Questions 8-11: Note that this question had a misspelled genotype. Also refer to

Question

Questions 8-11: Note that this question had a misspelled genotype. Also refer to the homework question for this. You have performed the following trihybrid test cross in maize with the coloriess (c), waxy (wx) and shrunken (sh) loci. You need to figure out the map order and distance between the loci. Cc Wxwx Shsh Colored, Starchy, Plump X cc wxwx shsh (tester) colorless, waxy, shrunken You record the following progeny Trihybrid Gamete Phenotype of # of Progeny Pro c wx sh C Wx sh C wx sh Colored, Starchy Plump colorless, waxy,shrunken colorless, waxy,Plump Colored, Starchy, shrunken colorless, Starchy, Plump Colored, waxy, shrunken colorless, Starchy, shrunken Colored, waxy, Plump Total 2708 2538 626 601 116 113 6708 11) Assume that the expected frequency of double recombinants is 0.00644. What is the coefficient of coincidence? 10) Which of the follow maps is correct sh c WX 3.5 Wx c a. 0.36 b. 0.66 c. 0.001 d. 0.14 18.4 sh b. 3.5 18.4 sh wx 18.4 3.5 sh c 3.5 21.7 sh 3.5 18.4 e. Is correct

Explanation / Answer

Answer:

10). e). c--3.5---sh-----------18.4------wx

11). Expected frequency of double recombinants = 0.0064 = 0.64%

Coefficient of Coincidence (COC) = 0.14

Explanation:

Trihybrid genotype = c wx Sh / C Wx sh

1).

If single crossover occurs between c & wx

Normal combination: c wx / C Wx

After crossover: c Wx / C wx

c Wx progeny= 626+116=742

C wx progeny = 601+113= 714

Total this progeny = 1456

The recombination frequency between c & wx = (number of recombinants/Total progeny) 100

RF = (1456/6706)100 = 21.71%

2).

If single crossover occurs between wx & Sh

Normal combination: wx Sh / Wx sh

After crossover: wx sh / Wx Sh

wx sh progeny= 2+601=603

Wx Sh progeny = 4+626=630

Total this progeny = 1233

The recombination frequency between wx & Sh = (number of recombinants/Total progeny) 100

RF = (1233/6708)100 = 18.38%

3).

If single crossover occurs between c & Sh..

Normal combination: c Sh / C sh

After crossover: c sh / C Sh

c sh progeny= 2+116=118

C Sh progeny = 4+113=117

Total this progeny = 235

The recombination frequency between c & Sh = (number of recombinants/Total progeny) 100

RF = (235/6708)100 = 3.50%

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

c---------3.5cM--------sh-----------18.38cM--------------wx

Expected double crossover frequency = (RF between c & sh) * (RF between sh & wx)

= 3.5%* 18.38% = 0.0064

The observed double crossover frequency = 4+2 / 6708 = 0.0009

Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency

= 0.0009 / 0.0064

= 0.14

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