Practice CH 15. Short Answer. Self-Quiz Questions

Transcription

1 Practice CH 15 Short Answer Self-Quiz Questions 1. A man with hemophilia (a recessive, sex-linked condition. has a daughter of normal phenotype) She marries a man who is normal for the trait. What is the probability that a daughter of this mating will be a hemophiliac? That a son will be a hemophiliac? If the couple has four sons, what is the probability that all four will be born with hemophilia? 2. Red-green color blindness is caused by a sex-linked recessive allele. A color-blind man marries a woman with normal vision whose father was color-blind. What is the probability that they will have a color-blind daughter? What is the probability that their first son will be color-blind? (Note: The two questions are worded a bit differently). 3. Determine the sequence of genes along a chromosome based on the following recombination frequencies: A- B, 8 map units; A-C, 28 map units; A-D, 25 map units; B-C, 20 map units; B-D, 33 map units. Multiple Choice Identify the choice that best completes the statement or answers the question. Media Activity Questions 4. The recombination frequency between gene A and gene B is 8.4%, the recombination frequency between gene A and gene C is 6.8%, and the recombination frequency between gene B and gene C is 15.2%. Which is the correct arrangement of these genes? a. ABC b. ACB c. BCA d. CAB e. CBA The following questions refer to the data and figures below. CROSS I. Purebred lines of wild-type fruit flies (gray body and normal wings) are mated to flies with black bodies and vestigial wings. F1 offspring all have a normal phenotype.

2 CROSS II. F1 flies are crossed with flies recessive for both traits (a testcross). Resulting Offspring Normal Percentage Gray body; normal wings Black body; vestigial wings Black body; normal wings Gray body; vestigial wings KEY: A. CROSS I results give evidence supporting the statement. B. CROSS I results give evidence against the statement. C. CROSS II results give evidence supporting the statement. D. CROSS II results give evidence against the statement. E. Neither CROSS I nor CROSS II results support the statement. 5. The genes for body color and wing shape are linked. a. A b. B c. C d. D e. E 6. There are 25 centimorgans (map units) between the genes for body color and wing shape. a. A b. B c. C d. D e. E 7. The following is a map of four genes on a chromosome: Between which two genes would you expect the highest frequency of recombination? a. A and W b. W and E c. E and G d. A and E e. A and G Refer to the figure below to answer the following questions.

3 8. In a series of mapping experiments, the recombination frequencies for four different linked genes of Drosophila were determined as shown in the figure. What is the order of these genes on a chromosome map? a. rb-cn-vg-b b. vg-b-rb-cn c. cn-rb-b-vg d. b-rb-cn-vg e. vg-cn-b-rb 9. The frequency of crossing over between any two linked genes is a. higher if they are recessive. b. different between males and females. c. determined by their relative dominance. d. the same as if they were not linked. e. proportional to the distance between them. Use the list of chromosomal systems below to answer the following questions. A. haploid-diploid B. X-0 C. X-X D. X-Y E. Z-W 10. What is the chromosomal system for sex determination in grasshoppers and certain other insects? a. A b. B c. C d. D e. E 11. Most calico cats are female because a. a male inherits only one of the two X-linked genes controlling hair color. b. the males die during embryonic development. c. the Y chromosome has a gene blocking orange coloration. d. only females can have Barr bodies. e. multiple crossovers on the Y chromosome prevent orange pigment production. Refer to the information below to answer the following questions.

4 An achondroplastic male dwarf with normal vision marries a color-blind woman of normal height. The man's father was six-feet tall, and both the woman's parents were of average height. Achondroplastic dwarfism is autosomal dominant, and red-green color blindness is X-linked recessive. 12. How many of their daughters might be expected to be color-blind dwarfs? a. all b. none c. half d. one out of four e. three out of four 13. How many of their sons would be color-blind and of normal height? a. all b. none c. half d. one out of four e. three out of four 14. A Barr body is normally found in the nucleus of which kind of human cell? a. unfertilized egg cells only b. sperm cells only c. somatic cells of a female only d. somatic cells of a male only e. both male and female somatic cells 15. A human individual is phenotypically female, but her interphase somatic nuclei do not show the presence of Barr bodies. Which of the following statements concerning her is probably true? a. She has Klinefelter syndrome. b. She has an extra X chromosome. c. She has Turner syndrome. d. She has the normal number of sex chromosomes. e. She has two Y chromosomes. 16. The karyotype shown below is associated with which of the following genetic disorders? a. Turner syndrome b. Down syndrome c. Klinefelter syndrome d. hemophilia e. male-pattern baldness 17. In humans, male-pattern baldness is controlled by a gene that occurs in two allelic forms. Allele Hn determines nonbaldness, and allele Hb determines pattern baldness. In males, because of the presence of testosterone, allele Hb is dominant over Hn. If a man and woman both with genotype HnHb have a son, what is the chance that he will eventually be bald? a. 0%

5 b. 25% c. 33% d. 50% e. 75% 18. Which of the following statements about mitochondria is false? a. Because of the role of the mitochondria in producing cellular energy, mitochondrial diseases often affect the muscles and nervous system. b. Because mitochondria are present in the cytoplasm, mitochondrial diseases are transmitted maternally. c. Like nuclear genes, mitochondrial genes usually follow Mendelian patterns of inheritance. d. Mitochondria contain circular DNA molecules that code for proteins and RNAs. e. Many mitochondrial genes encode proteins that play roles in the electron transport chain and ATP synthesis Practice CH 15 Answer Section SHORT ANSWER 1. ANS: 0; 1/2, 1/16 PTS: 1 2. ANS: 1/4 for each daughter (1/2 chance that child will be female 1/2 chance of a homozygous recessive genotype); 1/2 for first son. PTS: 1 3. ANS: D-A-B-C PTS: 1 MULTIPLE CHOICE 4. ANS: D PTS: 1 TOP: Web/CD Activity: Linked Genes and Crossing Over 5. ANS: D PTS: 1 TOP: Concept ANS: D PTS: 1 TOP: Concept ANS: E PTS: 1 TOP: Concept ANS: D PTS: 1 TOP: Concept ANS: E PTS: 1 TOP: Concept ANS: B PTS: 1 TOP: Concept ANS: A PTS: 1 TOP: Concept ANS: B PTS: 1 TOP: Concept ANS: C PTS: 1 TOP: Concept ANS: C PTS: 1 TOP: Concept 15.3

6 15. ANS: C PTS: 1 TOP: Concept ANS: C PTS: 1 TOP: Concept ANS: E PTS: 1 TOP: Concept ANS: C PTS: 1 TOP: Concept 15.5