Pigmented skin get married and have children

  1) Can two people with pigmented skin produce albino children? Yes, it is possible for two people with pigmented skin to have albino children. This can occur if both individuals are carriers of the recessive albinism allele (a), even though they themselves have pigmented skin. In this case, the genotype of both parents would be Aa, where A represents the dominant pigmented skin allele and a represents the recessive albinism allele. When these individuals have children, there is a 25% chance that each child will inherit two copies of the recessive albinism allele (aa), resulting in albinism. This is because each parent can pass on either the dominant pigmented skin allele (A) or the recessive albinism allele (a) to their offspring. The possible genotypes of their children would be: 25% chance of having pigmented skin (AA) 50% chance of being carriers of albinism (Aa) 25% chance of having albinism (aa) It is important to note that the occurrence of albinism depends on both parents being carriers of the recessive albinism allele. If one or both parents do not carry the albinism allele, they cannot have albino children. 2) Probability of not having mid-digital hair when a person heterozygous for mid-digital hair gets married to someone who does not have mid-digital hair When a person is heterozygous for mid-digital hair (Mm) and marries someone who does not have mid-digital hair (mm), there is a 50% chance that their children will not have mid-digital hair. This is because the presence of mid-digital hair is determined by the dominant allele (M). When a person is heterozygous (Mm), they have one copy of the dominant allele and one copy of the recessive allele. The recessive allele (m) does not determine the presence of mid-digital hair. The possible genotypes of their children would be: 50% chance of having mid-digital hair (Mm) 50% chance of not having mid-digital hair (mm) Therefore, there is a 50% probability that their children will not have mid-digital hair. 3) Chance of their kids having and not having Widow’s peaks when a person homozygous for Widow’s peak mates with someone who does not have a Widow’s peak If a person homozygous for a Widow’s peak (WW) mates with someone who does not have a Widow peak (ww), all their children have a Widow’s peak. This because the presence of a Widow’s peak is determined by the dominant allele (W). When a person is homozygous for Widow’s peak (WW), they have two copies of the dominant allele. The recessive allele (w) does not determine the presence of a Widow’s peak. Since all the offspring will inherit one copy of the dominant allele from the parent with Widow’s peak (WW) and one copy of the dominant allele from the parent without a Widow’s peak (ww), their genotypes will be Ww. This means that all their children will have a Widow’s peak. Therefore, there is a 100% chance that their kids will have Widow’s peaks and no chance that their kids will not have Widow’s peaks. 4) Number of kids with free earlobes in a family with parents heterozygous for free/attached earlobe trait If both parents are heterozygous for the free/attached earlobe trait (Ee), there a 75% chance each will have free earlobes (Ee or EEThis is because free earlobes are determined by the dominant allele (E). When a person is heterozygous (Ee), they have one copy of the dominant allele and one copy of the recessive allele. The presence of the dominant allele determines free earlobes, while the recessive allele (e) determines attached earlobes. The possible genotypes of their children would be: 25% chance of having attached earlobes (ee) 75% chance of having free earlobes (Ee or EE) Therefore, if a family has 16 kids and both parents are heterozygous for the free/attached earlobe trait, approximately 12 out of the 16 kids would be expected to have free earlobes. 5) Punnett Square showing how tongue-rolling can occur in your son and his mother but not in your daughter and you R r R RR Rr r Rr rr In this Punnett Square, “R” represents the dominant tongue-rolling allele, while “r” represents the recessive non-tongue-rolling allele. Based on the given information, your son and his mother can roll their tongues, so they must both be heterozygous for tongue-rolling (Rr). On the other hand, you and your daughter cannot roll your tongues, so you must both be homozygous for non-tongue-rolling (rr). When your son and his mother have offspring: There is a 25% chance that each child will inherit two copies of the dominant tongue-rolling allele and be able to roll their tongues (RR). There is a 50% chance that each child will inherit one copy of the dominant tongue-rolling allele and one copy of the recessive non-tongue-rolling allele, making them able to roll their tongues as well (Rr). There is a 25% chance that each child will inherit two copies of the recessive non-tongue-rolling allele and not be able to roll their tongues (rr). Therefore, this Punnett Square shows how it is possible for your son and his mother to roll their tongues while you and your daughter cannot.    

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