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47. In a region of chromosome 4 there are three genes, j, k, 1. 110 cM The coeff

ID: 259608 • Letter: 4

Question

47. In a region of chromosome 4 there are three genes, j, k, 1. 110 cM The coefficient of coincidence in this region is 0.8, and it is known that the interference is driven by j-k recombination where recombination between j and k interferes with recombination in the region between k and 1. 20 ?? A true -breeding k is mated with a true-breeding j 1 individual. The resultant triple heterozygote is test crossed. What classes (include phenotypes and proportions) are expected in the next generation if 1000 offspring?

Explanation / Answer

Answer:

Coefficient of coincidence = Observed double crossover frequency / Expected double crossover frequency

0.8 = ODCO / 0.02

Expectec double crossover frequency = (Single crossover frequency (SCO) b/w j&k) x (Single crossover frequency (SCO) b/w k&l)

= 10% x 20% = 0.02

Observed double crossover frequency = 0.8 * 0.02 = 0.016 =

+k+ / +k+ x   j + l/ j + l------P1

+k+ / j + l-------------F1

+k+ / j + x jkl/jkl ----------------Testcross

Class of gametes

Phenotype

Frequency of reciprocal pair

Numbers

Frequency of each class

Number of progeny frequency * 100 = %

Parental

+k+ /jkl

1- all recombinant progeny

1-(0.084+0.184+0.0) = 0.712

0.356

35.6%

j+l/jkl

0.356

35.6%

SCO 1

++l/jkl

RF in region 1 = SCO1 - DCO

SCO1 = 0.1 – 0.016 = 0.084

0.042

4.2%

jk+/jkl

0.042

4.2%

SCO 2

+kl/jkl

RF in region 2 = SCO2 - DCO

SCO2 = 0.2 – 0.016 = 0.184

0.092

9.2%

j++/jkl

0.092

9.2%

DCO

+++/jkl

(RF in region 1) * (RF in region 2)

0.1 * 0.2 = 0.02

0.01

1%

jki/jkl

0.01

1%

Class of gametes

Phenotype

Frequency of reciprocal pair

Numbers

Frequency of each class

Number of progeny frequency * 100 = %

Parental

+k+ /jkl

1- all recombinant progeny

1-(0.084+0.184+0.0) = 0.712

0.356

35.6%

j+l/jkl

0.356

35.6%

SCO 1

++l/jkl

RF in region 1 = SCO1 - DCO

SCO1 = 0.1 – 0.016 = 0.084

0.042

4.2%

jk+/jkl

0.042

4.2%

SCO 2

+kl/jkl

RF in region 2 = SCO2 - DCO

SCO2 = 0.2 – 0.016 = 0.184

0.092

9.2%

j++/jkl

0.092

9.2%

DCO

+++/jkl

(RF in region 1) * (RF in region 2)

0.1 * 0.2 = 0.02

0.01

1%

jki/jkl

0.01

1%

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