A man heterozygous for both free earlobes and ability to taste PTC marries a woman with attached earlobes who is heterozygous for the ability to taste PTC. E = free earlobes, e = attached earlobe, P = PTC taster, p = non PTC taster.
What are the genotypes of the parents? Mother? Father?
What are the possible gametes for each parent?
Draw a Punnett square to determine the genotype and phenotype of their children.
Answer and Explanation:
The mother has attached earlobes, this is a recessive trait which means it would have to be present together giving us an "ee" genoytype. The sentence also indicates she is heterozygous for ability to test PTC, so "Pp". The father is indicated as being heterozygous for both "Ee" and "Pp". Since there are two separate traits, a dihybrid cross would be the easiest and most effective way of determining the possible genotypic outcomes. Our cross will be the mothers genotype which is eePp and the father's which is EePp. The dominant "P" is underlined and in bold because it is difficult to distinguish between it and lower case "p"
Possible genotypes: 25% EePp, 25% eePp, 12.5% chance for genotypes EePP, Eepp, eePP, eepp.
Genotypes EePp and EePP would express the dominant free earlobes and PTC tasting ability. So there would be a 37.5% chance the offspring would have free earlobes and able to taste PTC
Genotypes eePP and eePp would have attached earlobes and able to taste PTC, this would give offspring a 37.5% chance of having this phenotype.
Genotype eepp would have attached earlobes and unable to taste PTC. Which would give offspring a 12.5% of expressing these traits.
Genotype Eepp would have free earlobes and unable to tasted PTC. The offspring would have a 12.5% chance of inheriting expressing this.
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from Biology 102: Basic GeneticsChapter 6 / Lesson 3