You are trying to determine the inheritance of coat color in dwarf hamsters. Coa
ID: 182184 • Letter: Y
Question
You are trying to determine the inheritance of coat color in dwarf hamsters. Coat color e 19. may be black, brown, gray or white. You have information from a number of families where only parental and offspring phenotypes are known, not genotypes. If you were to dross the female parent from family #6 with the male parent from family #3, what phenotypic distribution would you expect in the offspring? Explain your answer. Offspring blackx brown black x brown x brown blackx brown x white gray gray gray all brown ½black ½gray %brown '14 black all gray all brown ½gray ½white ½ gray 'a black 14 white ½black ½brown x white x gray x brown 20. A female Drosophila that is heterozygous for the recessive, sex-linked traits of hairy wing (h), yellow body () and white eyes (w) is mated to a male that is homozygous recessive for all three traits. The phenotypic distribution of 1000 male testcross progeny is shown below: wild type hairy yellow white 74 70 h y w 50 yellow white hairy yellow white hairy white yellow +y w h y + 368 388 h + w Show how you would determine the linkage relationships among the three genes. It is not necessary to have a final, numeric answer in map units, but be sure to put all numbers in the appropriate places (e.g., show how you would calculate RF value). Bonus: why do we only count male progeny?Explanation / Answer
Answer:
20).
Answer:
20).
Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.
Hence, the parental (non-recombinant) genotypes is +y+ / h+w.
1).
If single crossover occurs between h & +.
Normal combination: h+ / +y
After crossover: hy/++
hy progeny= 70+2= 72
++ progeny = 74+4 = 78
Total this progeny = 150
The recombination frequency between h&+ = (number of recombinants/Total progeny) 100
RF = (150/1000)100 = 15%
2).
If single crossover occurs between + & w.
Normal combination: +w/ y+
After crossover: ++/yw
++ progeny= 74+44=118
yw progeny = 70+50=120
Total this progeny = 238
The recombination frequency between +&w = (number of recombinants/Total progeny) 100
RF = (238/1000)100 = 23.8%
3).
If single crossover occurs between h & w.
Normal combination: hw / ++
After crossover: h+/+w
h+ progeny= 44+2=46
+w progeny = 50+4=54
Total this progeny = 100
The recombination frequency between h&w = (number of recombinants/Total progeny) 100
RF = (100/1000)100 = 10%
Recombination frequency (%) = Distance between the genes (map units, mu)
y-------------15mu-----------h------10mu-------w
RF between y&h = 15%
RF between h&w = 10%
RF between y&w = 23.8%
Males have only one X chromosomes and their genotype is pure with compares to phenotype. Females will have two X-chromosomes and their phenotype may not pure when genotype is heterozygous. Hence, the male progeny would be counted to calculate the distance between the genes.
19). Male and female parents were not indicated in the question. If gender indicates, it would be possible to answer the question.
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