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