Any help on this question please? (14) Evidence suggests that these three recess

Question

Any help on this question please?

(14) Evidence suggests that these three recessive mutations are on the same chromosome in crickets: as, nu, and pl. You cross males from a pure-breeding tripl pure-breeding normal strain. Heterozygous Fi female progeny are then crossed to male triple mutants. Here are the phenotypes observed in the F2 progeny of this t mutant lab strain to the est cross: wild type ast nu* pl as nu pl as nu pl ast nu pl+ as nur p ast nu pl as+ nu* pl as nu p 902 897 207 209 75 73 13 16 2392 triple mutant TOTAL a) What is the name for this type of test? (2 points) b) What is the order of the genes on the chromosome? In other words, which one of these is between the other two? (3 points) a) What are the map distances between these mutations? (5 points) (1 point) d) Is there interference on this chromosome?

Explanation / Answer

(a) What is the name for this type of test?

Solution: This is a test cross since the heterozygous F1 progeny is crossed with phenotypically recessive parent, which in this case is the male triple mutant.

(b) What is the order of the genes on the chromosomes? In other words, which one of these is between the other two?

Solution: Looking at the number of individuals with a different progeny, we find two with maximum number of individuals and two with the minimum number of individuals. The pair with the maximum number of individuals are the parental type; in this case, as+ nu+ pl+ (902) and as nu pl (897). The pair with the minimum number of individuals are the double cross over type; in this case, as+ nu+ pl (13) and as nu pl+ (16). Comparing the pair of parental type and the double cross over, we find that pl changed from the parental type so, pl must be in the middle. So, the correct arangement of genes on the chromosome is as pl nu or as+ pl+ nu+.

(c) What are the map distances between these mutations?

Solution: The distance between the genes are calculated as= [(single cross over + double cross over)/ total number of offspring] * 100

Distance between as and pl:

single cross over between as and pl= 75+73 = 148

double cross over between as and pl= 13+16 = 29

Total cross overs between as and pl= 177

Therefore, distance between as and pl= 177/2392 * 100= 7.39 map units

Distance between pl and nu:

single cross over between pl and nu= 207+209 = 416

double cross over between pl and nu= 13+16 = 29

Total cross overs between pl and nu= 445

Therefore, distance between pl and nu= 445/2392 * 100= 18.6 map units

(d) Is there interference on this chromosome?

Solution: Interference is determined by calculating the Co-efficient of co-incidence.

Co-efficient of co-incidence= (no. of double recombinants found in progeny/ no. of double recombinants expected in progeny)

If the co-efficient of coi-incidence =1, there is no interference.

If it is less than 1, Interference = (1- co-efficient of co-incidence)

We already know the number of double recombinants found in the progeny (determined in the previous questions) which is (13 + 16 = 29)

Now, in order to find the "expected" number of double recombinants in progeny we need to find the probablity of crossover between as- pl and pl- nu; which is the probability of crossover between as and pl (7.39 %, as calculated in the previous question) multiplied to probablity of crossover between pl and nu (18.6%, as calculated in the previous question).

So, the probability of both these event occuring = 0.0739 * 0.186 = 0.0137

And, the expected number of double recombinants in a total progeny of 2392= 0.0137 * 2392= 32.7

Now, calculating the co-efficient of co-incidence= 29/ 32.7 = 0.886

Clearly, there is interference, since the co-efficient is not 1.

Interference= 1- 0.886 = 0.114

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