An additional gene, gene W, was also examined. A test cross was made between true-breeding EEWW flies and EEWW flies. The result

Question

Nadusha1986 [10]

3 years ago

6

An additional gene, gene W, was also examined. A test cross was made between true-breeding EEWW flies and EEWW flies. The result

ing F_1 generation was then crossed with EEWW flies. 100 offspring in the F_2 generation were examined, and it was discovered that the E and W genes were not linked. Which is the correct genotype of the F_2 offspring if the genes were linked and if the genes were not linked? Linked: 50% EeWw and 50% EeWw; not linked: 25% EeWw, 25%, EeWw 25% EeWw, and 25% EeWw Linked: 25% EeWw, 50% EeWw, not linked: parental genotypes EeWw and EeWw Linked genotypes (EeWw and EeWw) and recombinant genotypes (EeWW and eeWw) In the F_2 generation are nearly the same irrespective of their linkage. Linked: mostly with parental genotypes. Eeww and eeWw; unlinked: 25% EeWw and eeww with 75% Eeww and eeWw.

Biology

1 answer:

Debora [2.8K] · Answer 1 · 2022-07-29T15:41:43+03:00

This question is incorrect but here is the correct question below;

An additional gene,gene W was also examined. a test cross was made between true breeding EEWW flies and eeww flies. The resulting F₁ generation was then crossed with eeww flies. 100 offspring in the F₂ generation were examined and it was discovered that the E and W genes were not linked.

Which is the correct genotype of the F₂ offspring if the genes were linked and if the genes were not linked?

a) Linked: 50% EeWw and 50% eeww; not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

b) Linked: 25% Eeww, 50% eeWw; not linked:parental genotypes EeWw and eeww.

c) Linked genotypes (EeWw and eeww) and recombinant genotype ( Eeww & eeWw) in the F₂ generation are nearly the same irrespective of their linkage.

d) Linked: mostly with parental genotypes, Eeww and eeWw; unlinked: 25% EeWw and eeww with 75% Eeww and eeWw.

Answer:

a) Linked: 50% EeWw and 50% eeww; not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

Explanation:

a test cross was made between true breeding EEWW flies and eeww flies

If EEWW self crossed, we have the following ( EW, EW, EW, EW)

Also, for eeww, we have ( ew, ew, ew, ew)

EW EW EW EW

ew EWew EWew EWew EWew

All offspring are (EWew)

The question goes further by saying "The resulting F₁ generation was then crossed with eeww flies".

And we are asked to find the correct genotype of the F₂ offspring if the genes were linked and if the genes were not linked

∴

To determine the offsprings of the linked genes we need to go by the definition and understand what linked genes are: Linked genes are genes that are physically close together on the same chromosomes. Effect of recombinantion on linked genes, results in gene swaps which occur in chromosomes that are homologous.

Having said that; If EWew × eeww

we have; EW & ew × ew & ew

EW ew

ew EeWw eeww

offspring that

are linked in ⇒ EeWw EeWw & eeww eeww

F₂ will be

$\frac{1}{2}$ = 50% of EeWw of the total 100 offspring in the F₂ cross

$\frac{1}{2}$ = 50% of eeww of the total 100 offspring in the F₂ cross

∴ Linked genes = 50% EeWw and 50% eeww.

For unlinked genes; If EWew × eeww

if rearrangement occurs in EWew and EWew self crossed, we have ( EW,Ew,eW,ew) as the traits needed for the unlinked gene F₂ crossing.

Also ewew will be (ew, ew, ew, ew).

EW Ew eW ew

ew EeWw Eeww eeWw eeww

We have the following results for the unlinked genes

$\frac{1}{4}$ = EeWw 25% of the total 100 offspring in the F₂ cross

$\frac{1}{4}$ = Eeww 25% of the total 100 offspring in the F₂ cross

$\frac{1}{4}$ = eeWw 25% of the total 100 offspring in the F₂ cross

$\frac{1}{4}$ = eeww 25% of the total 100 offspring in the F₂ cross

∴ not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.