} \times 400 = 100

Step-by-Step Solution

Determine the parental genotypes and gamete types.

The heterozygous rooster (

RrFf

) produces four gamete types in equal proportions:

RF

Rf

rF

, and

rf

. The homozygous recessive hen (

rrff

) produces only one gamete type:

rf

Mendel's Law of Independent Assortment states that alleles of different genes segregate independently into gametes during meiosis.

Derive the offspring genotypes and phenotypic proportions.

Combining gametes yields four distinct offspring genotypes:

RrFf

(rose comb, feathered legs),

Rrff

(rose comb, clean legs),

rrFf

(single comb, feathered legs), and

rrff

(single comb, clean legs) in a

1:1:1:1

ratio.

A dihybrid test cross always generates a 1:1:1:1 phenotypic frequency among offspring.

Calculate the expected count for the target phenotype.

Target phenotype proportion (

Rrff

) =

\frac{1}{4} = 0.25

. Expected count =

0.25 \times 400 = 100

Multiplying the phenotypic probability by the total sample size gives the expected frequency.

Key Concept

Dihybrid Test Cross Ratio

Estimated Time:2m 0s

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