1. Gene A converts compound A (green) to compound B (blue). Compound A is lethal

Question

1. Gene A converts compound A (green) to compound B (blue). Compound A is lethal to the organism if not broken down through this process. Gene B converts compound B to compound C (red). These compounds determine the color of the individual. In a self cross of A/a B/b individuals, what proportion of green to blue offspring should be seen?

2. In a cross of AaBbCcDdEe X AaBbccDDEe, what proportion will have the AaBbCcDdEe genotype?

3.In a cross of AaBbCcDdEe X AaBbccDDEe, what proportion will have the ABCDe phenotype?

4. white (w) is an X-linked recessive and tinman (tn) is an autosomal recessive mutation. What proportion of white, tinman females (relative to whole population) is expected in the F2 starting with a true breeding white female which is wild type for tinman mating with a true breeding male mutant only for tinman.

5. Color blindness is X-linked recessive and blood type is autosomal. If two parents who are both Type A and have normal vision produce a son who is color blind and type O, what is the probability that their next child will be a son who has normal vision and is blood type A?

6. In a cross of two double heterozygotes A/a B/b, what proportion will be dominant for a, recessive for b?

Explanation / Answer

1. A function gene A means that there is no lethality while a homozygous aa means that the offspring will not survive. In case of gene B there is no lethality involved, it is only a determinant of the colour.So selfing A/a B/b individuls will give rise to the following progenies:

1) aa BB (1), aa Bb (2), aa bb (1)

Due to the absence of functional gene A these individuals which are green in colour will not survive.

2) AA BB (1), AA Bb (2), Aa BB (2), Aa Bb (4)

These individuals will be red in colour due to the presence of functional B.

3) AA bb (1), Aa bb (2)

These individuals will be blue in colour due to presence of A.

Thus, the ratio of green to blue individuals is 0:9.

2. To find such a proportion a punnett square will be a very long method. If we pull out the gametes one by one and calculate the combination then based on the total product of proportion of gametes we can calculate the final proportion of the genotype.

Parent: AaBbCcDdEe X AaBbccDDEe

Offspring: AaBbCcDdEe

Aa+Aa= Aa proportion will be ½

Bb+Bb= proportion will be Bb ½

Cc+cc= Cc proportion will be ½

Dd+ DD= Dd proportion will be ½

Ee+ Ee= Ee proportion will be ½

Product comes out to be 1/32. Thus, the proportion of AaBbCcDdEe genotype will be 1/32 in a cross between AaBbCcDdEe X AaBbccDDEe.

3. In a cross of AaBbCcDdEe X AaBbccDDEe, ABCDe phenotype is not possible. So the proportion of this phenotype to occur will be 0.

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