Genetics Chapter 3

The probability that a daughter will inherit a particular allele from her mother is 50%; the probability that a son will inherit a particular allele from his mother is 100%

The probability that a daughter will inherit a particular allele from her mother is 100%; the probability that a son will inherit a particular allele from his mother is 50%

The probability that a daughter will inherit a particular allele from her mother is 50%; the probability that a son will inherit a particular allele from his mother is 50%

None of the above

Only the AABb genotype can produce Ab gametes, and the expected probability of Ab gametes produced will be 1/2. Because the parent has an equal probability (50%) of having either genotype (AaBB or AABb), the final probability of the parent producing an Ab gamete is (1/2) × (1/2) = 1/4 = 0.25 = 25%.

Only the AABb genotype can produce Ab gametes, and the expected probability of Ab gametes produced will be 1/4. Because the parent has an equal probability (50%) of having either genotype (AaBB or AABb), the final probability of the parent producing an Ab gamete is (1/4) × (1/4) = 1/2 = 0.50 = 50%.

Neither of the above

All of the above

For genotype AaBB, the probability of generating an AB gamete is 1/2. For genotype AABb, the probability of generating an AB gamete is 1/2. Again, because both genotypes have equal probabilities, the final probability of the parent generating an AB gamete is (1/2) × (1/2) + (1/2) × (1/2) = 2/4 = 0.5 = 50%.

For genotype AaBB, the probability of generating an AB gamete is 1/8. For genotype AABb, the probability of generating an AB gamete is 1/8. Again, because both genotypes have equal probabilities, the final probability of the parent generating an AB gamete is (1/8) × (1/8) + (1/8) × (1/8) = 2/32 = 0.031 = 3%.

Neither of them

All of the above

Cross 1 results in a 3:1 ratio of red-winged to clear-winged progeny, therefore the parents are most likely both Rr. Crosses 2 and 3 result in only red-winged progeny, therefore the parents are most likely all RR.

Cross 1 results in a 2:1 ratio of red-winged to clear-winged progeny, therefore the parents are most likely both Rr. Crosses 2 and 3 result in only red-winged progeny, therefore the parents are most likely all rr.

Neither

All of the above

Crosses 1 and 3 have a insufficient number of progeny, but the high number of progeny from cross 3 precludes making any conclusions.

Crosses 1 and 3 have a sufficient number of progeny, but the low number of progeny from cross 2 precludes making any conclusions.

Neither

All of the above

First, calculate the observed and expected numbers of progeny in the two categories. A 13:3 ratio is equivalent to an 0.8125 probability of having the larger class and a 0.1875 probability of being in the less frequent class. The total individuals are 200, so to calculate the expected number, multiply 0.8125 by 200 and 0.1875 by 200 to yield 162.5 and 37.5. Using the formula for the chi-square goodness-of-fit test, you should obtain a value of 0.402. By checking that value with a single degree of freedom on the chi-square table, you will see that the value is well below the significance threshold. (For your information, the probability is 0.526.) Therefore, we cannot conclude that there is a significant difference between the 13:3 ratio and the 159:41 ratio.

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Plugging the numbers into the chi-square equation: (63 – 60)2/60 + (17 – 20)2 /20 = 0.15 + 0.45 = 0.6 (the chi-square statistic). Degrees of freedom = the number of classes (phenotypes) – 1 = 2 – 1 = 1. Looking at the chisquare table for one degree of freedom and using the standard critical value of 5%, we do not reject the hypothesis and thus conclude that the differences between observed and expected progeny numbers are a result of chance. We can further conclude that both purple parents are heterozygous at the single gene locus controlling flower color (at least for purple and white).

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Sutton documented the fact that each homologous pair of chromosomes consists of one maternal chromosome and one paternal chromosome. Showing that these pairs segregate independently into gametes in meiosis, he concluded that this process is the biological basis for Mendel’s principles of heredity.

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Like flipping a fair coin, each birth is independent, so in both cases the probability is ½.

1/2 x 1/2 = 1/4

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Alleles

Loci

Genotypes

Chromosomes

Genomes

Penetrant

Homozygous

Polymorphic

Heterozygous

Mutually exclusive

Individuals with identical genotypes can exhibit different degrees of given phenotypes based on their sex, stress levels, lifestyle, nutritional status, and so forth. An individual’s genotype defines genetic potential; the expression of genetic potential can vary greatly depending on other factors that determine the overall context of gene expression. For example, men and women with the same genotype for pattern baldness exhibit very different phenotypes due to the biological context (i.e., male or female) in which the genotype exists in. The high relative level of testosterone produced in men provides a biological context whereby the genotype for baldness can be more fully expressed. The variation in the degree of pattern baldness in men and women with identical “baldness” genotypes is an example of partial expressivity. Note also that stress, nutritional deficiency, and so forth, can also result in loss of hair even in individuals not having a “baldness” genotype.

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