Independent Genes with a LETHAL Allele!

Alleles that cause an organism to die only when present in homozygous condition are called lethal alleles.

lethal alleles

The following exercise will help you know how to identify a lethal allele in Genetics’ problems.

Exercise:

The cross between two hybrids of F1 of genotypes AaBb gives the following phenotypes:

36 [AB], 12[Ab], 19[aB] and 6[ab].

1- Calculate the proportions of the descendants.

2- Knowing that the genes are independent, how can you explain such a result?

3- Make the factorial analysis verifying the above phenotypic proportions.

—————————————————————————————————

Answers:

1- Proportion Calculation:

  • Divide by the smallest number:

[AB] 36/6=6

[Ab] 12/6= 2

[aB] 18/6= 3

[ab] 6/6= 1

  • Find out the total: Total = 6+2+3+1=12
  • Write the proportions:

6/12 [AB]

2/12 [Ab]

3/12 [aB]

1/12 [ab]

2- Explanation of the Results:

The genes are independent, and this is a selfing between individuals of F1, this means that the resulting theoretical proportions of the phenotypes of F2 should be 9/16[AB]; 3/16[Ab]; 3/16 [aB]; 1/16[ab], the total of the genotypes will be 16.

        Theoretical Results Real Results

9/16[AB]

3/16[Ab]

3/16 [aB]

1/16[ab]

6/12 [AB]

2/12 [Ab]

3/12 [aB]

1/12 [ab]

Comparing the Theoretical to the  Real results, we got a change (less) proportions for the first 2 phenotypes, this means that a lethal allele caused the death of 4 out of the theoretical 16 genotypes (16-12= 4) in the first 2 phenotypes: [AB] and [Ab] phenotypes.

3- Factorial Analysis:

Genotypes of F1*F1: AaBb * AaBb

Gametes: ¼ AB, ¼ Ab, ¼ aB, ¼ ab for each parent

Table of cross:

¼AB ¼Ab ¼aB ¼ab
¼AB 1/16AABB 1/16AABb 1/16AaBB 1/16AaBb
¼Ab 1/16AABb 1/16AAbb 1/16AaBb 1/16Aabb
¼ aB 1/16AaBB 1/16AaBb 1/16aaBB 1/16aaBb
¼ab 1/16AaBb 1/16Aabb 1/16aaBb 1/16aabb

-The genotypes that will enable us to find out the lethal allele are:

We look at the phenotypes that have a change in the real proportions when compared to the theoretical ones:

  • 1/16AAbb, 2/16Aabb: all have [Ab] and in the real phenotypes there are 3-2=1 missing (dead) so the missing should be AAbb because if the dead was of Aabb, we would have been missing 2 not 1!
  • 1/16AABB, 2/16AaBB, 4/16 AaBb, 2/16 AABb: all have [AB] and in the real phenotypes there are 9-6=3 missing (dead) so with comparison to the above results, the missing genotypes should carry AA (1AABB, 2AABb).

So if we consider that the lethal allele is A in the homozygous state, the given results will be verified.

Best Regards,

Yusser CHIDIAC Smile

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