1. In Drosopbila, brittle (), cbony body () and gaucho ( are three recessive gen

Question

1. In Drosopbila, brittle (), cbony body () and gaucho ( are three recessive genes. If homozygous EBONY females are crossed with homozygous BRITTLE&GAUCHO males, the resulting F1 progeny are all wild-type. If heterozygous F1 females are mated with EBONY, BRITTLE&GAUCHO males, the following 1,000 progeny appear brittle, gaucho: 290 a) Derive a map for the three (3) genes. ebony: 280 brittle, ebony: 4 gaucho: 16 brittle: 132 ebony, gaucho: 128 wild-type: 70 brittle, gaucho, ebony: 80 "The assignment is worth 10 points. Please submit the assignment using the appropriate D2L Assignment Folder. ok Pro G H

Explanation / Answer

Let the genotype be:

Wild-type is dominant (BB or Bb, GG or Gg, and EE or Ee).

Brittle (bb), gaucho (gg), and ebony (ee) are always homozygous recessive.

The first cross is between an ebony female (eeBBGG) and a brittle gaucho male (EEbbgg). The ebony female can ONLY pass eBG to her offspring. The brittle gaucho male can only pass Ebg to his offspring. This is because they are each homozygous for each trait. You determine this because ALL of the F1 generation are wildtype (EeBbGg), and the ONLY way to get this combo is if each parent is homozygous for the opposite traits.

So the 2 genes that the offspring should each have are:

eBG from mother

Ebg from father

If there is no linkage involved, you would expect normal Mendelian ratios for each trait. Simply put, there should be 3:1 wildtype to each recessive trait. When we rearrange the data set, highest to lowest:

brittle, gaucho - 230

ebony - 220

=> These are two with the highest numbers. They would be Parental genotypes.

brittle - 120

ebony, gaucho - 110

brittle, gaucho, ebony - 92

wild-type - 88

=> These would be single-crossovers. It means they crossed over 1x in between 2 of the genes, resulting in an unusual genotype ratio.

gaucho - 12

brittle, ebony - 8

=> These are double-crossovers. They are the least frequent since the likelihood of crossing over between two genes of study is much lower than 1x (single crossovers) or 0x (parentals).

Comparing the Parentals with the F1 generation:

F1 generation resulted in heterozygous offspring

eBG from mother

Ebg from father

These are mated with a pure homozygous recessive (test cross): ebg, ebg.

Parentals:

brittle, gaucho = Eebbgg -> literally, Ebg.

ebony = eeBbGg -> literally, Ebg.

=> This is what we would expect if NO crossing over occurred. Therefore anything abnormal would mean that we have crossing over. The other results showed us that we do.

Now, looking at the double-crossovers:

gaucho = EeBbgg -> literally, EBg

brittle, ebony = eebbGg -> literally, ebG.

Comparing the parentals to the double-crossovers:

Parentals= Ebg, eBG

Double crossovers= EBg, ebG

=>Find the ONE gene that varies, you had a double-crossover around that gene. In this case, it is the B.

Shown:

E b g

x x

e B G

So the crossover took place between the E & B and between the B & G.

Now, we look at the single crossovers.

brittle: EebbGg -> literally, EbG

ebony, gaucho: eeBbgg -> literally, eBg

brittle, gaucho, ebony: bbggee -> literally, bge

wildtype: BbGgEe -> literally, BGE

Now, Add together ALL of our crossovers between E & B, 92+88+12+8=200. Divide by 1000 (total progeny) = 0.20 = 20 map units apart or 20 centimorgans apart.

Add together ALL of our crossovers between B & G. Again, 2 single crossovers + both double crossovers, 120+110+12+8=250. Divide by 1000 = 0.25 = 25 map units apart, 25 centimorgans.

So they are further apart, which is why the numbers for this group were a little lower.

Now the map will be:

E___20___B___25___G.

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