Question

In: Biology

In Drosophila, the mutant genes ct (cut wings), y (yellow body), and v (vermillion eye color)...

In Drosophila, the mutant genes ct (cut wings), y (yellow body), and v (vermillion eye color) are all X-linked. Females heterozygous for all three genes were mated with homozygous recessive males. The male progeny from the cross are given below. As is conventional with Drosophila, the wild-type allele of each gene is designated by a + sign.

ct         y          v                      4

ct         y          +                      93

ct         +          v                      54

ct         +          +                      349

+          y          v                      331

+          y          +                      66

+          +          v                      97

+          +          +                      6

Determine the allele arrangement in the heterozygous parents.

Determine the gene order of these three linked genes.

Construct a linkage map showing the map distances between the three genes.

Calculate interference.

a. Determine the allele arrangement in the heterozygous parents. b. Determine the gene order of these three linked genes. c. Construct a linkage map showing the map distances between the three genes. d. Calculate interference.

Solutions

Expert Solution

ct         y          v                      4

ct         y          +                      93

ct         +          v                      54

ct         +          +                      349

+          y          v                      331

+          y          +                      66

+          +          v                      97

+          +          +                      6

Total population = 1000

In any double cross, the maximum number of progeny is the progenies which have the same genotype as that of the parents and the progeny which have the least number is the progeny which are the result of a double crossover. So looking at the progenies we can easily conclude that ct         +          +     and   +          y          v    are the original parent type and ct y          v and + + + are the double recombinant.

This also gives us the idea about the gene which is present in the middle as only the middle gene is transferred to the other chromosome during the double crossover. So the middle gene is ct.

                             

The distance between ct and y.

If we want to find the distance between the ct and y then we need to find out the crossover frequency and divide it by the total number of progeny

ct         y          v                      4

ct         +          v                      54

+          y          +                      66

+          +          +                      6

Linkage distance = total number of recombinant * 100 / total number of progeny

130 * 100/ 1000

= 13

So, map distance between ct and y is 13 map units or 13cM

The distance between ct and v.

If we want to find the distance between the ct and v then we need to find out the crossover frequency and divide it by the total number of progeny

ct         y          v                      4

ct         y          +                      93

+          +          v                      97

+          +          +                      6

Linkage distance = total number of recombinant * 100 / total number of progeny

200*100/ 1000

= 20

So, map distance between ct and v is 20 map units or 20cM

Allelic arrangement is the first parent is +-ct-+

Allelic arrangement is the second parent is v-+-y

Since the flies have undergone the double recombination so we can expect the recombination interference

In order to find out the recombination interference we need to find out the coefficient of coincidence (c.o.c.)

Coefficient of coincidence is given by

the observed frequency of double recombinant/ expected frequency of double recombinant

expected frequency of double recombinant is given by

Map distance between ct and y /100 * map distance between ct and v/100 = 13/100*20/100 = 0.026

Observed frequency of double recombinant = 4+6 /1000

0.01

Coefficient of coincidence = 0.01/0.026= 0.38

Recombination interference is given by

1- C.o.C

1-0.38 =

0.62

gladiator answered 1 year ago
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