Solutions For An Introduction to Genetic Analysis Chapter 2 Problem 44P

2. The genotypes of John and Martha can be determined by the pedigree at least partially. The genotype of John and Martha can be G/-.

The probability of John carrying the allele for galactosemia is 2/3. His parents would be heterozygous because they were normal and have produced a child with recessive disorder. That is John's brother was homozygous (g/g). So, John's both parents were of genotype G/g.

Punnett square

The cross is Gg Gg

F₁ generation g/g John's brother

G/- John (G/g or G/G)

As John is not having Galactosemia, he can be G/g or G/G, which happens at the ratio 1:2. So, the probability of John carrying the g allele is 2/3.

The probability of Martha carrying the g allele is 1/4.

This can be explained as follows. Martha's great-grandmother has galactosemia and Martha's grandparent had got that allele. As the grandparent's phenotype was normal, the genotype is G/g.

The probability of Martha's grandparent passing the allele to Martha's parent is 1/2. The probability of Martha's parent passing the allele to Martha is 1/2. So, the probability that Martha's parent getting the allele and passed it to Martha is

1/2 1/2 = 1/4

The probability of John having G/G = 1/3 and G/g =2/3

The probability of Martha having G/G = 3/4 and G/g = 1/4

All these information does not fit easily into a Punnett square.

3. If the information is put into a branch diagram, it does not easily fit and may complicate the problem.

4. In the pedigree of this problem, the mating between John's mother and father illustrates Mendel's first law of segregation.

5. The scientific terms in the problem are autosomal, recessive, and galactosemia. The word autosomal refers to genes that are autosomes.

The word recessive means that if an allele should be expressed in the phenotype the recessive form only should be present in the organism.

The word galactosemia refers to a metabolic disorder. This disorder is characterized by the absence of the enzyme galactose-1-phosphate uridyl transferase and results in the accumulation of galactose. Other symptoms of this disorder are hepatomegaly, jaundice, vomiting, anorexia, and lethargy. If galactose free diet is not taken, this will lead to early death.

6. From this problem, we can assume that if the person is not affected in the phenotype or if nothing is expressed, the person is of normal phenotype. The other assumption is that a person who marries into these two families should not be carriers of this allele for galactosemia.

7. Parents of John and Martha's grandparent and parent who descended from her great-grandmother who was affected should be considered even though they are not mentioned.

8. The product rule is the major statistical rule needed to solve the problem. It is the 'AND' rule. This rule is used to calculate the probabilities of the unpacked solution in part 2. An example for this 'AND' rule is 'what is the probability that Martha's parent inherited the galactosemia allele AND passed that allele onto Martha AND Martha will pass that allele on to her child?'.

9. Autosomal recessive disorder is rare. Their frequency is equal in both males and females. The recessive disorder will be expressed only if the person is homozygous for the recessive genotype.

10. People who marry an individual from the family, which is being studied, do not carry the allele responsible for the disorder. This is the general assumption inferred in this problem.

11. The phenotype of John's parents, John's brother, Martha's great-grandmother, and Martha's grandmother has certain genotypes.

As John's parents are of normal phenotype, but produced a son with that disorder, both of them are heterozygous for this disorder. That is G/g.

As John's brother had that disorder, he is homozygous for that allele. That is g/g. As Martha's great-grandmother had that disorder, she is also homozygous for that recessive allele g/g.

Martha's grandmother was of normal phenotype, but was heterozygous. She has passed the allele to her daughter and Martha received it from her mother. So her genotype would be G/g.

All the other individuals John, Martha, Martha's parents, and her children have uncertain genotype.

12. John's parents are heterozygous for this allele and the probability of John being heterozygous is 2/3. In the case of Martha, whether either of her parents were carriers or not is unknown. So when we consider Martha, her probability of being a carrier of that disorder is to be calculated in series of probabilities.

13. Martha's sister was not taken into account in any of the calculations in this problem as it was not needed. So the information about her sister and her children is irrelevant to the problem.

14. This problem is similar in many ways to the other problems solved before. It is similar in drawing the pedigree, and finding out the genotypes and probabilities.

The different thing is that many assumptions had to be made in this problem. For example, every person who marries an individual from these families is of normal genotype.

15. Now we can make up a short story. Tom and Beth got married. No diseases or disorders were expressed in both of them. When they brought forth the third child they saw that the child was affected by this autosomal recessive disorder. They were much confused and were wondering how the child got this genetic disorder when both the parents are of normal phenotype. They were very much confused until they met a geneticist. The geneticist drew the pedigree of the family after hearing about all the family history. After that they were tracking back to the family tree and found the truth that both of them were carriers of the disease.

The solution to the problem is

The probability that John and Martha’s first child have galactosemia = probability of John carrying the allele probability of Martha becoming the carrier of the allele probability of both parents passing the recessive allele to the child (whether it is first or second the probability is the same)

2/3 1/4 1/4 = 1/24