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Suppose the tail length of two populations of jerboas is controlled by one gene. to determine the mode of inheritance, a homozygous short-tailed female is crossed with a homozygous long-tailed male.

Respuesta :

If the problem continues like this:

Then siblings from F1 are crossed, and the F2 decendents are counted, so the question is associate each counting with a mode of inheritance. 

Mode of inheritance:
a) Autosomal short-tail dominant
b) Autosomal short-tail recessive
c) Sex-linked short-tail dominant
d) Sex linked short-tail recessive

Countings:
a. 3 short-tailed, 1 long-tailed
b. 1 short-tailed, 3 long-tailed
c. all females short-tailed, males 1 short-tailed 1 long-tailed
d. 1 short-tailed, 1 long-tailed

Explanation:

a) SS (short-tailed) x ss (long-tailed)

F1: 100% Ss -> 100% short-tailed

F2: 25% SS, 50% Ss, 25% ss -> 75% short-tailed, 25% long-tailed (a 3:1 proportion, just like result a.)


b) LL (long-tailed) x ll (short-tailed)

F1: 100% Ll -> 100% long-tailed

F2: 25% LL, 50% Ll, 25% ll -> 75% long-tailed, 25% short tailed (a 3:1 proportion, just like result b.)


c) XsXs (female short-tailed) x XlY (male long-tailed)

F1: 50% XsXl, 50% XsY -> 100% short-tailed

F2: 25% XsXs, 25% XsY, 25% XsXl, 25% XlY -> all females short-tailed, males 50% short-tailed, 50% long-tailed (result c.)


d) XsXs x XlY

F1: 50% XlXs, 50% XsY -> all females long-tailed, all males short-tailed

F2: 25% XlXs, 25% XlY, 25% XsXs, 25% XsY -> females 50% short-tailed, 50% long-tailed, males 50% short-tailed, 50% long-tailed (a 1:1 proportion overall, just like result d.)

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