Genetics Review Problems Chps 11 And 12

Name: ________________________ Class: ___________________ Date: __________

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Genetics Review Problems Chps 11 and 12 Multiple Choice Identify the choice that best completes the statement or answers the question. ____

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1. Which of the following genotypes is homozygous? a. AaBB b. aABB c. aaBB d. aaBb e. AaBb 2. If all offspring of a cross have the genotype Aa, the parents of the cross are most likely a. AA × aa. b. Aa × Aa. c. Aa × aa. d. AA × Aa. e. none of these 3. If short hair (L) is dominant to long hair (l), then what fraction of the offspring produced by a cross of Ll x ll will be homozygous dominant? a. 1/2 b. 1/4 c. 1/3 d. none (no chance of this offspring) e. None of these answers is correct. 4. An individual with a genotype of Aa Bb CC is able to produce how many different kinds of gametes? a. 2 b. 3 c. 4 d. 7 e. 8 5. Some dogs have erect ears; others have drooping ears. Some dogs bark when following a scent; others are silent. Erect ears and barking are due to dominant alleles located on different chromosomes. If two dihybrids are crossed, a. the most common phenotype is drooping ears and barking. b. all droopy-eared, silent offspring are pure-breeding. c. the least common phenotype is drooping ears and barking. d. there will be no phenotypes or genotypes that resemble the original parents. e. there will be no offspring that resemble the F1 generation. 6. In cocker spaniels, black coat color (B) is dominant over red (b), and solid color (S) is dominant over spotted (s). If two dihybrids (Bb Ss) are crossed, the most common phenotype will be a. black and solid. b. black and spotted. c. red and solid. d. red and spotted. e. none of these

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ID: A

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7. Susan, a mother with type B blood, has a child with type O blood. She claims that Craig, who has type A blood, is the father. He claims that he cannot possibly be the father. Further blood tests ordered by the judge reveal that Craig is AA. The judge rules that a. Susan is right and Craig must pay child support. b. Craig is right and doesn't have to pay child support. c. Susan cannot be the real mother of the child; there must have been an error made at the hospital. d. it is impossible to reach a decision based on the limited data available. e. none of these ____ 8. The F2 phenotypic ratio of a monohybrid cross involving a gene with incompletely dominant alleles is a. 1:1. b. 2:1. c. 9:3:3:1. d. 1:2:1. e. 3:1. ____ 9. A woman is diagnosed to have a gene for the genetic disorder known as Huntington disease. It is a rare defect caused by an autosomal dominant allele. The chance for any one of her children to inherit the gene is a. dependent on the sex of the child. b. 1/3. c. 1/2. d. 3/4. e. 1.00 (or 100 percent). ____ 10. The Punnett square illustration below shows that

a. half of human sperm carry a Y chromosome. b. half of human zygotes are XY. c. all zygotes carry an X chromosome. d. gender depends upon which type of sperm fertilizes the egg. e. all of these ____ 11. Hemophilia A a. is rare in the human population. b. is more common among men. c. was common in the descendents of Queen Victoria. d. is an X-linked recessive trait. e. all of these

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Name: ________________________

ID: A

____ 12. A color-blind man and a woman with normal vision whose father was color blind have a son. Color blindness, in this case, is caused by an X-linked recessive gene. If only the male offspring are considered, the probability that their son is color blind is a. .25 (or 25 percent). b. .50 (or 50 percent). c. .75 (or 75 percent). d. 1.00 (or 100 percent). e. none of these ____ 13. A chromosome's gene sequence that was ABCDEFG before damage and ABCFG afterward is an example of a. inversion. b. deletion. c. duplication. d. translocation. e. aneuploidy. ____ 14. If a zygote is missing one chromosome, a. the chromosome number is expressed as 2n − 1. b. then one chromosome is without its homologue. c. the condition is called monosomy. d. the chromosome number is expressed as 2n − 1, and the condition is called monosomy. e. the chromosome number is expressed as 2n − 1, one chromosome is without its homologue, and the condition is called monosomy. ____ 15. Suppose that a hemophilic male (X-linked recessive allele) and a female carrier for the hemophilic trait have a nonhemophilic daughter with Turner syndrome. Nondisjunction could have occurred in a. both parents. b. neither parent. c. the father only. d. the mother only. e. none of these

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ID: A

Genetics Review Problems Chps 11 and 12 Answer Section MULTIPLE CHOICE 1. ANS: TOP: 2. ANS: TOP: 3. ANS: TOP: 4. ANS: TOP: 5. ANS: TOP: 6. ANS: TOP: 7. ANS: TOP: 8. ANS: TOP: 9. ANS: TOP: 10. ANS: TOP: 11. ANS: TOP: 12. ANS: TOP: 13. ANS: TOP: 14. ANS: TOP: 15. ANS: TOP:

C PTS: 1 DIF: Easy MENDEL, PEA PLANTS, AND INHERITANCE PATTERNS A PTS: 1 DIF: Moderate MENDEL'S THEORY OF SEGREGATION D PTS: 1 DIF: Easy MENDEL'S THEORY OF SEGREGATION C PTS: 1 DIF: Moderate MENDEL'S THEORY OF INDEPENDENT ASSORTMENT B PTS: 1 DIF: Moderate MENDEL'S THEORY OF INDEPENDENT ASSORTMENT A PTS: 1 DIF: Moderate MENDEL'S THEORY OF INDEPENDENT ASSORTMENT B PTS: 1 DIF: Difficult MORE PATTERNS THAN MENDEL THOUGHT D PTS: 1 DIF: Moderate MORE PATTERNS THAN MENDEL THOUGHT C PTS: 1 DIF: Difficult EXAMPLES OF AUTOSOMAL INHERITANCE PATTERNS E PTS: 1 DIF: Easy SEX DETERMINATION IN HUMANS E PTS: 1 DIF: Moderate EXAMPLES OF X-LINKED INHERITANCE PATTERNS B PTS: 1 DIF: Difficult EXAMPLES OF X-LINKED INHERITANCE PATTERNS B PTS: 1 DIF: Moderate HERITABLE CHANGES IN CHROMOSOME STRUCTURE E PTS: 1 DIF: Easy HERITABLE CHANGES IN CHROMOSOME NUMBER C PTS: 1 DIF: Difficult HERITABLE CHANGES IN CHROMOSOME NUMBER

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