In fruit flies (*Drosophila melanogaster*), the allele for normal wings (V) is completely … — JAMB UTME

In fruit flies (*Drosophila melanogaster*), the allele for normal wings ( $V$ ) is completely dominant over the allele for vestigial wings ( $v$ ). A monohybrid cross is carried out between two heterozygous normal-winged flies, producing a total of $320$ offspring in the $F_1$ generation. How many of these offspring are expected to be heterozygous normal-winged?

A
$80$
$160$ Answer
C
$240$
D
$320$

Answer

The expected number of heterozygous normal-winged offspring is

160

Crossing two heterozygous individuals (

Vv \times Vv

) yields a genotypic distribution of

1/4\,VV

1/2\,Vv

, and

1/4\,vv

. Taking half of the total population (

320

) gives

160

expected heterozygous (

Vv

) offspring.

Step-by-Step Solution

Determine parental genotypes and set up the Punnett square

Both parents are heterozygous (

Vv \times Vv

). Gametes produced by each parent are

V

and

v

Mendel's First Law (Law of Segregation) states that alleles segregate during gamete formation so that each gamete carries only one allele for each gene.

Determine the expected genotypic ratio of the

F_1

generation

Genotypes produced:

1/4\,VV

(homozygous dominant),

2/4\,Vv

(heterozygous dominant),

1/4\,vv

(homozygous recessive). Ratio is

1 : 2 : 1

Combining gametes randomly yields

25\%\,VV

50\%\,Vv

, and

25\%\,vv

Calculate the expected number of heterozygous offspring

\frac{2}{4} \times 320 = 160

offspring.

Multiply the fraction representing the heterozygous genotype (

1/2

50\%

) by the total offspring count (

320

Key Concept

Mendel's Law of Segregation and Monohybrid Genotypic Ratios

Estimated Time:1m 0s

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