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I have no idea what to do for this question. Please explain this question to me

ID: 203076 • Letter: I

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I have no idea what to do for this question. Please explain this question to me and give me the correct answer. A lot of the other people on Chegg got the answer wrong so please make sure it is right.

2/28/2018 11:55 PM 88.3/100 Gradebook 2/28/2018 06:31 PM Periadic Table #Attempts Score Print Calculator Question 6 of 9 Incorrect Mapoob Suppose a geneticist is using a three-point testcross to map three linked Drosophila recessive mutations called s, w, and b, where s is associated with abnormally slow movement, w is associated with wavy movement pattern; and b is associated with broad wings. The geneticist first crosses homozygous broad flies to homozygous slow, wavy flies. Next, he testcrosses the F1 progeny to slow, wavy, broad parents and obtains the following results. Based on this data, select the order of the three genes 95 F1 Testcross Progeny Phenotype broad slow, wavy wavy slow, broad slow wavy, broad slow, wavy, broad wild type Number 607 625 215 237 203 209 bsw s b w w b s w s b b ws s w b Incorrect At least one of the selected gene orders is incorrect. Focus on identifying the middle gene by comparing the parental and DCO classes. Also consider if directionality can be identified without map data to anchor two of the three genes Previous Give Up & View Solution # Try Again Next Exit

Explanation / Answer

Answer: w b s

Explanation:

Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.

Hence, the parental (non-recombinant) genotypes is SWb/swB

1).

If single crossover occurs between S&W

Normal combination: SW/sw

After crossover: Sw/sW

Sw progeny= 215+209= 424

sW progeny = 237+203=440

Total this progeny =864

Total progeny = 1500

The recombination frequency between S&W = (number of recombinants/Total progeny) 100

RF = (632/2113)100 = 41

2).

If single crossover occurs between W&b.

Normal combination: Wb / wB

After crossover: WB/wb

WB progeny= 203+8=211

wb progeny = 209+9=218

Total this progeny = 429

The recombination frequency between W & b = (number of recombinants/Total progeny) 100

RF = (429/2113)100 = 20%

3).

If single crossover occurs between S&b

Normal combination: Sb/sB

After crossover: SB/sb

SB progeny= 215+8=223

sb progeny = 237+9=246

Total this progeny = 469

The recombination frequency between S & b = (number of recombinants/Total progeny) 100

RF = (469/2113)100 = 22%

Recombination frequency (%) = Distance between the genes (cM)

s--------22cM-----b-----20cM-------w