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.

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.)