pierceessentials3e

WORKED PROBLEMS

Problem 1

In corn, purple kernels are dominant over yellow kernels, and full kernels are dominant over shrunken kernels. A corn plant with purple and full kernels is crossed with a plant with yellow and shrunken kernels and the following progeny are obtained:

purple, full	112
purple, shrunken	103
yellow, full	91
yellow, shrunken	94

What are the most likely genotypes of the parents and the progeny? Test your genetic hypothesis with a chi-square test.

Solution Strategy

What information is required in your answer to the problem?

The genotypes of parents and progeny. A chi-square test comparing the observed and expected results.

What information is provided to solve the problem?

Purple kernels are dominant over yellow kernels and full kernels are dominant over shrunken kernels.
The phenotypes of the parents.
The phenotypes and numbers of the different types of progeny of the cross.

For help with this problem, review:

Sections 3.3 and 3.4.

Solution Steps

The best way to begin this problem is to break the cross down into simple crosses for a single characteristic (seed color or seed shape):

Hint: A good strategy in a cross involving multiple characteristics is to analyze the results for each characteristic separately.

P	purple × yellow	full × shrunken
F₁	112 + 103 = 215 purple	112 + 91 = 203 full
	91 + 94 = 185 yellow	103 + 94 = 197 shrunken

In this cross, purple × yellow produces approximately ½ purple and ½ yellow (a 1:1 ratio). A 1:1 ratio is usually the result of a cross between a heterozygote and a homozygote. Because purple is dominant, the purple parent must be heterozygous (Pp) and the yellow parent must be homozygous (pp). The purple progeny produced by this cross will be heterozygous (Pp), and the yellow progeny must be homozygous (pp).

Now let’s examine the other character. Full × shrunken produces ½ full and ½ shrunken, or a 1:1 ratio, so these progeny phenotypes are also produced by a cross between a heterozygote (Ff) and a homozygote (ff); the full-kernel progeny will be heterozygous (Ff) and the shrunken-kernel progeny will be homozygous (ff).

Now combine the two crosses and use the multiplication rule to obtain the overall genotypes and the proportions of each genotype:

Recall: The multiplication rule states that the probability of two or more independent events occurring together is calculated by multiplying their independent probabilities.

P	purple, full	×	yellow, shrunken
	Pp Ff		pp ff
F₁ Pp Ff = ½ purple × ½ full = ¼ purple, full
Pp ff = ½ purple × ½ shrunken = ¼ purple, shrunken
Pp Ff = ½ yellow × ½ full = ¼ yellow, full
Pp ff = ½ yellow × ½ shrunken = ¼ yellow, shrunken

Our genetic explanation predicts that from this cross, we should see ¼ purple, full-kernel progeny; ¼ purple, shrunken-kernel progeny; ¼ yellow, full-kernel progeny; and ¼ yellow, shrunken-kernel progeny. A total of 400 progeny were produced; so ¼ × 400 = 100 of each phenotype are expected. Therefore, the observed numbers do not fit the expected numbers exactly.

Could the difference between what we observe and what we expected be due to chance? If the probability is high that chance alone is responsible for the difference between observed and expected values, we will assume that the progeny have been produced in the 1:1:1:1 ratio predicted for the cross. If the probability that the difference between observed and expected values is due to chance is low, we will assume that the progeny are not really in the predicted ratio, and that some other, significant factor must be responsible for the deviation from our expectations.

The observed and expected numbers:

Phenotype	Observed	Expected
purple, full	112	¼ × 400 = 100
purple, shrunken	103	¼ × 400 = 100
yellow, full	91	¼ × 400 = 100
yellow, shrunken	94	¼ × 400 = 100

To determine the probability that the difference between the observed and expected numbers is due to chance, we calculate a chi-square value using the formula

Hint: See Figure 3.12 for help on how to carry out a chi-square test.

Now that we have the chi-square value, we must determine the probability that this chi-square value is due to chance. To obtain this probability, we first calculate the degrees of freedom, which for a chi-square goodness-of-fit test are n − 1, where n equals the number of expected phenotypic classes. In this case, there are four expected phenotypic classes, so the degrees of freedom equal 4 − 1 = 3. We must now look up the chi-square value in a chi-square table (see Table 3.4). We select the row corresponding to 3 degrees of freedom and look along this row to find our calculated chi-square value. The calculated chi-square value of 2.7 lies between 2.366 (a probability of 0.5) and 6.251 (a probability of 0.1). The probability (P) associated with the calculated chi-square value is therefore 0.5 < P < 0.1. This is the probability that the difference between what we observed and what we expected is due to chance, which in this case is relatively high, and so chance is probably responsible for the deviation. We can conclude that the progeny do appear in the 1:1:1:1 ratio predicted by our genetic explanation.

Problem 2

Joanna has “short fingers” (brachydactyly). She has two older brothers who are identical twins; both have short fingers. Joanna’s two younger sisters have normal fingers. Joanna’s mother has normal fingers, and her father has short fingers. Joanna’s paternal grandmother (her father’s mother) has short fingers; her paternal grandfather (her father’s father), who is now deceased, had normal fingers. Both of Joanna’s maternal grandparents (her mother’s parents) have normal fingers. Joanna marries Tom, who has normal fingers; they adopt a son named Bill, who has normal fingers. Bill’s biological parents both have normal fingers. After adopting Bill, Joanna and Tom produce two children: an older daughter with short fingers and a younger son with normal fingers.

Using standard symbols and labels, draw a pedigree illustrating the inheritance of short fingers in Joanna’s family.
What is the most likely mode of inheritance for short fingers in this family?
If Joanna and Tom have another biological child, what is the probability (based on your answer to part b) that this child will have short fingers?

Solution Strategy

What information is required in your answer to the problem?

A pedigree to represent the family, drawn with correct symbols and labeling.
The most likely mode of inheritance for short fingers.
The probability that Joanna and Tom’s next child will have short fingers.

What information is provided to solve the problem?

The phenotypes of Joanna and Tom and their family members.

For help with this problem, review:

The information on pedigrees in Section 3.5.

Solution Steps

Hint: See Figure 3.13 for a review of symbols used in a pedigree.

In the pedigree for the family, use filled circles (females) and filled squares (males) to represent persons with the trait of interest (short fingers). Connect Joanna’s identical twin brothers to the line above by drawing diagonal lines that have a horizontal line between them. Enclose Bill, the adopted child of Joanna and Tom, in brackets; connect him to his biological parents by drawing a diagonal line and to his adopted parents by a dashed line.

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Hint: See Analysis of Pedigrees for a review of characteristics of dominant and recessive traits in pedigrees.

The most likely mode of inheritance for short fingers in this family is dominant. The trait appears equally in males and females and does not skip generations. When one parent has the trait, it appears in approximately half of that parent’s sons and daughters, although the number of children in the families is small.
If having short fingers is a dominant trait, Tom must be homozygous (bb) because he has normal fingers. Joanna must be heterozygous (Bb) because she and Tom have produced both short-fingered and normal-fingered offspring. In a cross between a heterozygote and a homozygote, half the progeny are expected to be heterozygous and half homozygous (Bb × bb→ ½ Bb, ½ bb); so the probability that Joanna’s and Tom’s next biological child will have short fingers is ½.