Biology Unit 5 AP CR Questions

A student crosses a pure-breeding line of red-flowered poinsettias with a pure-breeding line of white-flowered poinsettias. The student observes that all the plants in the F1F1 generation have pink flowers. The student then crosses the F1F1 plants with one another and records observations about the plants in the F2F2 generation. The student will use the F2F2 data to perform a chi-square goodness-of-fit test for a model of incomplete dominance. The setup for the student's chi-square goodness-of-fit test is presented in Table 1. Table 1. A chi-square goodness-of-fit test for incomplete dominance Phenotype - Observed - Expected Red flowers - 64 - 56 Pink flowers - 107 - 112 White flowers - 53 - 56 The critical value for a chi-square test with a significance level of p=0.05p=0.05 and 2 degrees of freedom is 5.99. Which of the following statements best completes the student's chi-square goodness-of-fit test? A The calculated chi-square value is 1.53, and the null hypothesis cannot be rejected. B The calculated chi-square value is 1.53, and the null hypothesis can be rejected. C The calculated chi-square value is 98, and the null hypothesis cannot be rejected. D The calculated chi-square value is 98, and the null hypothesis can be rejected.

A

Figure 1. Model of crossing over between homologous chromosomes, indicating crossing over rate of selected loci. During prophase II replicated homologous chromosomes pair up and undergo synapsis. What testable question is generated regarding synapsis and genetic variability by Figure 1 ? A Is the distance between two gene loci related to crossover rate? B Does crossing over occur more often in some chromosomes than in others? C Is crossing over inhibited by methylation? D Is crossing over promoted by methylation?

A

In a strain of tomato plants, short plant height and small fruit size are traits that display autosomal recessive patterns of inheritance. To investigate whether the traits segregate independently, researchers cross a pure-breeding line of tall tomato plants that have large fruits with a pure-breeding line of short tomato plants that have small fruits. The researchers observe that all the plants in the F1F1 generation are tall and have large fruits. The researchers cross the F1F1 plants with one another to generate an F2F2 generation. The researchers record observations for the F2F2 generation and will use the data to perform a chi-square goodness-of-fit test for a model of independent assortment. The setup for the chi-square goodness-of-fit test is shown in Table 1. Table 1. Observed and expected counts of tomato plants in the F2F2 generation Phenotype - Observed - Expected Tall plant with large fruits - 104 - 90 Tall plant with small fruits - 17 - 30 Short plant with large fruits - 26 - 30 Short plant with small fruits - 13 - 10 The researchers choose a significance level of p=0.05p=0.05. Which of the following best completes the chi-square goodness-of-fit test? A The calculated chi-square value is 9.24, and the critical value is 7.82. The null hypothesis of independent assortment can be rejected. B The calculated chi-square value is 9.24, and the critical value is 9.49. The null hypothesis of independent assortment cannot be rejected. C The calculated chi-square value is 13.13, and the critical value is 7.82. The null hypothesis of independent assortment can be rejected. D The calculated chi-square value is 13.13, and the critical value is 9.49. The null hypothesis of independent assortment cannot be rejected

A

In sweet pea plants, purple flower color is dominant over red flower color and long pollen grain shape is dominant over round pollen grain shape. Two sweet pea plants that are heterozygous for both flower color and pollen grain shape are crossed with one another. A geneticist is surprised to observe that there are far fewer round, purple offspring and long, red offspring than were predicted by the 9:3:3:19:3:3:1 expected phenotypic ratio. Which of the following statements provides the most reasonable prediction to account for the deviation from the expected results? A In sweet pea plants, the gene for flower color and the gene for pollen grain shape are genetically linked. B In sweet pea plants, the genes for flower color and for pollen grain shape both exhibit codominance. C Several mutations occurred spontaneously producing a deviation from the expected phenotypic ratios of the offspring. D The genes for flower color and pollen grain shape are inherited independently because of the law of independent assortment.

A

Insulin is a hormone produced by some pancreatic cells. Scientists have isolated the DNADNA sequence that codes for human insulin production. Which of the following best predicts the effect of inserting this gene into the DNADNA of a bacterial cell? A The recombinant bacterium will produce human insulin using its own transcription and translation machinery. B The recombinant bacterium will not transcribe the human insulin gene because this gene is not normally found in the bacterial genome. C The recombinant bacterium will transcribe the gene but will be unable to translate the mRNAmRNA. D The recombinant bacterium will die because it has been exposed to foreign DNADNA.

A

Scientists have found that DNADNA methylation suppresses crossing-over in the fungus Ascobolus immersus. Which of the following questions is most appropriately raised by this specific observation? A Is the level of genetic variation in the gametes related to the amount of DNADNA methylation observed? B Without crossing-over, will gametes be viable and be able to produce zygotes? C Does DNADNA methylation result in shorter chromosomes? D Is this species of fungus a diploid organism?

A

An experiment was performed to determine the mode of inheritance of two mouse genes, one for fur color and one for fur length. It is known that black fur (BB) is dominant over white fur (bb) and that long fur (LL) is dominant over short fur (ll). To determine how the genes are inherited, a cross was performed between two true-breeding mice, one with long black fur and one with short white fur. Their progeny, the F1F1 generation, all had long black fur. Five F1F1 male-female pairs were then crossed with one another. The F2F2 generation phenotypes for each cross are shown in Table·1. Table 1. Number of F2F2 generation phenotypes for five crosses Phenotype - Cross 1 - Cross 2 - Cross 3 - Cross 4 - Cross 5 Long black fur - 6 - 5 - 5 - 6 - 7 Long white fur - 1 - 1 - 0 - 2 - 2 Short black fur - 0 - 2 - 1 - 1 - 1 Short white fur - 2 - 3 - 3 - 2 - 2 Which of the following is the mean number per cross of F2F2 generation offspring that are the result of crossing over? A 1 B 2.2 C 2.4 D 5.8

B

The Russian blue is a rare breed of cat that is susceptible to developing cataracts on the eyes. Scientists hypothesize that this condition is inherited as a result of a mutation. Figure 1 shows a pedigree obtained in a study of cats owned by members of the Russian Blue Club in Sweden. Based on the inheritance pattern shown in Figure 1, which of the following best predicts the nature of the original mutation? A A recessive mutation on the X chromosome B A recessive mutation on a somatic chromosome C A dominant mutation on the X chromosome D A dominant mutation on a somatic chromosome

B

A model showing two possible arrangements of chromosomes during meiosis is shown in Figure 1. Figure 1. Two possible arrangements of chromosomes during meiosis Which of the following questions about genetic diversity could most appropriately be answered by analysis of the model in Figure 1 ? A Does crossing-over generate more genetic diversity than the fusion of gametes does? B Does DNADNA methylation prevent independent assortment during metaphase IIII? C How does the independent assortment of the two sets of homologous chromosomes increase genetic diversity? D Do daughter cells that are not genetically identical to parent cells produce viable zygotes?

C

A student carries out a genetics experiment with fruit flies to investigate the inheritance pattern of the white eye trait. The student crosses a homozygous white-eyed female with a wild-type male and records observations about the flies in the F1 generation. The student plans to use the F1 data to perform a chi-square goodness-of-fit test for a model based on an X-linked recessive pattern of inheritance. The student will use one degree of freedom and a significance level of p=0.05. The setup for the student's chi-square goodness-of-fit test is presented in Table 1. Table 1. Setup for the student's chi-square goodness-of-fit test Phenotype - Observed - Expected Red-eyed female - 53 - 50 White-eyed male - 47 - 50 The student calculates a chi-square value of 0.36. Which of the following statements best completes the student's chi-square goodness-of-fit test? A The critical value is 0.05, and the student cannot reject the null hypothesis. B The critical value is 0.05, and the student can reject the null hypothesis. C The critical value is 3.84, and the student cannot reject the null hypothesis. D The critical value is 3.84, and the student can reject the null hypothesis.

C

An African violet grower observes that genetically identical African violet plants growing near the walls of the greenhouse have white flowers, that plants growing farther away from the walls have pale blue flowers, and that plants growing nearest the center of the greenhouse have dark blue flowers. Which of the following best explains the differences in flower color of the African violets in the greenhouse? A Warmer temperatures result in genotypic alterations, which result in flower color differences. B The plants along the walls of the greenhouse are homozygous recessive and therefore have white flowers. C An enzyme responsible for flower color does not fold correctly in cooler temperatures, and the greenhouse is warmest in the center. D More light is available along the walls of the greenhouse, so the flowers need less pigment to absorb sunlight for photosynthesis.

C

Table 1 shows the stage and number of cells and chromosomes per cell at the end of the stage in a 2n=24 organism. Table 1. Cell and chromosome count during selected phases of meiosis Stage - Number of Cells - Number of Chromosomes per Cell Prophase II - 1 - 24 Metaphase II - 1 - 24 Anaphase II - 1 - 24 Telophase II - 1 - 24 Beginning of Prophase II - 2 - 12 A Which of the following statements correctly describes the chromosomes in each daughter cell at the end of meiosis II? Each daughter cell contains 12 chromatids. Each chromatid is one of two from a single chromosome with the other one of the pair found in the other daughter cell. B Each daughter cell contains 12 chromosomes, each composed of two chromatids. Since the chromosomes were randomly divided, one daughter cell may contain both of a pair of homologous chromosomes, while the other cell contains both of another pair of homologous chromosomes. C Each daughter cell contains 12 chromosomes, each composed of two chromatids. Each chromosome is one of a pair of homologous chromosomes from the parent cell, with the other homologue found in the other daughter cell. D Each daughter cell contains 24 separate chromatids. Since every two chromatids were originally joined, forming one homologous chromosome, the number of chromatids is divided by two to determine the number of chromosomes.

C

Trisomy 21 is a condition in which a child is born with an extra chromosome in pair 21. Researchers assessed the frequency of children born with trisomy 21 by age of the mothers at birth (maternal age) and primary cause of the error leading to trisomy 21. The findings are presented in Figure 1. Figure 1. Incidence and primary cause of trisomy 21 by maternal age-group Based on the data in Figure 1, which of the following is most likely the primary cause of the pattern of frequency of trisomy 21 births in the selected maternal age-groups? A At older maternal ages, there is an increase in the number of errors during mitosis, which leads to an increase in nondisjunction during egg production. B The incidence of nondisjunction errors in meiosis during sperm production is positively correlated with increasing maternal age. C At older maternal ages, the incidence of errors in meiosis during egg production increases, which leads to an increase in nondisjunction. D Errors in meiosis leading to nondisjunction are more likely to occur during meiosis I than during meiosis II.

C

A gene that influences coat color in domestic cats is located on the XX chromosome. A female cat that is heterozygous for the gene (XBXOXBXO) has a calico-colored coat. In a genetics experiment, researchers mate a calico-colored female cat (XBXOXBXO) with an orange-colored male cat (XOYXOY) to produce an F1F1 generation. The researchers record observations for the cats in the F1F1 generation and plan to use the data to perform a chi-square goodness-of-fit test for a model of XX-linked inheritance. The data for the chi-square goodness-of-fit test are presented in Table 1. Table 1. Data for the chi-square goodness-of-fit test Phenotype - Genotype - Observed - Expected - Calico-colored female - XBXO - 15 - 10 Orange-colored female - XOXO - 6 - 10 Black-colored male - XBY - 11 - 10 Orange-colored male - XOY - 8 - 10 The researchers calculate a chi-square value of 4.6 and choose a significance level of p=0.05p=0.05. Which of the following statements best completes the chi-square goodness-of-fit test? A The null hypothesis can be rejected because the chi-square value is greater than the critical value. B The null hypothesis can be rejected because the chi-square value is less than the critical value. C The null hypothesis cannot be rejected because the chi-square value is greater than the critical value. D The null hypothesis cannot be rejected because the chi-square value is less than the critical value.

D

A scientist studying phenotypic variation in a species of butterfly observed that genetically identical caterpillars grown in similar cages but exposed to different colored lights developed into butterflies with differences in wing color and body size, as shown in Table 1. Table 1. Effect of Exposing Identical Caterpillars to Specific Colors of Light Phenotype of Adult Butterfly - Caterpillars Exposed to Red Light - Caterpillars Exposed to Blue Light Wing color - Darker - Lighter Body size - Smaller - Larger Which of the following best explains the cause of the phenotypic variation observed in the butterflies? A Different mutations occurred in the caterpillars that were exposed to different colors of light. B The energy used to grow a larger body results in butterflies with lighter colored wings. C Individual caterpillars evolved adaptations to survive in each of the light conditions they were exposed to. D There was differential gene expression of wing color and body size in response to the colors of light the caterpillars were exposed to.

D

Australian dragon lizards have a ZW sex-determination system. The male genotype is homogametic (ZZ), and the female genotype is heterogametic (ZW). However, all eggs incubated at temperatures above 32°C tend to develop into females. Which of the following best explains how the development of phenotypic female Australian dragon lizards with a ZZ genotype occurs when incubation temperatures are above 32°C? A Lizard embryos with a ZZ genotype cannot develop at temperatures above 32°C. B At incubation temperatures above 32°C, Z chromosomes are mutated into W chromosomes. C At incubation temperatures above 32°C, crossing over transfers genes from the W chromosome to the Z chromosome, producing females. D Incubation temperatures above 32°C inhibit the genes on the Z chromosome that produce proteins necessary for male development

D

Both mitosis and meiosis begin with a parent cell that is diploid. Which of the following best describes how mitosis and meiosis result in daughter cells with different numbers of chromosomes? A In mitosis, the chromosomes consist of a single chromatid, which is passed to two haploid daughter cells. In meiosis, the chromosomes consist of two chromatids during the first round of division and one chromatid during the second round of division, resulting in two haploid daughter cells. B In mitosis, synapsis of homologous chromosomes results in four haploid daughter cells after one division. In meiosis, synapsis of homologous chromosomes occurs during the second division and results in four diploid daughter cells. C Mitosis produces one identical daughter cell after one round of division. Meiosis has two rounds of division and doubles the number of chromosomes in the second round of division, producing four diploid cells. D Mitosis produces two identical diploid daughter cells after one round of division. Meiosis produces four haploid daughter cells after two rounds of division.

D

Figure 1. X⁢ and Y chromosomes during meiosis I and meiosis II If the normal spermatogenesis is disrupted, the gametes can have different chromosomes than expected. Which of the following is the most likely cause of one of the four gametes having two X chromosomes and one having neither an X nor a Y chromosome? A Nondisjunction of the chromosomes during meiosis I B Nondisjunction of both the X and Y chromosomes during meiosis II C Nondisjunction of the Y chromosome during meiosis II D Nondisjunction of the X chromosome during meiosis I

D

In fruit flies, purple eyes and ebony body are traits that display autosomal recessive patterns of inheritance. In a genetics experiment, students cross wild-type flies with flies that have purple eyes and ebony bodies. The students observe that all the flies in the F1F1 generation have normal eyes and a normal body color. The students then allow the F1F1 flies to mate and produce an F2F2 generation. The students record observations about the flies in the F2F2 generation and use the data to perform a chi-square goodness-of-fit test for a model of independent assortment. The setup for the students' chi-square goodness-of-fit test is presented in Table 1. Table 1. The students' chi-square goodness-of-fit test for a model of independent assortment Phenotype - Observed - Expected Normal eyes, normal body - 187 - 171 Normal eyes, ebony body - 49 - 57 Purple eyes, normal body - 41 - 57 Purple eyes, ebony body - 27 - 19 The students choose a significance level of p=0.01p=0.01. Which of the following statements best completes the next step of the chi-square goodness-of-fit test? A The calculated chi-square value is 2.11, and the critical value is 7.82. B The calculated chi-square value is 2.11, and the critical value is 11.35. C The calculated chi-square value is 10.48, and the critical value is 7.82. D The calculated chi-square value is 10.48, and the critical value is 11.35.

D

In fruit flies, sepia eyes and ebony body are traits that display autosomal recessive patterns of inheritance. To investigate whether the traits are genetically linked, students cross wild-type flies with a line of flies that have sepia eyes and ebony bodies. The students observe that all the flies in the F1F1 generation have normal eyes and normal bodies. The students allow the flies in the F1F1 generation to mate and produce an F2 generation. The students then record observations for the flies in the F2 generation and use the data to perform a chi-square goodness-of-fit test for a model of independent assortment. The setup for the chi-square goodness-of-fit test is presented in Table 1. Table 1. Setup for the students' chi-square goodness-of-fit test Phenotype - Observed - Expected Normal eyes, normal body - 231 - 279 Normal eyes, ebony body - 86 - 93 Sepia eyes, normal body - 97 - 93 Sepia eyes, ebony body - 82 - 31 The students calculate a chi-squared value of 92.86 and compare it with a critical value of 7.82. Which of the following best completes the chi-square goodness-of-fit test? A The null hypothesis cannot be rejected, and the students should conclude that the data fit a model of independent assortment. B The null hypothesis cannot be rejected, and the students should conclude that the data may have resulted from genetic linkage. C The null hypothesis can be rejected, and the students should conclude that the data fit a model of independent assortment. D The null hypothesis can be rejected, and the students should conclude that the data may have resulted from genetic linkage.

D

Pigeons demonstrate ZW sex determination, such that a ZZ genotype produces a male and a ZW genotype produces a female. The gene for feather color is located on the Z chromosome, and the red allele is dominant over the brown allele. Three crosses between brown male pigeons and red female pigeons were performed, and the results are shown below. Table 1. Offspring from three separate crosses of a brown male pigeon and a red female pigeon Number of Offspring Phenotype - Cross 1 - Cross 2 - Cross 3 Red - 11 - 9 - 7 Brown - 9 - 11 - 13 Which of the following is the mean number of male offspring produced by the three crosses? A 27 B 20 C 11 D 9

D

Saccharomyces cerevisiae is a diploid yeast species that can reproduce either sexually or asexually. An experiment was performed to induce mitotically dividing S. cerevisiae cells in G2G2 to undergo meiosis. Which of the following best describes the steps these cells will follow to form gametes? A The first division will result in crossing over between homologous chromosomes, and the second division will reduce the original number of chromosomes by half in the daughter cells. B The first division will reduce the number of chromosomes by half for each daughter cell, and the second division will result in each daughter cell having one-fourth of the original number of chromosomes. C The first division will move single chromatids to each daughter cell, and the second division will double the number of chromosomes in each daughter cell. D The first division will reduce the number of chromosomes by half for each daughter cell, and the second division will move single chromatids to each daughter cell.

D