wild-type female fruit fly with the genotype +/bl +/vg +/bw is crossed with a ma
ID: 280061 • Letter: W
Question
wild-type female fruit fly with the genotype +/bl +/vg +/bw is crossed with a male fruit fly with the the F2 progeny. Draw a linkage map indicating the correct distances, arrangement of genes, coefficient of coincidence and type of interference. (16 points) genotype bl/bl vg'vg bw/bw. The data shown below are the Fi female's contribution to F2 genotype 373 ??1 bl + bw bl vg+ bl vg bw + vg bw bl++ t vg + + + bw 373 89 91 2 parentals ouo 49 89 Total 1040 a I 49 359.35 040 a) Write the parentals in the space below onJ D4D b) Write the double recombinants in the space below 8 low c) Which gene is in the middle? d) Which reciprocal set of gametes resulted from a single crossover between the first and secondExplanation / Answer
As per the data provided of F2 progeny and their count, if we tabulate the data and analyze it we get the following results:
Now, the genotype which has maximum number if F2 progeny is the parental type and those with minimum number are the double recombinants
Answer a) Parental types are (i) bl vg + and (ii) + + bw
Answer b) Double recombinants are (i) + + + and (ii) bl vg bw
Answer c) Gene in the middle: The gene which is in middle can be identified by comparing the parental genotype with the double recombinants. On comparing, bw has flipped its position between parentals and double recombinants i.e. in one of the parent recessive bl vg and dominant bw are present while in second dominant bl vg and recessive bw are present. The genes bl and vg are present in same fashion in double recombinants i.e. recessive bl vg (in one of double recombinant) and dominant bl vg (in other double recombinant) but bw in in flipped form. So, bw is present in the middle. And the arrangement is as bl bw vg.
Answer d) Single crossover between first and second
Now, we know the sequence, the first and second are bl and bw. In parents, bl recessive, bw dominant (one parent) and bl dominant, bw recessive (second parent) are present. Single crossovers or single recombinants will be both bl and bw recessive as well as both dominant. Such recombinants are : bl + bw (Entry no. 2) and + vg + (Entry no. 7)
Answer e) Single crossover between second and third
From the sequence, second is bw and third is vg. In parents, vg recessive, bw dominant (one parent) and vg dominant, bw recessive (second parent) are present. Recombinants will be both vg and bw recessive as well as both dominant. Such recombinants are: + vg bw (Entry no. 5) and bl + + (Entry no. 6).
Answer f) Gene map
Distance between bl and bw gene : Sum of numbers including bl and bw recombinant genes (Both single and double recombinants)/ total number
= Numbers in (Entry 1+ Entry 2+ Entry 4 + Entry 7)/1040
= (7+49+8 +52)/1040 = 0.1115 or 11.15 cM
Distance between bw and vg gene : Sum of numbers including vg and bw recombinant genes (Both single and double recombinants)/ total number
= Numbers in (Entry 1+ Entry 4+ Entry 5 + Entry 6)/1040
= (7+8 + 89+91)/1040 = 0.1875 or 18.75 cM
now, the map can be bl___________bw___________vg with bl bw distance of 11.15 and bw vg distance 18.75.
Distance between bl and vg is 29.90 cM
Answer g) Coefficient of coincidence:
[observed double recombinants/expected double recombinants]
Expected double recombinants is given by : 0.1115 x 0.1875 x 1040 = 21.74
So, c.o.c = [15/21.74) = 69%
Answer h) Interference = 1-c.o.c
= 1-69 = 31%
F2 progeny Number Inference Entry No + + + 7 Double Recombinant/ Double crossover 1 bl + bw 49 2 bl vg + 373 Parental type 3 bl vg bw 8 Double Recombinant/ Double crossover 4 + vg bw 89 5 bl + + 91 6 + vg + 52 7 + + bw 371 Parental type 8Related Questions
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