Mastering Biology Chp. 12 HW

PART E The column headings show the probability range for your χ2 number. What is the probability range that your data fit the expected 1:1:1:1 ratio of phenotypes?

0.50 < p < 0.70

PART D - Adjusting probabilities based on new information Individual IV-3 is born as a male affected with condition B but not condition A. His parents are bred again, and an ultrasound shows that the fetus is another male. The relevant portion of the pedigree is shown below. Use this new information to determine the parents' genotypes (indicated by red arrows). Then calculate the probabilities that the second male offspring will have each condition.

The new information allows you to be certain about two requirements that previously could only be assigned probabilities: 1. You now know that individual III-4 has the genotype X B X b , meaning that she received an X b chromosome from II-4. 2. You also now know that individual IV-4 is a male, meaning that he received a Y chromosome from III-5. This new information changes the probability calculations because you know that these two requirements have been met (thus, they have a probability of one -- certainty). The only requirement that is unchanged is that of III-4 passing on an X b chromosome to IV-4 -- the probability is still 1/2.

PART C - Interpreting the data The χ2 value means nothing on its own--it is used to find the probability that, assuming the hypothesis is true, the observed data set could have resulted from random fluctuations. A low probability suggests the observed data is not consistent with the hypothesis, and thus the hypothesis should be rejected. What is the hypothesis that you are testing?

The two genes are unlinked and are assorting independently, leading to a 1:1:1:1 ratio of phenotypes in the offspring.

PART B - A sex-linked gene for eye color in Drosophila The inheritance of eye color in Drosophila is controlled by genes on each of the fly's four chromosome pairs. One eye-color gene is on the fly's X chromosome, so the trait is inherited in a sex-linked manner. For this sex-linked trait, the wild-type (brick red) allele is dominant over the mutant vermilion (bright red) allele. A homozygous wild-type female fly is mated with a vermilion male fly. X⁺X⁺×XⁿY Predict the eye colors of F₁ and F₂ generations. (Assume that the F₁ flies are allowed to interbreed to produce the F₂ generation.) Drag the correct label to the appropriate location in the table. Labels can be used once, more than once, or not at all.

Both F₁ females and F₁ males will have the wild-type eye color, but because the trait is sex-linked, F₁ females will be heterozygous for the trait. As a result, F₂ males have a 1/2 chance of inheriting each allele, and thus of having that eye color. F₂ females, on the other hand, will all have the wild-type eye color because they inherit the dominant allele on the X chromosome from their fathers (the F₁males).

PART C - Experimental prediction: Comparing autosomal and sex-linked inheritance You now know that inheritance of eye color in fruit flies is sex-linked: The gene encoding eye color is located on the X chromosome, and there is no corresponding gene on the Y chromosome. How would the inheritance pattern differ if the gene for eye color were instead located on an autosome (a non-sex chromosome)? Recall that for autosomes, both chromosomes of a homologous pair carry the same genes in the same locations. Suppose that a geneticist crossed a large number of white-eyed females with red-eyed males. Consider two separate cases: -Case 1: Eye color exhibits sex-linked inheritance. -Case 2: Eye color exhibits autosomal (non-sex-linked) inheritance. (Note: In this case, assume that the red-eyed males are homozygous.)

CASE 1: Eye color exhibits sex-linked inheritance 1. If there were 100 female offspring, 100 would have red eyes and 0 would have white eyes. 2. If there were 100 male offspring, 0 would have red eyes and 100 would have white eyes. CASE 2: Eye color exhibits autosomal (non-sex-linked) inheritance 3. If there were 100 female offspring, 100 would have red eyes and 0 would have white eyes. 4. If there were 100 male offspring, 100 would have red eyes and 0 would have white eyes. [The Punnett squares below show the crosses and predicted results for Case 1 (sex-linked inheritance) and Case 2 (autosomal inheritance). Notice that the differences between these two cases occur only in the male offspring.]

PART B - Determining genotypes in pedigrees of X-linked conditions The pedigree from Part A is shown below. Fill in the most likely genotypes of the indicated individuals in the pedigree. Note that a dominant allele followed by an underscore (_) indicates that either the dominant or the recessive allele may be present at the second position.

Use these rules when assigning genotypes for X-linked recessive conditions: -Males have only one allele for every X-linked gene. Affected males have the recessive allele, and non-affected males have the dominant (wild-type) allele. -Unaffected females with affected sons are heterozygous (carriers). -Unaffected females with affected fathers are heterozygous. -All sons of affected females will be affected. -All daughters of affected females will be heterozygous. Use these rules when assigning genotypes for autosomal recessive conditions: -Unaffected parents with an affected offspring are both heterozygous. -Unaffected offspring of one affected parent will be heterozygous. -Unaffected offspring of two heterozygous parents may be homozygous dominant or heterozygous.

PART B - Experimental results: The F₂ generation In one of Morgan's experiments, he crossed his newly discovered white-eyed male with a red-eyed female. (Note that all of the females at that time were homozygous for red eyes because the allele for white eyes had not yet propagated through Morgan's flies.) All of the F1 flies produced by this cross (both males and females) had red eyes. Next, Morgan crossed the red-eyed F₁ males with the red-eyed F₁ females to produce an F₂ generation. The Punnett square below shows Morgan's cross of the F₁ males with the F₁ females.

When a homozygous red-eyed female was crossed with the white-eyed male (w⁺w⁺×wY), the resulting F₁ females were w⁺w and the F₁ males were w⁺Y. Crossing the F₁ males and F₁ females would yield these results: -All the F₂ females would have red eyes, although some would be homozygous (w⁺w⁺) and others would be heterozygous (w+w). -Half the F₂ males would have red eyes (w⁺Y), and half would have white eyes (wY).

PART D A standard cut-off point biologists use is a probability of 0.05 (5%). If the probability corresponding to the χ2 value is 0.05 or less, the differences between observed and expected values are considered statistically significant and the hypothesis should be rejected. If the probability is above 0.05, the results are not statistically significant; the observed data is consistent with the hypothesis. To find the probability, locate your χ2 value (2.14) in the χ2 distribution table below. The "degrees of freedom" (df) of your data set is the number of categories (here, 4 phenotypes) minus 1, so df = 3. Between which values on the df = 3 line does your calculated χ2 value lie?

between 1.42 and 2.37

PART A - Calculating the expected number of each phenotype In cosmos plants, purple stem (A) is dominant to green stem (a), and short petals (B) is dominant to long petals (b). In a simulated cross, AABB plants were crossed with aabb plants to generate F1 dihybrids (AaBb), which were then test crossed (AaBb X aabb). 900 offspring plants were scored for stem color and flower petal length. The hypothesis that the two genes are unlinked predicts the offspring phenotypic ratio will be 1:1:1:1.

900/4=225= 1:1:1:1 ratio

Imagine a human disorder that is inherited as a dominant, X-linked trait. How would the frequency of this disorder vary between males and females?

Females would display this disorder with greater frequency than males. [Men have only one X chromosome, whereas women have two. Thus, women have twice the chance of inheriting the dominant allele that causes this disorder. Remember that the disorder is caused by a dominant allele, so an individual who inherits a single dominant allele will display the disorder. Read about the inheritance of X-linked genes.]

PART F Based on whether there are non-significant differences (p > 0.05) or significant differences (p ≤ 0.05) between the observed and expected values, you can determine if the data are consistent with the hypothesis that the two genes are unlinked and assorting independently. Do your results support the hypothesis that the stem color and petal length genes are unlinked and assorting independently, or do the observed values differ from the expected values enough to reject this hypothesis?

The hypothesis is supported. Because 0.50 < p < 0.70, the differences between the observed and expected values are not statistically significant.

Which of these descriptions of the behavior of chromosomes during meiosis explains Mendel's law of segregation?

The two alleles for each gene separate as homologous chromosomes move apart during anaphase I. [The chromosomal basis of Mendel's law of segregation state that the two alleles for each gene separate during gamete formation.]

PART A - Experimental technique: Reciprocal crosses When Gregor Mendel conducted his genetic experiments with pea plants, he observed that a trait's inheritance pattern was the same regardless of whether the trait was inherited from the maternal or paternal parent. Mendel made these observations by carrying out reciprocal crosses: For example, he first crossed a female plant homozygous for yellow seeds with a male plant homozygous for green seeds, and then crossed a female plant homozygous for green seeds with a male plant homozygous for yellow seeds. Unlike Mendel, however, Morgan obtained very different results when he carried out reciprocal crosses involving eye color in his fruit flies. The diagram below shows Morgan's reciprocal cross: He first crossed a homozygous red-eyed female with a white-eyed male, and then crossed a homozygous white-eyed female with a red-eyed male.

When Morgan crossed a homozygous red-eyed female with a white-eyed male, all of the offspring had red eyes. However, in the reciprocal cross (homozygous white-eyed female with a red-eyed male), all of the females had red eyes while all of the males had white eyes.

PART A - Identifying X-linked inheritance in pedigrees You are investigating the inheritance of two rare conditions (A and B) in an extended family of thoroughbred racehorses. You have constructed the following pedigree for these conditions.

CONDITION A - autosomal recessive CONDITION B - X- linked recessive [Begin a pedigree analysis by determining the inheritance mode of the condition. -Is the condition recessive or dominant? Dominant conditions require that an affected individual have at least one affected parent. An affected offspring of two unaffected parents indicates a recessive condition. (This is the case for both conditions A and B.) -Is the condition autosomal or X-linked? An affected female offspring of two unaffected parents indicates an autosomal recessive condition. (This is the case for condition A.) If the condition were X-linked, the male parent would be affected. To distinguish between rare X-linked and autosomal recessive conditions, choose the simpler mode for the observed pattern—that is, the mode that requires fewer unrelated individuals to carry a rare allele. In this example, the autosomal recessive mode would require two unrelated horses (II-1 and II-5) to carry a rare recessive allele for condition B, whereas the X-linked recessive mode requires only one. Therefore, the X-linked recessive mode is a better choice for condition B.]

PART C - The inheritance of both a sex-linked trait and an autosomal trait in humans Red-green color blindness is due to an X-linked recessive allele in humans. A widow's peak (a hairline that comes to a peak in the middle of the forehead) is due to an autosomal dominant allele. Consider the following family history: -A man with a widow's peak and normal color vision marries a color-blind woman with a straight hairline. -The man's father had a straight hairline, as did both of the woman's parents. Use the family history to make predictions about the couple's children. Drag the correct label to the appropriate location in the table. Not all labels will be used.

For the hairline gene, the man's genotype is Ww and the woman's is ww. For the color vision gene, the man's genotype is Xⁿ Y and the woman's is Xⁿ Xⁿ. Because the genes are on different chromosomes, they assort independently. Also, because one gene is sex-linked, it exhibits a different inheritance pattern in males and females. You should use the multiplication rule to calculate the chances of two events (e.g., widow's peak and colorblindness) occurring together in a specific combination, paying attention to whether the offspring is male or female.

PART A - The inheritance of a skin condition in humans Consider the following family history: -Bob has a genetic condition that affects his skin. -Bob's wife, Eleanor, has normal skin. No one in Eleanor's family has ever had the skin condition. -Bob and Eleanor have a large family. Of their eleven children, all six of their sons have normal skin, but all five of their daughters have the same skin condition as Bob. Based on Bob and Eleanor's family history, what inheritance pattern does the skin condition most likely follow?

X-linked dominant [If the skin condition is caused by an X-linked dominant allele, a father would pass the allele on to all of his daughters, who would all have the skin condition. In contrast, the father would not pass the allele on to any of his sons because the sons would receive the father's Y chromosome, not his X chromosome. As a result, none of the sons would inherit the skin condition.]

PART C - Calculating probabilities in pedigrees of X-linked conditions The pedigree from Part B is shown below. Female III-4 is pregnant via male III-5. The owner of this breeding pair wants to know the probabilities of several possible outcomes for their offspring (IV-3).

A good method for calculating probabilities in pedigrees is to consider the requirements for a certain outcome. For question 1, the following are required for IV-3 to have condition A (aa): -II-5 has the genotype Aa (probability = 2/3). -If II-5 has the genotype Aa (accounted for by the above probability), then he passes an a allele to III-4 (probability = 1/2). -If III-4 has the genotype Aa (accounted for by the above probability), then she passes an a allele to IV-3 (probability = 1/2). -II-7 has the genotype Aa (probability = 2/3). -If II-7 has the genotype Aa (accounted for by the above probability), then he passes an a allele to III-5 (probability = 1/2). -If III-5 has the genotype Aa (accounted for by the above probability), then he passes an a allele to IV-3 (probability = 1/2). All of these requirements are needed in sequence, so you apply the product rule (2/3 x 1/2 x 1/2 x 2/3 x 1/2 x 1/2 = 1/36). Similarly, for question 2, the following are required for IV-3 to have condition B: -II-4 passes an X b chromosome to III-4 (probability = 1/2). -If III-4 has the genotype X B X b (accounted for by the above probability), then she passes an X b chromosome to IV-3 (probability = 1/2). -III-5 passes a Y chromosome to IV-3 (probability = 1/2). All of these requirements are needed in sequence, so you apply the product rule here, too (1/2 x 1/2 x 1/2 = 1/8). Once the individual probabilities are known, the sum and/or product rules can be used for various combinations (both conditions, either condition, etc.).