Humans vary in many ways from one another. Among many minor phenotypic differences are the following five independently assorting traits that (sort of) have a dominant and a recessive phenotype: (1) forearm hair (alleles F and f )—the presence of hair on the forearm is dominant to the absence of hair on the forearm; (2) earlobe form (alleles E and e)—unattached earlobes are dominant to attached earlobes; (3) widow's peak (alleles W and w)—a distinct 'V' shape to the hairline at the top of the forehead is dominant to a straight hairline; (4) hitchhiker's thumb (alleles H and h)—the ability to bend the thumb back beyond vertical is dominant and the inability to do so is recessive; and (5) freckling (alleles D and d)—the appearance of freckles is dominant to the absence of freckles. In reality, the genetics of these traits are more complicated than single gene variation, but assume for the purposes of this problem that the patterns in families match those of other single-gene variants.
If a couple with the genotypes Ff Ee Ww Hh Dd and Ff Ee Ww Hh Dd have children, what is the chance the children will inherit the following characteristics?

the same phenotype as the parents

Question

Humans vary in many ways from one another. Among many minor phenotypic differences are the following five independently assorting traits that (sort of) have a dominant and a recessive phenotype: (1) forearm hair (alleles F and f )—the presence of hair on the forearm is dominant to the absence of hair on the forearm; (2) earlobe form (alleles E and e)—unattached earlobes are dominant to attached earlobes; (3) widow's peak (alleles W and w)—a distinct 'V' shape to the hairline at the top of the forehead is dominant to a straight hairline; (4) hitchhiker's thumb (alleles H and h)—the ability to bend the thumb back beyond vertical is dominant and the inability to do so is recessive; and (5) freckling (alleles D and d)—the appearance of freckles is dominant to the absence of freckles. In reality, the genetics of these traits are more complicated than single gene variation, but assume for the purposes of this problem that the patterns in families match those of other single-gene variants.
If a couple with the genotypes Ff Ee Ww Hh Dd and Ff Ee Ww Hh Dd have children, what is the chance the children will inherit the following characteristics?

the same phenotype as the parents

If a couple with the genotypes Ff Ee Ww Hh Dd and Ff Ee Ww Hh Dd have children, what is the chance the children will inherit the following characteristics?

the same phenotype as the parents

Accepted Answer

Hi, everyone. Welcome back. Let's look at our next problem. It says an X linked genetic condition known as Duchenne Muscular Dystrophy. D MD is characterized by progressive muscle weakness and degeneration as a result of alterations in a protein known as dystrophin which aids in the preservation of muscle cells. What is the probability that each child will inherit D MD if an affected man marries a carrier woman And our answer choices are a 25%. B 50% C 75% or D 100%. Well, we know we have an excellent condition. Um excellent conditions can be dominant or recessive. But in this case, we're told that the woman is a carrier. So by that tells us that it's excellent recessive. If it were dominant, you wouldn't have carriers because having only one copy of the gene would be enough to be affected by the disease. So we have an excellent recessive trait and our parents here are going to be an affected father. So the father who is affected, um only has a recessive trait, but he only has one X chromosome. So we'll call him X and then superscript D to represent the gene for D MD. And then of course, there's other sex chromosomes of Y than the cross here. That's the father. And let's cross with the mother and put the female here is a carrier of the disease. She has two X chromosomes um and carries the one recessive allele. So her genotype must be X, just plain X. That's the unaffected X chromosome. And then the second X is X superscript D. So we'll make a little putt square to represent the offspring. We just have the two sex chromosomes. So it's a four square penne square and across the top, we'll put our father X D Y and then down the left side, we'll put our mother X and then X D. So when we fill in the offspring possibilities we have on the top left X and then X D, bottom left, X D X D, top, right X Y and bottom, right X D Y. So let's think about what each of these means, knowing that we're looking for the probability that a child will inherit the disorder. Again, it's linked recessive. So if there are two X chromosomes, the child must inherit two copies. But in the case of a male who has only one X chromosome, one copy is an F. So this top left, we have a female with one copy of the gene. So she'll be a carrier like her mother and not affected by the diseases in the bottom left. However, this female as is X D X D has inherited one affected X from each parent. So that female would be a it. Then the top right X Y A male who inherited the unaffected X from his mother and the bottom right X D Y A male who inherited the affected X chromosome from his carrier mother. So we end up with two out of four offspring affected by D MD. But our answer choices are expressed in 1%. So that would equal 50% as the probability of a child being affected. So let's go over to our answer choice and that is choice. B 50%. See you in the next video.

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Key Concepts

Mendelian Genetics

Independent Assortment

Genotype and Phenotype