e You are trying to determine the inheritance of coat color in dwarf hamsters. C

Question

e You are trying to determine the inheritance of coat color in dwarf hamsters. Coat color 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 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 ¼ black 14 white ½black ½brown x gray x white x gray x brown A female Drosophila that is heterozygous for the recessive, sex-linked traits of hairy wing 20. (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 +y w h y + 368 388 h + w yellow Show how you would determine the linkage relationships among the three genes. 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? It is not

Explanation / Answer

Hi,
The coat colour has 4 possible alleles or combination of alleles. the genes are not known nor are the inheritance pattern. From the given data, we can make following assumptions.
1. The brown color appears to be dominant over all other colours.
2. Grey and black share similar features with grey manifesting only when 2 alleles of same type come together.

When the female from family 6,i.e gray color mates wth male from family 3,i.e brown, we can expect that brown being dominant must form atleast 50% ofprogenies. From faimly 8 we can see that when gray and brown mate,black and brown would appear in equal proportions. So answer is 1/2 black and 1/2 brown.

20.
here, we need to find the gene in middle to determine the gene order. There is a simple way to do that. There are 3 genes in question, hairy wing(h), yellow body (y) and white eyes (w). in any cross, we get the parental types in highest numebr. But here the wild type and complete mutant are in 74 and 70. This type of cross is trans-recombination. Now, in any recombination event, chane of cross over to occur two time (double cross over) between the 3 genes is rare. So the least number ones are the DCOs,i.e 2 and 4.
Now compare the parental and DCO. the gene which has changed its allele is the gene in the middle.
+ + + 74
+ + w 4
here the gene w has changed from + to w. So the gene w is in middle. The gene order is h w y.

Let us see two parents: +++ and hwy.
when the genes cross from in the first place between the H and W, we ger +wy and h++. these correspond to number 44 and 50. recombination frequency RF = number of recombinant progeny / total progeny
= 44+50+2+4 / 1000 = 0.1
So map distance between the H and w is = RF x 100 = 0.1 x 100 = 10% = 10 cM.

Similarly the RF between W and Y = 368 +388+2+6/1000 = 0.76
so the distance = 0.76 x 100 = 76% = 76 cM

H -10--W---------------76-----------------Y

We usually cosider male to avoid sex linked genes.

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