(A) Type AB only. (B) Type A or Type B only. (C) Type A, AB, and B only. (D) All four types are possible: type A, AB, B or O.

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1 Genetics - Problem Drill 02:Mendelian Genetics and its Extensions No. 1 of In the case of a couple, where the husband has type A blood and the wife has type B, the blood types of their children should be. (A) Type AB only. (B) Type A or Type B only. (C) Type A, AB, and B only. (D) All four types are possible: type A, AB, B or O. (E) There is not enough information to deduce their children s blood type. If the husband has AO genotype, they may have type B children. No matter what kind of genotype the parents have in this case, there is possibility for their children to be type AB. If the parents have genotype AO and BO, then their children may have all four blood types. This happens only when the parents have AO and BO genotype; if they don t, for example, the father has type AA, mother has type BB, and then their children will be all type AB. E. Correct! The children s blood type is decided by parent s genotype; in this case, there is not enough information to predict the children s genotype or blood type. The blood type is determined by three alleles, A, B and O. A and B are dominant to O, while they are co-dominant to each other. Therefore, a person with AA or AO allele would have type A; BB or BO type would have type B; AB genotypes have AB type; and OO genotypes have O type. The genotype of the parents determines the children s genotype and phenotype (blood type). Type A and type B parents can have different combinations: AA x BB, AO x BB, AA x BO or AO x BO; each combination will have a different possibility for their children s blood type. (E) There is not enough information to deduce their children's blood type.

2 No. 2 of Below is a brain disease pedigree; which of the following statement is true? (A) This is a dominant trait, and II3 has a genotype of AA. (B) This is a recessive trait, and II3 has a genotype of aa. (C) This is a recessive trait, II6 has a genotype of AA. (D) This is a dominant trait, II6 has a genotype of aa. (E) This is a recessive trait, III4 has a genotype of aa. I2 I4 II3 II6 III This is a dominant trait because it does not skip a generation. However, II3 genotype must be Aa because his children are all normal. This is not a recessive trait because some children from two affected individuals (I3 and I4) have normal phenotype (not affected). If it is recessive, I3 and I4 would have genotype of aa and aa. Consequently, all children from this marriage would be affected. This is not a recessive trait. Therefore, the normal II6 cannot have a genotype of AA. II6 has aa recessive genotype. D. Correct! This is a dominant trait; the unaffected individuals should have genotype aa, so does II6. This is not a recessive trait, and III4 should have a genotype of Aa, not aa. One way of distinguishing between a dominant trait and a recessive trait is to see if this trait skips a generation. This disease does not skip a generation, so it is most likely dominant. Although the marriage from II3 and II6 generated all healthy children, it is by chance (II3 has genotype of Aa and II6 has aa, they have 50% to be normal). (D) This is a dominant trait, II6 has a genotype of aa.

3 No. 3 of Cross 1: blue-eyed, long-toothed mouse X brown-eyed, short-toothed mouse gives F1: all blue-eyed, short-toothed. Cross 2: blue-eyed, short-toothed F1 X blue-eyed, short-toothed F1 gives F2: 94 blue-eyed short-toothed; 31 blue-eyed long-toothed; 28 brown-eyed shorttoothed; 8 brown-eyed long-toothed. Which of the following statements is NOT correct? (A) Blue is dominant to brown eye, and short-toothed is dominant to longtoothed. (B) The parental genotypes are BBss and bbss (B for blue, b for brown, S for short-toothed and s for long-toothed). (C) Among the F2 94 blue-eyed short-toothed mice, there are 1/3 that are homozygous for B allele (BB). (D) Among F2 31 blue-eyed long-toothed, there are 1/4 of the mice which are homozygous for B allele (BB). (E) All brown-eyed long-toothed F2 individuals have bbss genotype. This statement is correct, so A is not the right choice. Blue is dominant to brown, and short-toothed is dominant to long-toothed because F1 is all blue eyed and short-toothed. This statement is correct, so B is not the right choice. Since the F1 all have the same phenotype, their parents should have been homozygous for each allele. The F2 blue-eyed, short-toothed have B_S_ genotype; the correct answer is 1/3 of them have BB genotype (homozygous for BB). D. Correct! The ratio is wrong; there should be 1/3 of these mice, which have the BB genotype. This statement is true; all brown-eyed long-toothed mice have bbss genotype. For F2 blue-eyed short-toothed mice, the genotype is 1 BBSS 2 BBSs 4 BbSs 2 BbSS Therefore, homozygous BB for these mice account for 3/9 = 1/3. (D) Among F2 31 blue-eyed long-toothed, there are 1/4 of the mice which are homozygous for B alele l (BB).

4 No. 4 of To identify the genotype of yellow-seeded pea plants as either homozygous dominant (YY) or heterozygous (Yy), you could do a test cross with plants of genotype? (A) YY (B) Yy (C) yy (D) either B or C (E) None of the above A YY plant cross with any other plants will only generate yellow seeds and you wouldn t be able to tell the difference between Yy and YY. This is because Y is dominant and will be expressed, regardless of the other allele present. Yy cross with YY generates all yellow seeds, while cross with Yy generate 3:1 yellow:green seeds. However, this is not a test cross because a test cross is always done with a recessive homozygous parent. C. Correct! A test cross is done with a recessive homozygous parent. When YY cross with yy, all progeny will be Yy (yellow0; if it is Yy, then the progeny will segregate yellow: green = 1:1. As pointed out in B and C, YY or Yy cannot work. There is one correct answer above. For any genetic testcross, it is desirable to choose homozygous recessive genotype as the testing type because this allows the offspring phenotype ratio to equal the gamete ratio of the tested individual. (C) yy

5 No. 5 of A genetic cross between two F1-hybrid pea plants having yellow seeds will yield what percent green-seeded plants in the F2 generation? Yellow seeds are dominant to green. (A) 25% (B) 50% (C) 75% (D) 100% (E) None of the above A. Correct! F1 would have genotype of Yy. When Yy is crossed with another Yy, there would be ¼ of yy, 2/4 of Yy and another quarter of YY. Therefore, the answer is 25% (1/4). 50% (1:1 ratio) usually appears in a backcross, not an intercross between F1. 75% is ¾; this ratio normally appears in a dominant trait, but green is recessive here. Only when a homozygous dominant allele is crossed to other genotypes, 1:0 ratio (100%) can occur. Here, it is a heterozygous cross; therefore, it is not 100%. There is one correct answer above. To solve any genotype-phenotype problem, it is always useful to write out the genotype, the gametes, and gamete ratio and then deduce the offspring phenotype. (A)25%

6 No. 6 of Which of the following statements about genetic terms is true? (A) Homologs are a pair of nearly identical chromosomes, one inherited from the mother and one inherited from the father. (B) Homologs are a pair of absolutely identical chromosomes, one inherited from the mother and one inherited from the father. (C) An allele is defined as the copy of a gene on a chromosome that is always identical to all other copies. (D) The allele of a gene (from a 2N cell) can only be written as either: AA or Aa. (E) The allele of a gene (from a 2N cell) can only be written as either: AA or aa. A. Correct! Homologs are a pair of nearly identical chromosomes, one inherited from the mother and one inherited from the father. Homologs are a pair of nearly identical chromosomes, one inherited from the mother and one inherited from the father. An allele is defined as the copy of a gene on a chromosome that can either be the same or different than the other copies. The allele of a gene can be written as: AA, aa, Aa. The allele of a gene can be written as: AA, aa, Aa. Homologs are a pair of nearly identical chromosomes, one from the father and the other from the mother. Allele: each homologous chromosome has one copy of a gene (one allele). A 2N cell, having a pair of homologous chromosomes, carries two alleles, one for each gene. These alleles can be identical or different: AA, aa or Aa. (A) Homologs are a pair of nearly identical chromosomes, one inherited from the mother and one inherited from the father.

7 No. 7 of A gamete cell. (A) Is haploid and has two complete sets of chromosomes. (B) Is haploid and has only one set of chromosomes. (C) According to the law of independent assortment, is formed when the genes from the same chromosome assort independently and combine randomly. (D) According to the law of independent assortment, is formed when the genes from different chromosomes assort independently and combine in a predetermined manner. (E) Can exist anywhere in the body and represents the major cell type in human tissue. A gamete cell is haploid and has only one set of chromosomes. B. Correct! A gamete cell is haploid and has only one set of chromosomes. According to the law of independent assortment, is formed when the genes from different chromosomes assort independently and combine randomly. According to the law of independent assortment, is formed when the genes from different chromosomes assort independently and combine randomly. A gamete cell is made in the gonads, and a sperm cell or egg cell only exists inside the genitourinary tract. Gametes are formed during meiosis and exist as either sperm or egg cells in humans. When formed from a diploid organism, they are haploid with a single set of chromosomes. (B) Is haploid and has only one set of chromosomes.

8 No. 8 of The disease sickle cell anemia is an example of co-dominance. Which of the following statements is true? (A) When two alleles are co-dominant, the phenotype of each allele is not masked by the other. (B) When two alleles are co-dominant, the phenotype of one allele is masked by the other. (C) In an individual with sickle cell anemia, it is not possible to have 50% normal and 50% abnormal red blood cells. (D) In an individual with sickle cell anemia, only the abnormal phenotype can be expressed. (E) The genotype of sickle cell anemia patients would be Hbb:Hbb. A. Correct! When two alleles are co-dominant, the phenotype of each allele is not masked by the other. When two alleles are co-dominant, the phenotype of each allele is not masked by the other. In an individual with sickle cell anemia, it is possible to have 50% normal and 50% abnormal red blood cells. In an individual with sickle cell anemia, both the abnormal phenotype and the normal red blood cell phenotype can be expressed. Although some of the cells can be normal, there can also be abnormal red blood cells with the genotype HbS. When two alleles are co-dominant, the phenotype of each allele is not masked by the presence of the other alleles. Sickle cell anemia is an example of co-dominant alleles. Both phenotypes are expressed. (A) When two alleles are co-dominant, the phenotype of each allele is not masked by the other.

9 No. 9 of The chi-square test (calculated with the formula below). χ 2 = Σ d 2 e (A) Measures the probability that there is no significant difference between the expected frequency of an occurrence and the observed frequency of an occurrence. (B) Measures the probability that there is a small difference between the expected frequency of an occurrence and the square root of the observed frequency of an occurrence. (C) Is a goodness-of-fit test to see how well the expected data fits into the original hypothesis. (D) Is a formula that includes d 2, which is equal to the expected number multiplied by the expected number. (E) Is a formula that includes d 2, which is equal to the observed number multiplied by the observed number. A. Correct! The chi-square test measures the probability that there is no significant difference between the expected frequency of an occurrence and the observed frequency of an occurrence. The chi-square test measures the probability that there is no significant difference between the expected frequency of an occurrence and the observed frequency of an occurrence. The chi-square test is a goodness-of-fit test to see how well the experimental data fits into the original hypothesis. The chi-square formula includes d 2, which is equal to the (observed number minus the expected number) 2. The chi-square formula includes d 2, which is equal to the (observed number minus the expected number) 2. The chi square test measures the probability that there is no significant difference between the expected frequency of an occurrence and the observed frequency of that occurrence. It is a goodness-of-fit test to see how good the experimental data can fit into the hypothesis. Probability can be found in chi square table with a defined degree of freedom (df), which equals to n-1 (n class of object). (A)Measures the probability that there is no significant difference between the expected frequency of an occurrence and the observed frequency of an occurrence.

10 No. 10 of Which of the following statements about gene epistasis is true (see figure below)? (A) Epistasis is when one gene s function is modified by all the other genes. (B) Using the figure above, if Gene A is inactivated due to a mutation, then there is blue color produced. (C) Using the figure above, if Gene A is inactivated due to a mutation, then there is no blue color produced. (D) If gene B was mutated and inactivated, based on the above figure, blue color would still be produced. (E) If gene B was mutated but Gene A was not, based on the above figure, blue color would still be produced. Epistasis is when one gene s function is modified by one or more other genes. If Gene A is inactivated due to a mutation, then there is no blue color produced. C. Correct! If Gene A is inactivated due to a mutation, then there is no blue color produced. If gene B was mutated and inactivated, based on the above figure, blue color would not be produced. If gene B was mutated and inactivated, based on the above figure, blue color would not be produced. Epistasis is the interaction between genes, whereby the effects of one gene are modified by one or more other genes. In the figure given in this question, blue color would not be produced if either Gene A or Gene B were mutated and inactivated. If only Gene A were mutated, even though Gene B was not its function, it would be masked by the change in function of Gene A. (C)Using the figure above, if Gene A is inactivated due to a mutation, then there is no blue color produced.