3. (105 pts) Three Point Mapping Problem recessive to normal spiny bristles (s);

Question

3. (105 pts) Three Point Mapping Problem recessive to normal spiny bristles (s); and yellow body (y) is recessive to grey body (y). A recessive fly with the genotype cc ss yy ccssyy X cc ss yy The cross produces the following progeny: In Drosophila curved-wing pattern (c) is recessive to straight wings (c); spineless bristles (s) are rosophila fly with the genotype cvev ss y y is crossed with a triple homozygous Number Phenotype straight, spineless, yellow curved, spineless, grey curved, spineless, yellow curved, spiny, yellovw straight, spiny, grey straight, spineless, grey straight, spiny, yellow curved, spiny, grey Gamete genotype 37 1. c'sy 3. 4. 5. 6. 7. 8. 93 1501 112 1421 4 29 3198

Explanation / Answer

(a) 2. c s y+ 3. c s y 4. c s+ y 5. c+ s+ y+ 6. c+ s y+ 7. c+ s+ y 8. c s+ y+

(b) The frequency of parental combination is far more than recombinants, indicating that the genes are situated closer to each other; are linked.

(c) The combination with highest frequency indicate the parental types. class 6 ( c+ s y+ ) and class 4 (c s+ y)

(d) The combination with lowest frequency indicate the double cross over progenies. class 2 (c s y+) and class 7 (c+ s+ y).

(e) The combinations with single cross overs are: class 1 ( c+ s y), class 3 ( c s y ), class 5 (c+ s+ y+) and clss 8 (c s+ y+).

(f) Linkage phase between c and s is of coupling type, s and y is of repulsion type, whereas between c and y is of repulsion type. ( c s y+/ c+ s+ y).

(g) It is evident from the double cross over observation that gene for spiny/spineless bristles is present in the middle. In the parental ( c s y+/ c+ s+ y) arrangement, double cross will result in ( c s+ y+ and c+ s y) individuals, only which corresponds to double cross over. so, the order of genes will be -c -s-y.

(h) crossover between c and s (classes 3,5,2,7)

no. of resultant recombinant progenies= ( 93+ 112+ 1+4)= 210.

so, distance between c and s is, = 210/3198= 0.06 cM

crossover between s and y (classes 1,8,2,7)

no. of resultant recombinant progenies= (37+29+1+4)= 71.

so, distance between s and y is= 71/3198= 0.02 cM

(i) recombination distance between c and y is (0.06+0.02)= 0.08cM

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