Suppose the tail length of two populations of jerboas is controlled by one gene. to determine the mode of inheritance, a homozygous

Question

Suppose the tail length in two populations of jerboas is controlled by a single gene. To determine the mode of inheritance, a homozygous short-tailed female is crossed with a homozygous long-tailed male. The resulting offspring (F1 generation) are observed, and siblings from this F1 generation are then crossed with each other. The number of short-tailed and long-tailed animals in the F2 generation is counted to determine the phenotypic ratio. Match the phenotypic ratio expected in the offspring from a cross of two heterozygous individuals to the appropriate mode of inheritance for the short-tail allele.

Anonymous · Accepted Answer

a) If it's autosomal recessive mode of inheritance, the proportion is 3/4 long tail and 1/4 short tail.b) If the mode of inheritance is autosomal dominant, the proportion is 3/4 short tail and 1/4 long tail.c) if the mode of inheritance is sex linked dominant, and the man has the  x mutant chromosome, all women in F2 generation will has short tail and 1/2 of men has short tail and 1/2 of men has long tail.  So in order to explain the inheritance in sex linked clarify how is the male chromosomes xy is needed.Explanation: In the Po generation, we have a female homozygous mutant with short tail mu/mu which is crossed  with a male homozygous normal with long tail +/+ who get a son +/mu in the F1 generation. If the F1 generation is crossed with a heterozygous we can obtain the next possibilities in the case of autosomal mode of inheritance.

Earlene Connelly · Answer

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 dominantb) Autosomal short-tail recessivec) Sex-linked short-tail dominantd) Sex linked short-tail recessiveCountings:a. 3 short-tailed, 1 long-tailedb. 1 short-tailed, 3 long-tailedc. all females short-tailed, males 1 short-tailed 1 long-tailedd. 1 short-tailed, 1 long-tailedExplanation:a) SS (short-tailed) x ss (long-tailed)F1: 100% Ss -> 100% short-tailedF2: 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-tailedF2: 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-tailedF2: 25% XsXs, 25% XsY, 25% XsXl, 25% XlY -> all females short-tailed, males 50% short-tailed, 50% long-tailed (result c.)d) XsXs x XlYF1: 50% XlXs, 50% XsY -> all females long-tailed, all males short-tailedF2: 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.)

Cody Rohan

Suppose the tail length of two populations of jerboas is controlled by one gene. to determine the mode of inheritance, a homozygous

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