7.4 Application: Practicing Inheritance Probabilities

1. The children would all have partially attached earlobes.
2. All the female children will have unattached earlobes, and all the male children will have attached earlobes.
3. All of their children would have unattached earlobes.
4. All of their children would have attached earlobes.

1. 50%
2. 75%
3. 0%
4. 100%

1. ee
2. X^eY
3. EE
4. Ee

1. ee
2. X^eY
3. EE
4. Ee

The pedigree in Fig. 7.5b tracks Duchenne Muscular Dystrophy (DMD) through several generations. DMD is an X-linked recessive trait.

If individuals I-1 and I-2 had another son, what is the chance that he would have DMD? Check your answer in footnote^[5]

1. 0%
2. 25%
3. 50%
4. 100%

1. Individual II-3 has a genotype of X^DY.
2. If individuals II-4 and II-5 have a fourth child, there is a 50% chance that it will not have DMD.
3. Individual II-1 is a carrier for DMD.
4. If individual III-1 marries an unaffected, non-carrier female, none of their offspring will have DMD.

1. X^dY
2. X^dX^d
3. X^DD^d
4. X^DX^D

1. 100%
2. 50%
3. 25%
4. 0%

The pedigree in Fig. 7.5c tracks the presence of dimples through a family’s generation. Having dimples is an autosomal dominant trait.

If individuals II-1 and II-2 have a fourth child, what is the probability that the child will have dimples? Check your answer in footnote^[9]

1. 25%
2. 50%
3. 75%
4. 100%

1. Dimples
2. No dimples
3. dd
4. DD

1. I-1 → Dd
2. II-3 → dd
3. II-2 → DD
4. III-2 → Dd

1. 0%
2. 25%
3. 75%
4. 100%

1. Because the trait we are tracking (attached earlobes) is autosomal recessive, shaded individuals, like III-6, will have a homozygous recessive genotype (ee).
2. If III-6 (ee) were to have a child with a man who was homozygous for unattached earlobes (EE), then all of the children would be heterozygous – getting one E from their father and one e from their mother.
   Attached earlobes is a recessive trait and will only occur in ee genotypes. Heterozygotes (Ee) will have unattached earlobes, as that is the dominant condition.
3. The correct answer is
   All of their children would have unattached earlobes.

1. Because the trait we are tracking (attached earlobes) is autosomal recessive, shaded individuals, like I-2, will have a homozygous recessive genotype (ee).
   I-1 must have a heterozygous genotype because he is able to pass on a recessive allele to some of his offspring (II-2 and II-4).
2. If I-1 and I-2 had another child, the cross would be:
   

   –	E	e
   e	Ee	ee
   e	Ee	ee

   Only offspring with ee genotypes will have attached earlobes (2/4 boxes).

   2÷4=0.5=50%

3. The correct answer is 50%

1. Individual I-1 is represented by a non-shaded square, indicating that it is a male with unattached earlobes.
2. Because the trait we are tracking, attached earlobes, is autosomal recessive, shaded individuals will have a homozygous recessive genotype (ee).
   Individuals that are non-shaded will have at least one E allele.
3. I-1 has children with attached earlobes (II-2 and II-4 are ee), meaning he must be able to pass on at least e allele. However, he shows the dominant condition, so he must also have one E allele.
   Therefore, his genotype is Ee.
4. The correct answer is Ee

1. Individual II-3 is represented by a non-shaded square, indicating that it is a male with unattached earlobes.
2. Because the trait we are tracking, attached earlobes, is autosomal recessive, shaded individuals will have a homozygous recessive genotype (ee).
   Individuals that are non-shaded will have at least one E allele.
3. II-3 has a mother with attached earlobes (ee), meaning he must get one e allele from her. However, he shows the dominant condition, so he must also have one E allele.
   Therefore, his genotype is Ee.
4. The correct answer is Ee

1. Individual I-2 is represented by a shaded circle, indicating that it is an affected female.
   Therefore, she must have a homozygous recessive genotype of X^dX^d
2. Because males always get their X chromosome from their mother, all of the sons that individual 2 has will receive a recessive X^d allele.
   
3. Males will also receive their Y chromosome from their father, giving any son of individuals I-1 and I-2 a genotype of X^dY
4. The correct answer is 100%.

1. Unaffected males, such as individual II-1 have a genotype of X^DY.
   On the other hand, affected males, such as individual II-3, have a genotype of X^dY.
   Since males only have one X chromosome, they cannot be carriers.
2. Individuals II-4 and II-5 are both shaded in, indicating that they are affected.
   In order to be affected, they must have the recessive genotypes X^dY and X^dX^d. This means that any child they have will have DMD because each parent can only pass on a recessive DMD allele.
3. Individual III-1 is an unaffected male, meaning that he has a genotype of X^DY.
   If he mates with an unaffected, non-carrier female (X^DX^D), there is no chance that the children will inherit the DMD allele.
4. The correct answer is
   If individual III-1 marries an unaffected, non-carrier female, none of their offspring will have DMD.

1. Individual II-2 is represented by a non-shaded circle, indicating that it is an unaffected female.
2. In order for individual II-2 to have a normal phenotype, but also produce an affected son, she must be a carrier for DMD. This means that she has one of each allele, X^DX^d.
3. The correct answer is X^DX^d

1. Individual I-3 is represented by a shaded square, indicating that it is an affected male. Therefore, he must have a genotype of X^dY.
   If he has a child with a DMD carrier (X^DX^d), the cross would be:

   –	XD	Xd
   Xd	XDXd	XdXd
   Y	XDY	XdY

    

2. In order for a daughter to be affected, her genotype must be X^dX^d.
   Only one box has this genotype, so the chances of having an affected daughter is: ¼ =25%
3. The correct answer is 25%

1. Because the trait we are tracking, dimples, is autosomal dominant, any shaded individuals will have at least one dominant allele (D).
   Any unshaded individuals will have the recessive genotype (dd).
2. II-2 has dimples, meaning she must have at least one D allele. In addition, she has a recessive parent, and one of her children has no dimples (dd), so she must also have at least one d allele. This makes her genotype Dd.
   Individual II-1 has no dimples, meaning that he must have a homozygous recessive genotype (dd).
3. Now that we know their genotypes, if we perform a cross between individuals II-1 and II-2 we find:
   

   –	D	d
   d	Dd	dd
   d	Dd	dd

   2÷4=50%

4. The correct answer is 50%.

1. The trait that we are tracking, dimples, appears to be dominant, as all offspring who have the trait have an affected parent.
   Having dimples also does not skip a generation, which suggests that it is likely dominant.
2. Shaded individuals have dimples, meaning that they must have at least one D allele.
   Unshaded individuals, like I-1, do not have dimples, meaning that they must have the homozygous recessive genotype dd.
3. An individual with dimples can be either DD or Dd.
   Individuals like II-2, II-3, and III-2 all have at least one recessive parent. Since the recessive parent can only donate a d, each of them must have a d in their genotype.
   Because they all have dimples, we know they must have one D allele as well, giving them all the genotype of Dd.
4. The correct answer is III-2 → Dd

1. Because the trait we are tracking (dimples) is autosomal dominant, any shaded individuals have at least one dominant allele (D).
   Any unshaded individuals have the recessive genotype (dd).
2. Individual III-3 must be heterozygous (Dd) because he has an unshaded father (dd).
   If he was to have a child with a woman who was heterozygous for dimples (Dd), then the cross would be:

   –	D	d
   D	DD	Dd
   d	Dd	dd

3. Dimples is a dominant trait and will occur whenever a dominant allele is present in the genotype.
   Homozygous recessive individuals (dd) will have no dimples, as that is the recessive condition.
   Looking at the Punnett square, we find that ¾ =75% offspring will have at least one D allele.
4. The correct answer is 75%