The correct answer is c A. Answer a is incorrect. The white-eye gene must be recessive since heterozygous females have red eyes.

Transcription

1 1. Why is the white-eye phenotype always observed in males carrying the white-eye allele? a. Because the trait is dominant b. Because the trait is recessive c. Because the allele is located on the X chromosome and males only have one X d. Because the allele is located on the Y chromosome and only males have Y chromosomes A. Answer a is incorrect. The white-eye gene must be recessive since heterozygous females have red eyes. B. Answer b is incorrect. The white-eye gene is recessive; however, this does not explain why it is always observed in males. Because the allele is located on the X chromosome and males only have one X C. Answer c is correct. Because the gene is located on the X chromosome, and the Y chromosome lacks most X chromosome genes, any male carrying the white-eye allele will show the phenotype. D. Answer d is incorrect. The allele is located on the X chromosome as demonstrated by the testcross. The Y chromosome does not carry many genes that are expressed. 2. An autosome is a chromosome that a. contains genetic information to determine the sex of an organism b. determines all other traits of an organism other than sex c. is only inherited from the mother (maternal inheritance) d. has no matching chromosome within an organism s genome A. Answer a is incorrect. The sex of an organism is determined by the sex chromosomes. determines all other traits of an organisms other than sex B. Answer b is correct. Autosomes are the chromosomes that carry the majority of genetic information that determines the overall phenotype of an organism. The sex chromosomes, in contrast, determine the sex of the organism. C. Answer c is incorrect. The autosomes are inherited from both parents. D. Answer d is incorrect. The autosomes exist as homologous pairs. 3. Sex linkage in humans occurs when a. an allele is located on both the X and Y chromosomes b. all allele is located on the X chromosome c. an allele is located on an autosome

2 d. a phenotype is only observed in females A. Answer a is incorrect. An allele is sex-linked if it is present on only one of the sex chromosomes. Few of the genes found on the Y chromosome are expressed. an allele is located on the X chromosome B. Answer b is correct. The phenotype of genes present on the X chromosome will be observed in human males since there is not a second X available. Males cannot be heterozygous for a recessive allele located on the X chromosome. C. Answer c is incorrect. Autosomal inheritance is the usual mode of inheritance and does not include sex-linked traits. D. Answer d is incorrect. Female humans are XX. There are two alleles for any gene present on the X chromosome; this will prevent the expression of a recessive phenotype in heterozygous individuals. 4. What are Barr bodies? a. Barr bodies are X chromosomes inactivated to prevent overexpression of the alleles found on the X chromosome in females. b. Barr bodies are highly condensed Y chromosomes in males. c. Barr bodies are X chromosomes inactivated to allow for expression of the malesspecific phenotype. d. Barr bodies are inactive autosomal chromosomes specific to females. Barr bodies are X chromosomes inactivated to prevent overexpression of the alleles found on the X chromosome in females. A. Answer a is correct. Inactivation of one X chromosome in females allows for dosage compensation. B. Answer b is incorrect. Although Y chromosomes are highly condensed, a Barr body is an inactivated X chromosome. C. Answer c is incorrect. X inactivation only occurs in females to allow for dosage compensation. D. Answer d is incorrect. Inactivation only applies to the X chromosome, not to autosomal chromosomes. 5. How does maternal inheritance of mitochondrial genes differ from sex linkage? a. Mitochondrial genes do not contribute to the phenotype of an individual. b. Because mitochondria are inherited from the mother, only females are affected.

3 c. Since mitochondria are inherited from the mother, females and males are equally affected. d. Mitochondrial genes must be dominant. Sex-linked traits are typically recessive. A. Answer a is incorrect. Mitochondrial genes influence the phenotype of an individual since they can alter the function of the individual s mitochondria. B. Answer b is incorrect. Maternal inheritance means that both males and females receive mitochondria from their mothers. Since mitochondria are inherited from the mother, females and males are equally affected. C. Answer c is correct. The mitochondria found in the fertilized egg become the mitochondria for the new individual, independent of gender. D. Answer d is incorrect. The idea of dominant and recessive does not apply in this case since there are no homologous mitochondria. 6. What cellular process is responsible for genetic recombination? a. Independent assortment b. Separation of the homologues in meiosis 1 c. Separation of the chromatids during meiosis II d. Crossing over A. Answer a is incorrect. Independent assortment refers to the distribution of homologous chromosomes during meiosis. It does not result in genetic recombination. B. Answer b is incorrect. Genetic recombination refers to changes in the genes present on a single chromosome. C. Answer c is incorrect. Anaphase of meiosis II does not alter the genetic makeup of the sister chromatids. Crossing over D. Answer d is correct. Crossing over results in a physical change in the genes present on a chromosome. 7. The number of map units between two genes is determined by a. the recombination frequency b. the frequency of parental types c. the total number of genes within a given piece of DNA d. the number of linked genes within a chromosome

4 the recombination frequency A. Answer a is correct. Map units are defined as percent recombination. B. Answer b is incorrect. The recombination frequency is the value used to determine map units. C. Answer c is incorrect. The number of genes on a chromosome is not relevant to the determination of map units. D. Answer d is incorrect. The number of linked genes is not relevant to the determination of map units. 8. How many map units separate two alleles if the recombination frequency is 0.07? a. 700 cm b. 70 cm c. 7 cm d. 0.7 cm A. Answer a is incorrect. One centimorgan (cm) is equal to a recombination frequency of 1%. The recombination frequency would have to be 700% for this answer to be correct. B. Answer b is incorrect. One centimorgan (cm) is equal to a recombination frequency of 1%. The recombination frequency would have to be 70% for this answer to be correct. 7 cm C. Answer c is correct. One centimorgan (cm) is equal to a recombination frequency of 1%. A recombination frequency of 0.07% equals 7 cm. D. Answer d is incorrect. One centimorgan (cm) is equal to a recombination frequency of 1%. The recombination frequency would have to be for this answer to be correct. 9. Multiple crossovers lead to a. restoration of the paternal combination of genes b. increased genetic diversity c. increased numbers of recombinant progeny d. anueploidy restoration of the paternal combination of genes A. Answer a is correct. Two crossovers will return the alleles to their original location. B. Answer b is incorrect. Crossovers represent exchanges between homologous chromosomes; they do not alter the genes themselves.

5 C. Answer c is incorrect. Multiple crossovers can restore the parental combination, therefore even though crossovers occurred, the progeny will have the parental phenotype. D. Answer d is incorrect. Anueploidy occurs when there is too much or too little genetic information. Crossovers only exchange information. 10. The disease sickle cell anemia is a caused by a. an altered expression of the Hbβ gene b. a change of a single amino acid in the protein hemoglobin c. a change in the Hbβ gene d. both b and c A. Answer a is incorrect. This disease is a consequence of a change in the gene and the gene product, not in the expression of the gene. B. Answer b is incorrect. Although there is a single amino acid change in the sickle cell hemoglobin, the change in the amino acid is the result of a change in the gene. C. Answer c is incorrect. The change in the Hbβ gene alters the Hbβ protein. It is the altered hemoglobin protein that represents the disease phenotype. both b and c D. Answer d is correct. The disease is the result of the combined effects of a change in a gene leading to a change in a protein 11. What determines whether an individual is a genetic mosaic? a. The presence of different alleles on the autosomal chromosomes b. The inactivation of an allele on an autosomal chromosome c. The inactivation of an allele on the X chromosome of a heterozygous female d. The inactivation of an allele on the X chromosome of a homozygous male A. Answer a is incorrect. An individual that is heterozygous is not a genetic mosaic. B. Answer b is incorrect. Autosomal chromosomes do not undergo inactivation. The inactivation of an allele on the X chromosome of a heterozygous female C. Answer c is correct. The cells of a female have only one active X chromosome. They are a mosaic since the specific X chromosome that has been inactivated varies between cells.

6 D. Answer d is incorrect. The X chromosome does not become inactive in a male. 12. Down syndrome is the result of a. a single-base substitution on human chromosome 21 b. nondisjunction of chromosome 21 during meiosis c. inactivation of chromosome 21 d. nondisjunction of chromosome 21 during mitosis in early development A. Answer a is incorrect. Down syndrome is an example of a trisomy. Trisomy results when there is an extra copy of a chromosome occurs in a cell. nondisjunction of chromosome 21 during meiosis B. Answer b is correct. The failure of the homologues of chromosome 21 to separate during anaphase of meiosis I leads to the trisomic condition. C. Answer c is incorrect. Down syndrome is the result of too much genetic information. Inactivation removes genetic information from a cell. D. Answer d is incorrect. Nondisjunction must occur during meiosis (the production of gametes) in order for the organism to become trisomic. 13. Which of the following examples of nondisjunction of sex chromosomes is lethal? a. XXX b. XXY c. OY d. XO A. Answer a is incorrect. Triple X females are viable. They have a single active X chromosome and two Barr bodies. B. Answer b is incorrect. Klinefelter males are viable. OY C. Answer c is correct. An embryo that lacks an X chromosome cannot survive. D. Answer d is incorrect. These individuals are viable. 14. What is imprinting? a. The blending of a phenotype due to the genetic contribution of both parents b. The expression of a dominant allele c. The development of a phenotype in response to interactions between distinct alleles

7 d. The expression of different phenotypes dependent upon the parent from which the allele was inherited A. Answer a is incorrect. Imprinting occurs when one allele is inactive. The phenotype would appear to be homozygous. B. Answer b is incorrect. Imprinting results in variable expression of an allele, depending on whether that allele was inherited from the mother or father, not on whether the allele is dominant or recessive. C. Answer c is incorrect. Interactions between distinct alleles leading to an altered phenotype are called epistasis. The expression of different phenotypes dependent upon the parent from which the allele was inherited D. Answer d is correct. Specific alleles can be inactivated by methylation of the DNA. For some genes the methylation of an allele follows a pattern based on whether the source of the alleles is maternal or paternal 15. Which of the following is NOT a method used in genetic counseling? a. Ultrasound b. Chorionic villi sampling c. Amniocentesis d. Pedigree analysis Ultrasound A. Answer a is correct. Ultrasound can provide a visual image of a developing fetus; however, it does not provide any information about genotype. B. Answer b is incorrect. Chorionic villi sampling is a method that provides fetal cells that can be used to examine chromosome number and other biochemical tests. C. Answer c is incorrect. Amniocentesis provides a sample of fetal cells that can be examined using karyotyping. D. Answer d is incorrect. Pedigree analysis is a noninvasive method used by genetic counselors to predict the possibility of a genetic defect being inherited by a child based on family history.

8 Challenge Questions 1. Color blindness is caused by a sex-linked, recessive gene. If a woman, heterozygous for the color blind allele, marries a man with normal color vision, what percentage of their children will be color blind? What sex will the color blind children be? Answer Use a simple Punnett square to calculate the percentage of color blind children. X (color blind) X (normal) X (normal) Y Theoretically, 25% of the children from this cross will be color blind. All of the color blind children will be male and 50% of the males will be color blind. 2. What conditions would have to exist to produce a color blind female? Answer A female would have to inherit the color blind gene from both parents. For this to happen the father would have to be color blind and the mother would have to be heterozygous for the color blind allele. XX (normal)(color blind) XY (color blind) XX (normal)(normal) XY (normal) X (color blind) X (normal) X (color blind) Y XX (color blind)(color blind) XY (color blind) XX (color blind)(normal) XY (normal) 3. Imagine that the genes for seed color and seed shape are located on the same chromosome. A cross was made between two, true-breeding plants. One plant produces green wrinkled seed (rryy) and the second parent produced round yellow seeds (RRYY). A testcross is made between the F 1 generation with the following results: green, wrinkled 645 green, round 36 yellow, wrinkled 29 yellow, round 590 Calculate the distance between the two loci.

9 Answer To answer this question it is first necessary to determine which of the progeny are expressing the parental phenotypes and which are the recombinants (that is, the progeny with phenotypes that are different from the parental). In this problem the original parents had the phenotypes green, wrinkled and yellow, round. That means that the recombinants are the green, round and yellow, wrinkled. The next step is to add up the total number of recombinant ( = 65) and then divide that number by the total number of progeny ( = 1300) 65/1300 = Each 1% of recombination is a map unit. For this problem the number of map units between the two genes is 5 cm. 4. Is it possible to have a calico cat that is male? Why or why not? Answer Male calico cats are very rare. The coloration that is associated with calico cats is the product of X inactivation. X inactivation only occurs in females as a response to dosage levels of the X-linked genes. Females that are heterozygous for coat color will have different regions or patches of colored fur, representing the differential inactivation of the black or orange pigment genes. The only way to get a male calico is to be heterozygous for the color gene and to be the equivalent of a Klinefelters male (XXY).