Use the following data to answer questions 45 In Drosophila

Use the following data to answer questions 4-5. In Drosophila, the genes y, f and v are all X-linked. y f v females are crossed to wild-type males and the F1 females are test- crossed. The F2 are distributed as follows: Which of the following linkage maps correctly shows the order and distance between the y, f and v genes? f_____35 mu_____y_____15 mu _____v f _____22 mu_____ y _____15 mu _____v y_____35 mu_____f_____22 mu _____v y_____22 mu_____v_____15 mu _____f. y_____15 mu_____v_____22 mu _____f What is the coefficient of coincidence in this region? 0 0.2 0.4 0.6 0.8

Solution

4. Wild type is represented by +++ and mutant types are given to be yfv. So, here we are talking about 3 genes present on the same chromosome because in the F2 generation we have drastically different proportions of parental phenotypes i.e. +++ and yfv and very less frequencies of recombinant types. The F1 females will be yfv/+++. The test cross is always performed with the recessive or mutant alleles. So, they will be crossed to a male having yfv/Y genotype.

Now, to solve any of these types of questions, first we need to identify the parental genotypes and double crossovers. We already know the parental genotypes from the above information but to be doubly sure, the parental genotypes will have highest numbers and double cross overs will have lowest number. So, from the above table, yfv and +++ are parental types. The lowest frequency is for the genotypes yf+ and ++v. So, they are obtained by double crossover.

Now consider this condition. Let us consider there are 3 genes ABC in this order and +++ is the wild type. Then, on double cross over,

ABC -> A+C

+++ +B+

So, only the middle gene is changed on double crossover. Therefore, the middle gene is v and the order is yvf. Note that yvf is same as fvy. Now, we have the order of genes. Lets calculate the recombination frequencies for this to find out the distance between genes.

distance between y and v = recombination frequency between y and v

To find out recombination frequency between y and v, we first need to know the cases that will result from such a recombination. The two double crossover cases will need crossing over between y and v. For other cases, lets again see the example of ABC and +++ for single crossover between A and B.

ABC -------> A++

+++ +BC

So, if the recombination is in between y and v, then the genotypes will be y++ and +vf having numbers 678 and 690 respectively. So,

recombination frequency between y and v = recomination/Total progeny = (678 + 690 + 72 + 60) * 100 / 10000 = 15%

Therefore, distance between y and v is 15mu.

Similarly, for recombination between v and f the recombined cases will be yv+ and ++f for which numbers are 1024 and 1044 respectively. So,

distance between v and f = (1024 + 1044 + 72 + 60) * 100 / 10000 = 22 mu.

Hence, the option E is correct.

5. Coefficient of coincidence is the ratio of actual double crossovers observed and expected double crossovers. Double crossover means that there is recombination between y and v as well as v and f. The probability of recombination between y and v is given by (678 + 690 + 72 + 60) / 10000 = 0.15 and the probability of recombination between v and f is given by (1024 + 1044 + 72 + 60) / 10000 = 0.22. So, the expected probability of double crossover will be the product of these 2 probabilities because for double crossover, one needs a crossover at both the places. So, expected probability of double crossover is 0.15 * 0.22 = 0.033. However, observed probability is (72 + 60) / 10000 = 0.0132. So,

coefficient of covariance = observed double crossovers / Expected double crossovers = 0.0132/0.033 = 0.4.

So, option C is the correct answer.