Biology Forum › Genetics › Genetic interaction
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- December 10, 2011 at 12:18 pm #15825
MSB2012ParticipantI’m having trouble solving this question hope someone can help me 🙂
A dominant gene V in humans causes certain areas of the skin to become depigmented, a condition called "vitiligo." Albinism is the complete lack of pigment production and is produced by the recessive genotype aa. The albino locus is epistatic to the vitiligo locus. Another gene locus, the action of which is independent of the previously mentioned loci, is known to be involved in a mildly anemic condition called "thalassemia." (a) When the adult progeny from parents both of whom exhibit vitiligo and a mild anemia is examined, the following phenotypic proportions are observed : 1/16 normal : 3/16 vitiligo : 1/8 mildly anemic : 1/12 albino : 3/8 vitiligo and mildly anemic : 1/6 albino and mildly anemic. What is the mode of genetic action of the gene for thalassemia? (b) What percentage of the viable albino offspring in part (a) would carry the gene for vitiligo? (c) What percentage of viable offspring with symptoms of mild anemia also show vitiligo?
- December 10, 2011 at 5:30 pm #108670CatParticipant
Please, post your work first.
P.S. You need to draw a trihybrid cross Punnett square to solve this problem like AaBbCc x AaBbCc…
- December 10, 2011 at 7:35 pm #108673MSB2012Participant
The main problem is that I don’t know how to write genetic constitution of parents 🙁
- December 12, 2011 at 9:05 pm #108688CatParticipant
Like I said, you need to draw a trihybrid cross Punnett square to solve this problem.
Both parents are heterozygous and here is why:
Both parents are vitiligo with mild anemia.
1. Vitiligo is dominant. So, it can be due VV or Vv genotype. If either parent is VV, then none of the progeny would be normal. That is not the case. Thus, both parents must be heterozygous Vv.
2. Albino is due to homozygous recessive aa genotype. So, parents who are not albino can be either AA or Aa. If either parent is AA, then all progeny would be normal – no albino. That is not the case. Thus, both parents must be heterozygous Aa.
3. Both parents have thalassemia. If is is due to homozygous recessive tt, then both parents are tt and all progeny must be tt. That is not the case (some progeny don’t have anemia). If it is due to dominant allele T, parental genotypes can be TT or Tt. If at least one parent is TT, then all progeny would be anemic. That is not the case. Thus the only possible case is both patents are heterozygous Tt. Please, note that this conclusion does not say that thalassemia gene is dominant, just that is not recessive. Just that both parents are heterozygous.
4. To find if thalassemia gene is dominant or incomplete dominant you need to examine the progeny. So, add all anemic = 1/8 mildly anemic + 3/8 vitiligo and mildly anemic : 1/6 albino and mildly anemic = 2/3 anemic. Now add all not anemic = 1/16 normal + 3/16 vitiligo + 1/12 albino = 1/3. So the ratio is 2 anemic: 1 normal. Normal ratio for dominant allele is 3:1 which does not fit. Incomplete dominant however is 1:2:1 which would fit if you had one more phenotype. Now consider adding a phenotype that you will not be able to see in the progeny (you are given a hint in the problem to help identifying it). In case you are not sure, it’s lethal. So, now you have 1 lethal (TT) : 2 anemic (Tt) : 1 normal (tt). Please, keep in mind that in this case even though I designated a lower case letter to the “normal” allele of thalassemia gene, this allele is NOT recessive but also incomplete dominant.
Thus you need to do a cross of VvAaTt x VvAaTt to calculate answers to the rest of the questions…
- December 13, 2011 at 12:57 pm #108700MSB2012Participant
Thank you so much your post was very helpfull
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