Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. |
| 1. | The “father” of genetics was a. | T. A. Knight. | c. | Gregor Mendel. | b. | Hans Krebs. | d. | None of the above | | | | |
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| 2. | Mendel obtained his P generation by allowing the plants to a. | self-pollinate. | c. | assort independently. | b. | cross-pollinate. | d. | segregate. | | | | |
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| 3. | What is the probability that the offspring of a homozygous dominant individual and a homozygous recessive individual will exhibit the dominant phenotype? |
| 4. | True-breeding pea plants always a. | are pollinated by hand. | b. | produce offspring with either form of a trait. | c. | produce offspring with only one form of a trait. | d. | are heterozygous. | | |
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| 5. | The first filial (F1) generation is the result of a. | cross-pollination among parents and the next generation. | b. | crosses between individuals of the parental generation. | c. | crosses between the offspring of a parental cross. | d. | self-fertilization between parental stock. | | |
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| 6. | Which of the following is the designation for Mendel’s original pure strains of plants? |
| 7. | A genetic trait that appears in every generation of offspring is called a. | dominant. | c. | recessive. | b. | phenotypic. | d. | superior. | | | | |
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| 8. | To describe how traits can disappear and reappear in a certain pattern from generation to generation, Mendel proposed a. | the law of independent assortment. | b. | the law of segregation. | c. | the law of genotypes. | d. | that the F2 generation will only produce purple flowers. | | |
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| 9. | When Mendel crossed pea plants with two contrasting traits, such as flower color and plant height, a. | these experiments led to his law of segregation. | b. | he found that the inheritance of one trait did not influence the inheritance of the other trait. | c. | he found that the inheritance of one trait influenced the inheritance of the other trait. | d. | these experiments were considered failures because the importance of his work was not recognized. | | |
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| 10. | If an individual has two recessive alleles for the same trait, the individual is said to be a. | homozygous for the trait. | c. | heterozygous for the trait. | b. | haploid for the trait. | d. | mutated. | | | | |
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| 11. | An individual heterozygous for a trait and an individual homozygous recessive for the trait are crossed and produce many offspring that are a. | all the same genotype. | c. | of three different phenotypes. | b. | of two different phenotypes. | d. | all the same phenotype. | | | | |
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| | In humans, having freckles (F) is dominant to not having freckles (f). The inheritance of these traits can be studied using a Punnett square similar to the one shown below. |
| 12. | Refer to the illustration above. The genotype represented in box “1” in the Punnett square would a. | be homozygous for freckles. | b. | have an extra freckles chromosome. | c. | be heterozygous for freckles. | d. | have freckles chromosomes. | | |
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| 13. | Refer to the illustration above. The genotype in box “3” of the Punnett square is a. | FF. | c. | ff. | b. | Ff. | d. | None of the above | | | | |
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| 14. | How many different phenotypes can be produced by a pair of codominant alleles? |
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| 15. | Refer to the illustration above. The phenotype represented by the cell labeled “1” is a. | green, inflated. | c. | yellow, inflated. | b. | green, constricted. | d. | yellow, constricted. | | | | |
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| 16. | 2,000 yellow seeds : 8,000 total seeds :: a. | 1 : 6 | c. | 1 : 3 | b. | 1 : 8 | d. | 1 : 4 | | | | |
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| | In rabbits, black fur (B) is dominant to brown fur (b). Consider the following cross between two rabbits. |
| 17. | Refer to the illustration above. Both of the parents in the cross are a. | black. | c. | homozygous dominant. | b. | brown. | d. | homozygous recessive. | | | | |
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| 18. | Refer to the illustration above. The phenotype of the offspring indicated by box “3” would be a. | brown. | c. | a mixture of brown and black. | b. | black. | d. | The phenotype cannot be determined. | | | | |
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| 19. | What is the expected phenotypic ratio resulting from a homozygous dominant ´ heterozygous monohybrid cross? |
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| 20. | Refer to the illustration above. The phenotype represented by the cell labeled “1” is a. | round, yellow. | c. | wrinkled, yellow. | b. | round, green. | d. | wrinkled, green. | | | | |
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| 21. | An organism that has inherited two of the same alleles of a gene from its parents is called a. | hereditary. | c. | homozygous. | b. | heterozygous. | d. | a mutation. | | | | |
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| 22. | In pea plants, yellow seeds are dominant over green seeds. What would be the expected genotype ratio in a cross between a plant with green seeds and a plant that is heterozygous for seed color? |
| 23. | codominance : both traits are displayed :: a. | probability : crosses | b. | heterozygous : alleles are the same | c. | homozygous : alleles are the same | d. | Punnett square : chromosomes combine | | |
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| 24. | The difference between a monohybrid cross and a dihybrid cross is that a. | monohybrid crosses involve traits for which only one allele exists, while dihybrid traits involve two alleles. | b. | monohybrid crosses involve self-pollination, while dihybrid crosses involve cross-pollination. | c. | monohybrid crosses involve one gene; dihybrid crosses involve two genes. | d. | dihybrid crosses require two Punnett squares; monohybrid crosses need only one. | | |
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| 25. | What fraction of the offspring resulting from a heterozygous ´ heterozygous dihybrid cross are heterozygous for both traits? |