★ ap bio : unit 5 heredity mcqs

derived from the unit 5 progress check - excludes question 12 + includes anwser explaination (mostly written by me) ★ - view all ap bio figures here - https://tinyurl.com/54yt9p3a

Which of the following best explains how the sweet pea plants in the parental generation produce F1 offspring with 14 chromosomes based on Figure 5-1 and Table 5-1?

A) Meiosis II and IIII lead to the formation of cells with 14 chromosomes. When two cells combine during fertilization, extra chromosomes are randomly broken down, leading to offspring with 14 chromosomes.

B) Meiosis I and II lead to the formation of cells with 14 chromosomes. When two cells combine during fertilization, extra chromosomes with recessive traits are broken down, leading to offspring with 14 chromosomes.

C) Meiosis I and II lead to the formation of cells with 7 chromosomes. During meiosis I, homologous chromosomes separate. During meiosis II, sister chromatids separate. Two cells combine during fertilization to produce offspring with 14 chromosomes.

D) Meiosis I and II lead to the formation of cells with 7 chromosomes. During meiosis I, sister chromatids separate. During meiosis II, homologous chromosomes separate. Two cells combine during fertilization to produce offspring with 14 chromosomes.

C) Meiosis I and II lead to the formation of cells with 7 chromosomes. During meiosis I, homologous chromosomes separate. During meiosis II, sister chromatids separate. Two cells combine during fertilization to produce offspring with 14 chromosomes.

Meiosis produces haploid cells with half the number of chromosomes of the parent. When cells combine during fertilization, the diploid chromosome number is restored. Before meiosis, chromosomes replicate, forming duplicated chromosomes, each made of two sister chromatids. During meiosis I, homologous duplicated chromosomes line up in pairs, forming a set of 4 chromatids. The homologous chromosomes separate, forming cells with half the normal number of chromosomes, but the chromosomes are still duplicated. In meiosis II, the sister chromatids separate, forming cells with half the normal number of unduplicated chromosomes.

Which of the following questions would be most useful to researchers trying to determine the role of meiosis in the F2 phenotypic frequencies based on Figure 5-1 and Table 5-1?

A) What is the molecular mechanism underlying the dominance of erect petals and long pollen?

B) Which phenotypes give pea plants the highest level of fitness: erect or hooded petals and long or round pollen?

C) How do the phases of meiosis differ between sweet pea plants and other organisms?

D) What is the recombination frequency between the genes for petal shape and pollen shape?

D) What is the recombination frequency between the genes for petal shape and pollen shape?

The F2 generation in this experiment showed a greater-than-expected frequency of phenotype combinations from the parental generation. The erect petals, round pollen, and hooded petals, long pollen phenotype combinations are the result of crossing over of homologous chromosomes during meiosis I. The closer genes are to each other, the lower the chance of them being separated from each other by crossing over and the lower the recombination frequency

How many degrees of freedom should be used when looking up the critical value for a chi-square analysis of the ratios of phenotypes observed among the F2 offspring versus the expected phenotypic ratio assuming independent assortment based on Table 5-1?

A) 2

B) 3

C) 4

D) 5

B) 3

degrees of freedom = number of values - 1

In this experiment, 4 - 1 = 3

For sexually reproducing diploid parent cells, which of the following statements best explains the production of haploid cells that occurs in meiosis but not in mitosis?

A) Separation of chromatids occurs once, and there is one round of cell division in meiosis.

B) Separation of chromatids occurs twice, and there are two rounds of cell division in mitosis.

C) Separation of chromatids occurs once, and there are two rounds of cell division in meiosis.

D) Separation of chromatids occurs twice, and there is one round of cell division in mitosis.

C) Separation of chromatids occurs once, and there are two rounds of cell division in meiosis.

Separation of chromatids occurs only once in anaphase I, and there are two rounds of cell division.This ensures that the haploid gamete cells are formed in sexually reproducing diploid organisms.

Which of the following best explains a distinction between metaphase I and metaphase II? (You can use Figure 5-2 as a reference here)

A) The nuclear membrane breaks down during metaphase I but not during metaphase II.

B) Chromosomes align at the equator of the cell during metaphase II but not during metaphase I.

C) The meiotic spindle is needed during metaphase I but not during metaphase II.

D) Homologous pairs of chromosomes are aligned during metaphase I, but individual chromosomes are aligned during metaphase II.

D) Homologous pairs of chromosomes are aligned during metaphase I, but individual chromosomes are aligned during metaphase II.

During metaphase I, homologous pairs of chromosomes are aligned along the metaphase plate. During metaphase II, however, single chromosomes (each composed of two chromatids) are aligned along the metaphase plate.

A compound that prevents the separation of the homologous chromosomes in anaphase I is being studied. Which of the following questions can be best answered during this study based on Figure 5-3?

A) Will the cells produced at the end of meiosis still be genetically identical to each other in the presence of this compound?

B) Will the long-term development of the individual be affected by this meiotic error?

C) When do the centrosomes start to move apart during meiosis I as compared to meiosis II?

D) Is there a pattern to the movement of homologous chromosomes in the presence of this compound?

D) Is there a pattern to the movement of homologous chromosomes in the presence of this compound?

Since the compound prevents separation of homologous chromosomes, both the maternal and paternal chromosomes will migrate along the spindle fibers together. Patterns and other aspects of the movement of these homologous chromosomes in the presence of this compound could be recognized during this study.

Based on Figure 5-3, which of the following questions could best be addressed?

A) Does synapsis of homologous chromosomes in the parent cell contribute to an increase in genetic diversity in the daughter cells?

B) Do sister chromatids separate and form diploid daughter cells?

C) Do chromatids from nonhomologous chromosomes rearrange to produce identical daughter cells?

D) Does synapsis of homologous chromosomes produce daughter cells that are identical to the parent cell?

A) Does synapsis of homologous chromosomes in the parent cell contribute to an increase in genetic diversity in the daughter cells?

The model shows that the synapsis of homologous chromosomes allows the opportunity for crossing over to occur, which can result in chromosomal rearrangement and increased genetic variation in the daughter cells.

Which of the following is closest to the calculated chi-square (χ2) value for the data presented in Table 5-2?

A) 8,35

B) 72.01

C) 98.00

D) 2,546.00

B) 72.01

This is calculated using the chi square formula.

Based on the data in Figure 5-4, which of the following is the best prediction of the mode of inheritance of red eyes in Japanese koi?

A) The allele for red eyes is inherited in an autosomal dominant pattern.
B) The allele for red eyes is inherited in an autosomal recessive pattern.

C) The allele for red eyes is inherited in an X-linked recessive pattern.

D) The allele for red eyes is inherited in an X-linked dominant pattern.

The allele for red eyes is inherited in an autosomal dominant pattern.

The pedigree shows a autosomal dominant pattern as the trait appears frequent and does not skip a generation. In the second generation, the two koi with red eyes produced black eye offspring— this is only possible if both parents are heterozygous with one dominant allele.

Using a significance level of p=0.05, which of the following statements best completes a chi-square goodness-of-fit test for a model of independent assortment based on Table 5-3?

A) The calculated chi-square value is 0.66, and the critical value is 0.05. The null hypothesis can be rejected.

B) The calculated chi-square value is 0.66, and the critical value is 3.84. The null hypothesis cannot be rejected.

C) The calculated chi-square value is 3.91, and the critical value is 5.99. The null hypothesis can be rejected.

D) The calculated chi-square value is 3.91, and the critical value is 7.82. The null hypothesis cannot be rejected.

D) The calculated chi-square value is 3.91, and the critical value is 7.82. The null hypothesis cannot be rejected.

The chi-square value is calculated using the chi square formula- this should give you 3.91. Since p = 0.05, and the degree of freedom is 3, the critical value is 7.82. Since the chi-square value is less than the critical value, the null hypothesis cannot be rejected.

For this condition, which of the following modes of inheritance is most consistent with the observations based on Figure 5-5?

A) autosomal dominant

B) autosomal recessive

C) X-linked dominant

D) X-linked recessive

A) autosomal dominant

The only way for two affected parents to produce offspring that do not have this trait is if the two parents are heterozygous, passing over their recessive alleles to the unaffected offspring.

Glycolysis is a metabolic pathway that converts glucose into pyruvate and is observed in each of the three domains. The hexokinase family of enzymes is required during glycolysis to phosphorylate six-carbon sugars. Researchers designed a general hexokinase inhibitor that is effective in the neurons of rats.

Which of the following best predicts the effect of adding this inhibitor to a culture of plant cells?

A) Plant cells will be unaffected by the inhibitor as they do not perform glycolysis.

B) Plant cells will be unable to perform glycolysis due to the inhibitor and will die.

C) Plant cells will be unable to perform photosynthesis due to the inhibitor and will die.

D) Plant cells will still be able to perform glycolysis since plant hexokinase is not structurally similar to animal hexokinase.

B) Plant cells will be unable to perform glycolysis due to the inhibitor and will die.

Glycolysis is a necessary process for the plant cells to create energy. If this process is inhibited, the plant cells cannot perform glycolysis and will die.

The students plan to use a significance level of p=0.01. Based on Table 5-4, which of the following is the most appropriate critical value for the students to use in their chi-square goodness-of-fit test?

A) 7.82

B) 11.34

C) 13.28

D) 326.7

B) 11.34

Knowing that p=0.01 and the degrees of freedom is 3, the critical value should be 11.34.

Based on Table 5-5, the mean map distance between gene R and gene L is closest to which of the following?

A) 0.28 map units

B) 28 map units

C) 0.14 map units

D) 14 map units

B) 28 map units

1% of recombination frequency is equal to the map distance. Knowing this you can simply determine the mean be adding all the values together and dividing it be the number of trials (4).

The researchers calculated a chi-square value of 29.25. If there are three degrees of freedom and the significance level is p=0.05, which of the following statements best completes the chi-square test based on Table 5-6?

A) The critical value is 0.05, and the null hypothesis cannot be rejected because the calculated chi-square value is greater than the critical value.

B) The critical value is 0.05, and the null hypothesis can be rejected because the calculated chi-square value is greater than the critical value.

C) The critical value is 7.82, and the null hypothesis cannot be rejected because the calculated chi-square value is greater than the critical value.

D) The critical value is 7.82, and the null hypothesis can be rejected because the calculated chi-square value is greater than the critical value.

D) The critical value is 7.82, and the null hypothesis can be rejected because the calculated chi-square value is greater than the critical value.

Knowing that p=0.05 and the degrees of freedom is 3, the critical value is 7.82. Since the chi-square value is greater than the critical value, the null hypothesis can be rejected.

Which of the follow indicates the mean number per cross of F2 plants producing medium-red grain and correctly explains the distribution of the phenotypes based on Table 5-7?

A) The mean number of medium-red phenotypes per cross is 20.8. The distribution of phenotypes suggests that grain color is under environmental control.

B) The mean number of medium-red phenotypes per cross is 20.8. The distribution of phenotypes suggests that multiple genes are involved in grain color determination.

C) The mean number of medium-red phenotypes per cross is 104. The distribution of phenotypes suggests that grain color is under environmental control.

D) The mean number of medium-red phenotypes per cross is 104. The distribution of phenotypes suggests that multiple genes are involved in grain color determination.

B) The mean number of medium-red phenotypes per cross is 20.8. The distribution of phenotypes suggests that multiple genes are involved in grain color determination.

The mean is calulated by adding all the medium red crosses together and dividing it be the number of crosses (5). The phenotype disturbution shows plants of different phenotypes under the same conditions, showing that the grain color is contolled by multiple alleles.

Based on Table 5-8, the mean number of fruit flies per student that are homozygous recessive for both genes is closest to which of the following?

A) 89.75

B) 29.0

C) 22.75

D) 18.5

B) 29.0

The mean is found by adding all offsprings that are homozygous recessive for both genes (eess) and dividing it by the number of students who seperately gathered this data (4).

Himalayan rabbits are a breed of rabbits with highly variable fur color. If genetically similar rabbits are raised in environments that have different temperature conditions, the rabbits can have different color patterns.

Which of the following statements best explains how the fur color can be different in Himalayan rabbits raised under different temperature conditions?

A) The genotype does not contribute to coat color in Himalayan rabbits.

B) The phenotype determines the genotype of coat color in Himalayan rabbits.

C) Different environments cause specific mutations in the genes controlling pigment production.

D) The environment determines how the genotype is expressed.

D) The environment determines how the genotype is expressed.

Gene activity can be influenced by environmental factors, which includes temperature, the most significant environmental factor.

The tadpoles of Mexican spadefoot toads are known to exhibit phenotypic plasticity depending on food availability. Tadpole mouthparts can vary significantly, prompting researchers to categorize them as either omnivore-morph or carnivore-morph. Carnivore-morph tadpoles are larger and have mouthparts that are better suited for predation. Remarkably, carnivore-morph tadpoles can change into omnivore-morph tadpoles when the food supply changes.

Which of the following best describes an advantage of the phenotypic plasticity displayed by the tadpoles?

A) It allows the tadpoles to change their genome in response to environmental pressures.

B) It enables the tadpoles to develop into a distinct species of toads.

C) It gives the tadpoles increased versatility with respect to diet.

D) It allows the tadpoles to delay metamorphosis until there is maximal food available for the adults.

C) It gives the tadpoles increased versatility with respect to diet.

Phenotypic plasticity allows the tadpoles to produce distinct phenotypes in response to changes in the environment, allowing them to best exploit the available resources under various environments.

When a mustard plant seedling is transferred to an environment with higher levels of carbon dioxide, the new leaves have a lower stomata-to-surface-area ratio than do the seedling's original leaves.

Which of the following best explains how the leaves from the same plant can have different stomatal densities when exposed to an elevated carbon dioxide level?

A) Increased photosynthesis leads to larger leaves that need more stomata for photosynthesis, leading to an increase in stomatal density.

B) Leaf growth is promoted through increased photosynthesis, but the genetically regulated rate of stomatal production is not altered, leading to a decrease in stomatal density.

C) Leaf growth is inhibited by decreased photosynthesis, and the genetically regulated rate of stomatal production remains the same, leading to an increase in stomatal density.

D) Leaf growth is inhibited by decreased photosynthesis, and the genetically regulated rate of stomatal production remains the same, leading to a decrease in stomatal density.

B) Leaf growth is promoted through increased photosynthesis, but the genetically regulated rate of stomatal production is not altered, leading to a decrease in stomatal density.

Elevated CO2 levels will promote leaf growth due to increased photosynthetic activity. As the leaf grows larger, the stomatal density will decrease if the number of stomata produced remains the same.

Which of the following best predicts the effect of the chromosomal segregation error shown in Figure 5-7?

A) All of the resulting gametes will have an extra chromosome.

B) All of the resulting gametes will be missing a chromosome.

C) Half of the resulting gametes will have an extra chromosome and the other half will be missing a chromosome.

D) Half of the resulting gametes will have the correct number of chromosomes, and the other half will have an incorrect number of chromosomes.

D) Half of the resulting gametes will have the correct number of chromosomes, and the other half will have an incorrect number of chromosomes.

The nondisjunction of sister chromatids during meiosis II will produce two gametes with the correct number of chromosomes (n). The remaining two gametes that are produced will be n+1 and n−1.

Based on Figure 5-8, If the woman and a man with normal clotting function have children, what is the probability of their children exhibiting hemophilia A? A) 50 percent for daughters, 0 percent for sons

B) 50 percent for sons, 0 percent for daughters

C) 50 percent for all children

D) 0 percent for all children

B) 50 percent for sons, 0 percent for daughters

X-linked traits are likely inherited to males as they only have one X chromosone versus females. Since the father gives the male a Y chromosone, the mother will either give the male a unaffected allele or an affected allele.

Based on Table 5-9, which of the following is most likely the immediate cause of the first appearance of Huntington's disease in a person?

A) A point mutation occurs in the HTT gene

B) The first appearance of CAG repeat occurs in the HTT gene.

C) An allele with more than 39 CAG repeats was inherited by the affected person.

D) The person inherited two alleles that each contained 20 CAG repeats.

C) An allele with more than 39 CAG repeats was inherited by the affected person.

A person with more than 39 CAG repeats likely happens if the person inherited an allele that will cause symptoms of Huntington’s disease.