Which layer of Earth’s mantle is the hottest?

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

Debora [2.8K]

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.

3 0

2 years ago

Can someone text someone for me

Rus_ich [418]

yeah whatsup. what u need?

6 0

3 years ago

Read 2 more answers

Will award 40 points if correct!

professor190 [17]

Answer:

Hello, The DNA fingerprinting technique and development and comparison of physical fingerprints can be the methods used for the purpose of solving crimes and identification of missing person. The DNA fingerprinting can be done from the physical evidences such as blood, sweat, hair, tissues, teeth and other material obtain at crime scene. These can be compared with the suspects, victim and missing persons so as to establish their identity. The physical fingerprints can be developed on the scene of crime and compared with the requested specimen from suspect, victim and the missing person or in case of impersonation to establish the desired identity.

<em>Hope This Helps!</em>

<h2 />

5 0

2 years ago

Read 2 more answers

Need help quickly plz

musickatia [10]

Explanation:

I'm not sure but look online for the answers

7 0

2 years ago

Consider the Color locus. At this locus there are two alleles - red and green. The red allele is dominant to the green allele. I

Anuta_ua [19.1K]

Answer:

0.7

Explanation:

Using Hardy-Weinberg equation of genetic variation being constant when disturbing factors such has mutation and others are removed.

p² + pq + q² = 1 and p + q = 1

where p² is the frequency of the homozygous dominant genotype (RR) and q² is the frequency of the homozygous recessive genotype (rr) and 2pq is the frequency of heterozygous genotype (Rr). p represent the frequency of "R" and q represent "g". since the coefficient against the green/green homozygote is 0.30 then

the fitness of the green/green homozygote = 1 - 0.3 = 0.7

7 0

3 years ago