Red eyes, an ebony body and a striped wing are produced by recessive genes \'r\'
ID: 67486 • Letter: R
Question
Red eyes, an ebony body and a striped wing are produced by recessive genes 'r', 'e' and 's' on the third chromosome of Drosophilia. When the homozygous red eyed, ebony body females were mated with homozygous striped wing males and the resulting tihybrid F1 females were test crossed to produce the F2 generation , it was noticed that 2765 were red eyed with ebony bodies, 2920 had only striped wings, 456 were red eyed with striped wings, 378 had the ebony body only, 227 were red eyed only, 215 had striped wings and an ebony body, 21 had all three characteristics and 18 were wild type. Using the data presented above determine how the genes are arranged on the parent chromosome, the distance between the genes and the level of interference.
Explanation / Answer
2765 were red eyed with ebony bodies (r, e, S),
2920 had only striped wings (R, E, s),
456 were red eyed with striped wings (r, E, s),
378 had the ebony body only (R, e, S),
227 were red eyed only (r, E, S),
215 had striped wings and an ebony body (,
21 had all three characteristics (r, e, s)
18 were wild type (R, E, S).
To find gene order compare Parental offsprings with double recombinanta
Parental offsprings are always in large numbers whereas double recombinants are in low numbers. Double recombinants are formed when the middle gene in the parental genotype shuffles. Looking carefully, at the phenotypes we can find that striped gene must be in middle.
So, gene order is R - S - E
Genetic distance between R and S is calculated by its recombination frequency which results from single cross over.
Single cross over between R and S will produce progency of phenotypes: red eyed only; striped wings, ebony body
Recombination frequency (%) between R and S = (number of recombinants / total progeny) x 100
= (227 + 215 / 7000) x 100
= 6.3 %
Single cross over between R and S will produce progency of phenotypes: red eyed with striped wings; ebony body only.
Recombination frequency (%) between S and E = (456 + 378 / 7000) x 100
= 12.0 %
Recombination frequency between double recombinants = (18 + 21 / 7000) x 100
= 0.56 %
Actual recombination frequency between R and S = 6.3 + 0.56 = 6.86 = 7.0 % = 7.0 mU (genetic distance)
Actual recombination frequency between S and E = 12 + 0.56 = 12.56 = 12.6 % = 12.6 mU (genetic distance)
Total distance between R and E = 7.0 + 12.6 = 19.6 mU.
To measure Interference (I), we first calculate the coefficient of coincidence (c.o.c.) which is the ratio of observed to expected double recombinants. Interference is then calculated as 1 - c.o.c.
I = 1 - c.o.c
= 1 - [observed double recombinants / expected double recombinants]
= 1 - [(18 + 21) / (0.07 x 0.126 x 7000)]
= 1 - [39 / 61.7]
= 1 - 0.63
= 0.37 (which means that 37 % interference is occuring in the area)
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.