You performed matings with other flies from question 3 and found some unusual re

Question

You performed matings with other flies from question 3 and found some unusual results. The 1600 F2 progeny are the result of crossing the F1 progeny to each other.

P1: thick wings, white bristles, long antenna female x skinny wings, red bristles, aba male

F1: all thick wings, red bristles, long antenna flies

F2:       910 thick wings, red bristles, long antenna flies

            290 skinny wings, red bristles, aba flies

            110 skinny wings, white bristles, aba flies

            280 thick wings, white bristles, long antenna flies

                3 thick wings, red bristles, aba flies

                4 skinny wings, red bristles, long antenna flies

                1 thick wings, white bristles, aba flies

                2 skinny wings, white bristles, long antenna flies

A. What is the inheritance pattern of the traits?

B. Which, if any, of the genes are linked? What are the map distances between the linked genes?

Explanation / Answer

B. There is incomplete linkage has occurred.

910 thick wings (T), red bristles (R), long antenna flies (L) – T R L (wild type)

110 skinny wings (t), white bristles (r), aba flies (l) – t r l (wild type)

290 skinny wings, red bristles, aba flies – t R l (double cross over [D.C.O])

280 thick wings, white bristles, long antenna flies – T r L (double cross over [D.C.O])

4 skinny wings, red bristles, long antenna flies – t R L (single cross over [S.C.O] I)

2 skinny wings, white bristles, long antenna flies – t r L (single cross over [S.C.O] I)

3 thick wings, red bristles, aba flies – T R l (single cross over [S.C.O] II)

1 thick wings, white bristles, aba flies – T r l (single cross over [S.C.O] II)

Total progeny – [ 910 + 110 + 290 + 280 + 4 + 2 + 3 + 1] = 1600

Frequency of double cross over – [(570/1600)*100] = 35.625

Frequency of S.C.O I – [(6/1600)*100] = 0.375

Frequency of S.C.O II – [(4/1600)*100] = 0.25

Map distance between T and R is – S.C.O I + D.C.O = 35.625 + 0.375 = 36 m.u

Map distance between R and L – S.C.O II + D.C.O = 35.625 + 0.25 = 35.875 m.u

T------------------------------36 m.u--------------------------R--------------------------------35.875 m.u--------------------------L

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