4)A female Drosophila heterozygous for black body color (b), curved wings (c) an
ID: 170731 • Letter: 4
Question
4)A female Drosophila heterozygous for black body color (b), curved wings (c) and purple eye color (p) is test crossed to a black bodied, curved winged, purple eyed male.The following data were obtained:
Phenotype Number
Black, purple
1,383
Wild-type
5,701
Purple, curved
388
Purple
60
Black
367
Curved
1,412
Black, curved
72
Black, purple, curved
5,617
Total
15,000
a)What is the order of the genes on the chromosome?(3 points)
b)Write the genotypes of the parents to show linkage phase.(2 points)
c)Draw a map showing the linkage relationships of these three loci?(6 points)
d)Calculate the coefficient of coincidence and the interference values.(3 points)
5. You have obtained the following linkage map for three gene loci:
A. What is the distance between the L and S loci? (2 points)
B. Based on the linkage map, calculated the expected number of double crossovers you would expect to observe if you scored 1000 progeny. (2 points)
C. If you observed 30 double crossovers among the 1000 progeny, calculate the interference. (2 points)
Black, purple
1,383
Wild-type
5,701
Purple, curved
388
Purple
60
Black
367
Curved
1,412
Black, curved
72
Black, purple, curved
5,617
Total
15,000
Explanation / Answer
a. The order of genes on the chromosome is b--------p---------c
b. The genotypes of the parents b p c /b p c and + + +/ Y
c. linkage map b------p------c / b-------p-------c
d. Coefficient of coincidence
Total progeny for Between black and purple: 2927
Total progeny for between purple and curved: 887
2927 / 15000 x 887 / 15000 = 0.195 x 0.059 = 0.011 or 165 per 15000
No of double recombinants = 132
132 / 150 = 0.88
Interference = 1 - coc = 1- 0.88 = 0.12
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