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1 Name: 1. Compared to human cells resulting from mitotic cell division, human cells resulting from meiotic cell division would have A) twice as many chromosomes B) the same number of chromosomes C) one-half the number of chromosomes D) one-quarter as many chromosomes Unit 11 Test: Genetics Date: /Period: 4. Which process is represented in the diagram below? 2. The distribution of chromosomes in one type of cell division is shown in the diagram below. A) budding B) fertilization Which process is represented in the diagram? A) vegetative propagation B) asexual reproduction C) meiosis D) mitosis 3. Asexual reproduction primarily involves the process of C) mitosis D) meiosis 5. Mendel studied the inheritance traits in A) four-o'clock flowers B) roan cattle C) pea plants D) fruit flies A) spermatogenesis B) pollination C) ovulation D) mitosis Version 1 Page 1

2 6. A sperm cell of an alligator has 16 chromosomes. What is the total number of chromosomes normally present in a stomach cell of this alligator? A) 8 B) 16 C) 32 D) 48 Unit 11 Test: Genetics 10. In pea plants, tall stems are dominant over short stems. If two heterozygous tall plants are crossed, what percentage of the offspring would be expected to have the same phenotype as the parents? A) 25% B) 50% C) 75% 7. Hereditary information for traits is located in A) the lysosomes in the cytoplasm B) the mitochondria of gametes C) genes found on chromosomes D) chromosomes found on genes 8. In owls, red feathers are dominant over gray feathers. If two heterozygous red-feathered owls are mated, what percentage of their offspring would be expected to have red feathers? A) 25% B) 50% C) 75% 9. An organism that has two identical genes for a trait is said to be A) homozygous for the trait 11. In watermelon, solid green fruit (G) is dominant over the allele for striped fruit (g). A homozygous green watermelon is crossed with a heterozygous green watermelon. What percent of the offspring of this cross will be striped watermelons? A) 0% B) 25% C) 50% 12. If the parents do not show a particular trait but it appears in their offspring, the genotypes of the parents are A) monoploid B) pure recessive C) heterozygous D) homozygous B) incompletely dominant for the trait C) heterozygous for the trait D) hybrid for the trait Version 1 Page 2

3 13. The principles of genetics were first described by Unit 11 Test: Genetics A) Lamarck B) Darwin C) Mendel D) Watson and Crick 14. The process of mitotic cell division normally results in the production of A) two cells with the same number of chromosomes as the parent cell B) one cell with a replicated set of homologous chromosomes C) two cells with only one chromosome from each set of homologous chromosomes D) four cells with half the number of chromosomes as the parent cell Version 1 Page 3

4 Base your answers to questions 15 through 17 on on the diagram below which shows the offspring of a white mouse and a gray mouse. 15. What must be the gene combination of the parents to produce the offspring that are all gray? A) Gg GG B) Gg Gg C) gg GG D) gg Gg 16. What is the gene combination of the offspring from the parents shown above? A) 50% GG; 50% Gg B) 25% GG; 75% Gg C) 100% Gg D) 25% GG; 50% Gg; 25% gg 17. If two gray mice (Gg) mated what percentage of their offspring would be white? A) 25% B) 50% C) 0% D) 75% Version 1 Page 4

5 Base your answers to questions 18 through 20 on the diagram below which shows a model of inheritance for height in pea plants. 18. What height would each parent be? A) The male would be tall and the female short. B) The male would be short and the female tall. C) The male would be short and the female short. D) The male would be tall and the female tall. 19. If these two parents were to produce 50 offspring, what percentage would have the genetic makeup to be tall plants? A) 0% B) 25% C) 50% 20. Identify what two offspring from the cross shown above could produce offspring of short plants. A) TT TT B) TT Tt C) Tt Tt D) It wouldn't be possible. Version 1 Page 5

6 Base your answers to questions 21 through 24 on on the diagram below which shows the offspring of a rabbit with short ears with a rabbit with big ears. The gene for big ears is dominant. 21. Using the Punnett Square shown above show the gene combinations of the parents and the resulting offspring that produced the baby rabbits shown above. 22. Show all the possible gene combinations (ie. EE, Ee or ee) that could be present in the following rabbits. Short ear rabbit: Long ear rabbit: 23. State one advantage the rabbit has by having big ears. 24. State one advantage that a species that produces sexually has over a species that produces asexually. Version 1 Page 6

7 Answer Key 1. C 2. C 3. D 4. D 5. C 6. C 7. C 22. Short ear rabbit: ee Long ear rabbit: EE or Ee 23. Big ears will let the rabbit hear better or move around to listen in different directions. 24. It allows for different traits to occur which may lead an advantage for the species. 8. C 9. A 10. C 11. A 12. C 13. C 14. A 15. C 16. C 17. A 18. D 19. D 20. C 21. Version 1

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