Question

In Drosophila curved-wing pattern (c) is recessive to straight wings (c): spineless bristles (s) are recessive to normal spiny bristles (s); and yellow body (y) is recessive to grey body (y). A trihybrid Drosophila fly with the genotype cvev ss y y is crossed with a triple homozygous recessive fly with the genotype cc ss yy cessyy X ce ss y The cross produces the following progeny: Number Phenotype straight, spineless, yellovw curved, spineless, grey curved, spineless, yellow curved, spiny, yellow straight, spiny, grey straight, spineless, grey straight, spiny, yellow curved, spiny, grey Gamete genotype 37 c sy 2. 93 1501 4. 112 1421 6. 7. 29 3198

PLEASE SHOW WORK

Answer 1

Gamate Genotype Ct S C S csY csty C+ Phenotype Straight, spineless, yellow Curved, spineless, grey Curved, spineless, yellow Curved, spiny, yellow Straight, spiny, grey Straight, spineless, grey Straight, spiny, yellow Curved, spiny, grey Number Type of gamate 37 Single cross over between sand y 1 Double cross over 93 Single cross over between cand s 1501 Parental 112 Single cross over between c and s st y+ c+sy+ 1420 Parental 4 Double cross over 29 Single cross over between sand y Total 3198

The genotype is filled as above according to the information given in the problem.

The number of progenies does not follow the normal 1:1:1:1:1:1:1:1 ratio of phenotypes. So this indicates that linkage is occuring at the three loci.

The parental genotype will have the maximum number of progenies. Because they will have least recombinations, thus higher frequency. So c s+ y and c+ s y+ are the parental genotype.

The double cross over genotype will have the minimum number of progenies. Because they will have most recombinations, thus lowest frequency. So c+ s+ y and c s y+ are the double cross over genotype.

The single cross over phenotypes between 2 genes will result in phenotype that is different from the parental phenotype for that particular 2 genes. For example, cross over between c and s results in phenotypes having c s and c+ s+, that is c s y, c s y+, c+ s+ y and c+ s+ y+. In similar ways the remaining is cross overs are found as following.

Parental genotype cand s cross over Number csY Cs y+ ctsty sand y cross over cs+ y+ Number 29 112 93 37 cand y cross over Cs y+ cs+ y+ ct sy Number 93 st y+ csY ct S 29 37 4 c+ sy+ 4 112 Total 210 271 71

Distance between c and s = Number of progenies with cross over between c and s*100 / Total number of progenies

= (93+1+4+112)*100/3198 = 6.566 cM

Distance between c and s = 6.566 cM

Distance between s and y = 271*100/3198 = 8.474 cM

Distance between s and y = 8.474 cM

Distance between c and y = 71*100/3198 = 2.22 cM

Distance between c and y = 2.22 cM

From the distance calculated, we can interpret that the c and y are located closer to each other as they have lowest distance between them. Similarly, s and y are located farthest to each other as they have highest distance between them and c and s has intermediate distance. So c and y are located next to each other and s and y are separated by c. Given these facts, we can say that the genes are located in the below order.

6.566 2.22 8.474