The correct answer is c B. Answer b is incorrect. True breeding involves self-fertilization, not fertilization between two different plants.

Transcription

1 1. A true-breeding plant is one that a. produces offspring that are different from the parent b. forms hybrid offspring through cross-pollination c. produces offspring that are always the same as the parent d. can only reproduce with itself A. Answer a is incorrect. A true breeding plant will produce offspring that are identical to the parent B. Answer b is incorrect. True breeding involves self-fertilization, not fertilization between two different plants. produces offspring that are always the same as the parent C. Answer c is correct. The offspring are genetically identical to the parent. D. Answer d is incorrect. True breeding only means that the offspring produced by selffertilization are identical. These plants can be crossed with other true-breeding plants. 2. What property distinguished Mendel s investigation from previous studies? a. Mendel used true-breeding pea plants. b. Mendel quantified his results. c. Mendel examined many different traits. d. Mendel examined the segregation of traits. A. Answer a is not correct. Other early researchers had used pea plants. Mendel quantified his results. B. Answer b is correct. The fact that Mendel determined the ratio of different phenotypic traits led to our current understanding of inheritance. C. Answer c is not correct. Other early researchers had examined a variety of traits. D. Answer d is not correct. The segregation of traits had been observed by others. 3. A monohybrid cross a. is the same as self-fertilization b. examines a single variant of a trait c. produces a single offspring d. examines two variants of a single trait A. Answer a is not correct. Monohybrid crosses can be made between two individuals.

2 B. Answer b is not correct. A monohybrid cross examines two versions of a single trait. C. Answer c is not correct. The term monohybrid refers to the trait being examined, not the number of offspring. examines two variants of a single trait D. Answer d is correct. A monohybrid cross examines a single trait with two versions (e.g., purple versus white flowers). 4. What was the appearance of the F 1 generation of a monohybrid cross of purple (PP) and white (pp) flower pea plants? a. All the F 1 plants had white flowers. b. The F 1 plants had a light purple or blended appearance. c. All the F 1 plants had purple flowers. d. The most of the F 1 (3/4) had purple flowers, but 1/4 of the plants had white. A. Answer a is incorrect. The purple flower color is the dominant trait, so the F 1 generation (Pp) cannot appear white. B. Answer b is incorrect. The phenotype for flower color did not blend. Rather, it provided evidence for the principle of segregation. All the F 1 plants had purple flowers. C. Answer c is correct. The genotype of the F 1 generation would be Pp, meaning that these individuals would be heterozygous and would have the appearance of the dominant phenotype. D. Answer d is incorrect. A plant with white flowers would have to be homozygous recessive. All of the F 1 individuals from this monohybrid cross would be heterozygous. 5. The F 1 plants from the previous question are allowed to self-fertilize. What will the phenotypic ratio be for this F 2? a. All purple b. 1 purple:1 white c. 3 purple:1 white d. 3 white:1 purple A. Answer a is incorrect. The F 1 is all purple, but segregation will occur in the F 2. B. Answer b is incorrect. The phenotypic ratio for a test cross of the F 1 heterozygote would be 1 purple:1 white, but not for a self-cross.

3 3 purple:1 white C. Answer c is correct. Self-crossing a heterozygous F 1 would produce a ratio of 3 dominant to 1 recessive. D. Answer d is incorrect. The ratio is correct in terms of numbers (3:1), but the white trait is recessive and the purple trait is dominant (see previous problem) so the ratio is reversed. 6. Which of the following is NOT a part of Mendel s five-element model? a. Traits have alternative forms (what we now call alleles). b. Parents transmit discrete traits to their offspring. c. If an allele is present it will be expressed. d. Traits do not blend. A. Answer a is incorrect. The fact that traits have alternative forms was critical for Mendel s observations. B. Answer b is incorrect. Mendel concluded that parents transmitted discrete units of information that determined the traits of the offspring. If an allele is present it will be expressed. C. Answer c is correct. Alleles are not always expressed. Some alleles are dominant while others are recessive. D. Answer d is incorrect. Mendel did not observe blending of traits. 7. A heterozygous individual is one that carries a. two completely different sets of genes b. two identical alleles for a particular gene c. only one functional allele d. two different alleles for a given gene A. Answer a is incorrect. The concept of heterozygosity relates to a single gene, not the entire genome. B. Answer b is incorrect. A heterozygous individual has two different versions of a single gene. C. Answer c is incorrect. A heterozygous individual has two alleles; however, one is dominant while the other is recessive. two different alleles for a given gene

4 D. Answer d is correct. A heterozygous individual has a dominant and a recessive copy of a single allele. 8. An organism s is determined by its. a. genotype; phenotype b. phenotype; genotype c. alleles; phenotype d. F 1 generation; alleles A. Answer a is incorrect. The genotype is the controlling factor that determines phenotype. phenotype; genotype B. Answer b is correct. The phenotype of an individual is determined by the alleles that make up that individual s genotype. C. Answer c is incorrect. The alleles are the genes that determine the phenotype of the individual D. Answer d is incorrect. The F 1 generation is the product of reproduction and so would depend on the genetic makeup of the gametes. 9. Which of the following represent the phenotype for the recessive human trait, albinism? a. Absence of the pigment melanin b. Presence of a nonfunctional allele for the enzyme tyrosinase c. Absence of the enzyme tyrosinase from the individual s cells d. Both a and c A. Answer a is incorrect. Although albinism is associated with a lack of pigment, that is not the only phenotypic trait. B. Answer b is incorrect. The allele for the enzyme tyrosinase is nonfunctional; however, this represents the genotype of the individual, not phenotype. C. Answer c is incorrect. The absence of the enzyme is part of the phenotype; however, it does not represent the total phenotype. Both a and c D. Answer d is correct. The phenotype represents all observed effects of the genotype; that is, the absence of the enzyme and the resulting absence of pigment. 10. A dihybrid cross between a plant with long smooth leaves and a plant with short hairy leaves produces a long smooth F 1. If this F 1 is allowed to self-cross to produce an F 2, what would you predict for the ratio of F 2 phenotypes?

5 a. 9 long smooth:3 long hairy:3 short hairy:1 short smooth b. 9 long smooth:3 long hairy:3 short smooth:1 short hairy c. 9 short hairy:3 long hairy:3 short smooth:1 long smooth d. 1 long smooth:1 long hairy:1 short smooth:1 short hairy A. Answer a is incorrect. The double-recessive phenotype (short hairy) is switched with recessive for length and dominant for texture (short smooth). 9 long smooth:3 long hairy:3 short smooth:1 short hairy B. Answer b is correct. The phenotypic ratio should be 9 dom dom:3 dom rec:3 rec dom:1 rec rec. C. Answer c is incorrect. The double-dominant phenotype (long smooth) is switched with the double recessive (short hairy). D. Answer d is incorrect. This 1:1:1:1 ratio would be seen in a test cross not in a self-cross of the F A testcross is used to determine if an individual is a. homozygous dominant or heterozygous b. homozygous recessive or homozygous dominant c. heterozygous d. true breeding homozygous dominant or heterozygous A. Answer a is correct. The ratio of phenotypes produced from a testcross allows for the determination of whether an individual is homozygous dominant or heterozygous. B. Answer b is incorrect. The phenotypes of these two individuals would be very different. No testcross would be needed. C. Answer c is incorrect. The phenotypic ratio resulting from a testcross would establish whether an individual was heterozygous, but it can also be used to distinguish between heterozygous and homozygous dominant individuals. D. Answer d is incorrect. A testcross is used to determine whether an individual is heterozygous or homozygous dominant. The test for whether an individual is true breeding is whether self-fertilization results in offspring that are identical to the parent. 12. What is a polygenic trait? a. A set of multiple phenotypes determined by a single gene. b. A single phenotypic trait determined by a two alleles. c. A single phenotypic trait determined by more than one gene.

6 d. The collection of traits possessed by an individual. A. Answer a is incorrect. A polygenic trait is a single phenotypic trait. B. Answer b is incorrect. An allele is just a variation of a given gene. A polygenic trait involves multiple genes. A single phenotypic trait determined by more than one gene. C. Answer c is correct. Polygenic means that more than one gene is involved in determining a phenotypic trait. D. Answer d is incorrect. A polygenic trait refers to a single phenotypic trait, not many traits. 13. When a single gene influences multiple phenotypic traits the effect is called a. codominance b. epistasis c. incomplete dominance d. pleiotropy A. Answer a is incorrect. Codominance occurs when the influence of different alleles are equally expressed in the heterozygous individual. B. Answer b is incorrect. Epistasis occurs when the phenotype of an individual is influenced by the interactions between gene products. C. Answer c is incorrect. Incomplete dominance occurs when the phenotype of the heterozygous individual represents a blending of the dominant and recessive traits. pleiotropy D. Answer d is correct. Some genes can influence multiple phenotypic traits. These are pleiotropic genes. 14. What is the probability of obtaining an individual with the genotype bb from a cross between two individuals with the genotype Bb? a. 1/2 b. 1/4 c. 1/8 d. 0 A. Answer a is incorrect. For an individual with the genotype Bb the probability of forming a b gamete is 1/2, so the probability of bb offspring is (1/2)(1/2) = 1/4. Overall, in this cross: the progeny should be 1/4 BB, 1/2 Bb, and 1/4 bb.

7 1/4 B. Answer b is correct. The probability of receiving a b allele from each parent is 1/2. 1/2 (male) x 1/2 (female) = 1/4. Overall, in this cross: the progeny should be 1/4 BB, 1/2 Bb, and 1/4 bb. C. Answer c is incorrect. For an individual with the genotype Bb the probability of forming a b gamete is 1/2, so the probability of bb offspring is (1/2)(1/2) = 1/4. Overall, in this cross: the progeny should be 1/4 BB, 1/2 Bb, and 1/4 bb. D. Answer d is incorrect. For an individual with the genotype Bb the probability of forming a b gamete is 1/2, so the probability of bb offspring is (1/2)(1/2) = 1/4. Overall, in this cross: the progeny should be 1/4 BB, 1/2 Bb, and 1/4 bb. 15. What is the probability of obtaining an individual with the genotype CC from a cross between two individuals with the genotypes CC and Cc? a. 1/2 b. 1/4 c. 1/8 d. 1/16 1/2 A. Answer a is correct. The probability of the homozygous dominant parent generating a gamete carrying a C allele is 1. The probability of producing a C gamete for the heterozygous parent is 1/2. B. Answer b is incorrect. The probability of the homozygous dominant parent generating a gamete carrying a C allele is 1. The probability of producing a C gamete for the heterozygous parent is 1/2. The product rule applies so 1 x 1/2 = 1/2. C. Answer c is incorrect. The probability of the homozygous dominant parent generating a gamete carrying a C allele is 1. The probability of producing a C gamete for the heterozygous parent is 1/2. The product rule applies so 1 x 1/2 = 1/2. D. Answer d is incorrect. The probability of the homozygous dominant parent generating a gamete carrying a C allele is 1. The probability of producing a C gamete for the heterozygous parent is 1/2. The product rule applies so 1 x 1/2 = 1/2.

8 Challenge Questions 1. Create a Punnett square for the following crosses and use this to predict phenotypic ratio for dominant and recessive traits. Dominant alleles are indicated by uppercase letters and recessive are indicated by lowercase letters. a. A monohybrid cross between individuals with the genotype Aa and Aa b. A dihybrid cross between two individuals with the genotype AaBb c. A dihybrid cross between individuals with the genotype AaBb and aabb The approach to solving this type of problem is to identify the possible gametes. Each individual parent will donate one allele to the offspring. Separate the possible gamete combinations into the boxes along the top and side. Fill in the Punnet square by carrying each letter representing each allele into the boxes below and to the right for each column and row. As a general rule the dominant allele (represented by a capital letter) is always listed first. Notice that if an individual only makes one type of gamete, it is not necessary to use more than one row or column for this individual. Answer for 1a A a A AA Aa a Aa aa Phenotypic ratio: 3 dominant to 1 recessive Answer for 1b AB Ab ab ab AB AABB AABb AaBB AaBb Ab AABb AAbb AaBb Aabb ab AaBB AaBb aabb aabb ab AaBb Aabb aabb aabb Phenotypic ratio: 9 dominant dominant to 3 dominant recessive to 3 recessive dominant to 1 recessive recessive Answer for 1c AB Ab ab ab

9 ab AaBb Aabb aabb aabb Phenotypic ratio: 1 dominant dominant to 1 dominant recessive to 1 recessive dominant to 1 recessive recessive 2. Use probability to predict the following. a. What is the probability of obtaining an individual with the genotype rr from the self-fertilization of a plant with the genotype Rr? b. What is the probability that a testcross with a heterozygous individual will produce homozygous recessive offspring? c. A plant with the genotype Gg is self-fertilized. Use probability to determine the proportion of the offspring that will have the dominant phenotype. d. Use probability to determine the proportion of offspring from a dihybrid cross (GgRr x GgRr) that will have the phenotype ggr_. Answer for 2a One half of the gametes produced by this plant will have the recessive allele r. The probability of obtaining an rr individual is the product of the probabilities of finding a gamete carrying the r allele. 1/2 x 1/2 = 1/4. Answer for 2b The testcross involves the use of a homozygous recessive individual to cross with an unknown individual. The problem states that the second parent in heterozygous. The probability of obtaining a recessive allele from the homozygous recessive individual is 1. The probability of obtaining a recessive allele from a heterozygous individual is 1/2. 1 x 1/2 = 1/2. The probability is that 1/2 of the offspring from a testcross with a heterozygous individual will be homozygous recessive. Answer for 2c The dominant phenotype (G_) will occur for individuals for individuals with the genotypes GG, Gg, or gg. There is a 1/4 probability of combining a G gamete with another G gamete. Likewise, there is a 1/4 probability of combining G with g or g with G. The rule of addition applies 1/4 + 1/4 + 1/4 = 3/4. Answer for 2d When considering dihybrid crosses it is important to remember that the dominant phenotype will occur 3/4 of the time, and the recessive phenotype only appears 1/4 of the time (the ratios of a simple monohybrid cross). Assuming that the alleles behave independently, then the rule of multiplication applies. There is a 1/4 chance of obtaining the combination gg because this is the recessive phenotype. There is a 3/4 chance of obtaining the combination R_ since this is the dominant phenotype. 3/4 x 1/4 = 3/16.