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In Drosophila, the two genes w and sn are X-linked and 25 map units apart. A female fly of genotype w+ sn+/w sn is crossed to a wild-type male. What proportion of the male progeny will be w+ sn?

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marianaegarciaperedo

Answer and Explanation:

Available data:

  • Two genes w and sn are X-linked and 25 map units apart.
  • A female fly of genotype w+ sn+/w sn is crossed to a wild-type male w+ sn+ Y.

Cross:

Parental)      w+ sn+/w sn        x         w+ sn+ / Y

Gametes) w+sn+ (parental)                    

                 w+sn (recombinant)              

                 w sn+ (recombinant)

                 w sn (parental)

Punnet square)

                    w+sn+               w+sn                wsn+                 wsn

w+sn+   w+sn+/w+sn+  w+sn+/w+sn   w+sn+/wsn+    w+sn+/wsn

    Y            w+sn+/Y            w+sn/Y            wsn+/Y             wsn/Y

F1) 50% of the progeny will be males.

     50% of the progeny will be females.

There are two possible answers, that depends on how we might analyze the situation:

  • If we consider the whole progeny (males + females), the proportion of male individuals w+sn/Y will be 1/8=0.125=12.5%

This is:

8 ------- 100% of male and female individuals

1 ------- X = 12.5% of w+sn/Y individuals

From the 8 genotype possibilities, there is only one possible w+ sn/Y genotype.

  • But if we consider only the male progeny, the 50% of the total progeny that belongs to males, represents 100% of male individuals. In this case, the proportion of male individuals w+sn/Y will be 1/4=0.25=25%

This is:

4 ------- 100% of males

1 ------- X = 25% of w+sn/Y males

From the 4 male genotype possibilities, there is only one possible w+ sn/Y genotype.                              

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