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5 October, 07:51

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.

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  1. 5 October, 11:44
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    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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