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

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. a short tailed jerboa. a long tailed, long eared jerboa. then, siblings from the f1 are crossed, and the number of short‑ and long‑tailed animals are counted to determine the phenotype ratio. match the phenotypic ratio anticipated in the offspring from a cross of two heterozygous individuals to the appropriate mode of inheritance for the short‑tail allele.

Answers

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.


Suppose the tail length of two populations of jerboas is controlled by one gene. to determine the mo

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.


Suppose the tail length of two populations of jerboas is controlled by one gene. to determine the mo

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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