Question

Ear lobes in people may be free hanging or completely attached to the side of the face. This is determined by a single gene locus; the free hanging allele, E, is dominant and the attached allele (e) is recessive.A person has the heterozygous genotype Ee. With respect to this gene locus, how many kinds of gametes, eggs or sperm, will this person produce? What will the percent or frequency of each kind of gamete be out of the total possible?A man and a woman who are both heterozygous for ear lobe condition have children. What is the probability that a child will have free ear lobes? What is the probability that a child will have attached ear lobes?A man with attached ear lobes and a woman with free ear lobes have three children; two have free ear lobes and one has attached ear lobes.What is the man’s genotype?What is the woman’s genotype?What are the genotypes of the three children?

Answer 1

Answer and explanation:

This person can produce two types of gametes, E and e, each one in the same proportion (50 % or 1/2)

It is expected that a monohybrid cross between two heterozygotes for this gene produces a 1 (homozygous dominant) : 2 (heterozygous) : 1 (homozygous recessive) genotype ratio. Since E allele is dominant, it is expected a phenotype ratio of 3 : 1

Answer 2

Answer:

See the answer below

Explanation:

Free hanging ear lobe's allele = E

Attached ear lobe's allele = e

E is dominant over e.

1. A heterozygous person with Ee genotype will produce E and e egg/sperm during gametogenesis.

      E gamete = 50%

       e gamete = 50%

    2. A man and a woman both heterozygous for earlobe

         Ee   x   Ee

        Offspring = EE, 2Ee, ee

         Probability of a child having free ear lobe = 3/4

         Probability of a child having attached earlobe = 1/4

    3. A man attached ear lobe with a woman with free earlobe produced     three children out of which two have free earlobes and one has attached ear lobes.

     Genotype of man with attached earlobe = ee

    Genotype of woman with free earlobes = Ee (since they produced both   free and attached earlobe children)

    ee   x   Ee

Genotype of children = Ee, ee, Ee and ee

   Ee = free earlobes

   ee = attached earlobe