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:

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?

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):

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:

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:

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

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.

Solution Strategy

What information is required in your answer to the problem?

What information is provided to solve the problem?

For help with this problem, review:

The information on pedigrees in Section 3.5.

Solution Steps