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Chapter 5 Extensions and Exceptions to Mendel's Laws Answers to Review Questions 1. a. Lethal alleles eliminate a progeny class that Mendel's laws predict should exist. b. Multiple alleles create the possibility of more than two phenotypic classes. c. Incomplete dominance introduces a third phenotype for a gene with two alleles. d. Codominance introduces a third phenotype for a gene with two alleles. e. Epistasis eliminates a progeny class when a gene masks another's expression. f. Incomplete penetrance produces a phenotype that does not reveal the genotype. g. Variable expressivity can make the same genotype appear to different degrees. h. Pleiotropy can make the same genotype appear as more than one phenotype because subsets of effects are expressed. i. A phenocopy mimics an inherited disorder, but is an environmental effect. j. Conditions with the same symptoms but caused by different genes (genetic heterogeneity) will not recur with the frequency that they would if there was only one causative gene. 2. Dominant and recessive alleles are of the same gene, whereas epistasis is an interaction of alleles of different genes. 3. It can skip generations in terms of phenotype. 4. The IA allele is codominant with the IB allele; both are completely dominant to i. 5. The parents with albinism could have mutations in different genes. 6. Variable expressivity means different degrees of symptoms. Pleiotropy means several symptoms (fever, weakness, abdominal pain, red urine). Genetic heterogeneity means that mutations in different genes in the heme biosynthetic pathway can cause porphyria. 7. Smoking 8. Unlike nuclear DNA, mtDNA is present in many copies, cannot repair itself, and is passed from mothers only. 9. Maternal inheritance describes transmission of mitochondrial genes, which sperm do not usually contribute to oocytes and therefore these traits are always passed from mothers only. Linked genes are transmitted on the same chromosome. Mendel's second law applies to genes transmitted on different chromosomes. 10. Only females transmit maternally inherited traits. All of a woman's children inherit a mitochondrial trait, but a male does not pass the trait to his children. 11. Twenty-four linkage groups, including 22 pairs of autosomes, and the X and the Y. Answers to Applied Questions 1. a. D b. C c. A d. E e. H f. C g. H h. F i. G 2. Haplotypes can reveal if people without symptoms have the haplotype associated with the condition. These people are non-penetrant. If the genotype is lethal, individuals with the haplotype containing the lethal allele should not exist. 3. a. 1/2 b. 1/2 c. 0 d. 0 4. 1/4 5. a. This alters the phenotype. b. Bombay phenotype 6. It would be Mendelian because the gene that causes the condition when mutated is in the nucleus. 7. .45 .05 = .0225 8. The sperm are all of genotype hhsese. Oocytes are of genotype HhSese, with alleles in cis: H ___ Se h ------- se All sperm are hse. Oocytes: HSe 49.5% hse 49.5% Hse 0.5% hSe 0.5% Chance offspring like father = 1 (hse) 49.5 (hse) = 49.5 9. a. Male: RrHhTt Female: rrhhtt b. Parental progeny classes: Round eyeballs, hairy tail, 9 toes; square eyeballs, smooth tail, 11 toes Recombinant progeny classes: Round eyeballs, hairy tail, 11 toes; square eyeballs, smooth tail, 9 toes; round eyeballs, smooth tail, 11 toes; square eyeballs, hairy tail, 9 toes c. Crossover frequency between eyeball shape (R) and toe number (T): Round eyeballs, 11 toes = 6 + 4; square eyeballs, 9 toes = 6 + 4; crossover frequency = 20/100 = 20% 10. Ehlers-Danlos syndrome, FG syndrome, Kabuki syndrome, VATER association 11. Kearn-Sayre syndrome, heart failure, infections, anemia, hearing loss, pancreatic failure, visual loss 12. chromosome 3: genes for sodium channel, voltage-gated, type V, alpha and contactin 3 and chondroitin sulfate proteoglycan 5 13. Disorders that are caused by more than one mutation include muscular dystrophy, hemophilia, myotonic dystrophy, and epidermolysis bullosa. 14. a. alcohol, infection, starvation, hormonal changes, drugs b. sunlight, alcohol, estrogens c. drugs d. drugs e. sunlight 15. a. a, b, c, d b. People with one copy have abnormalities so mild that they are not noticeable without a test. People with two copies can be severely affected. c. The parents are heterozygotes, so each of their offspring has a 1 in 4 probability of inheriting two wild type alleles, like Tina. d. 16. Phenotype a. Designation of "recessive" was based on what we could observe. Biochemical tests enable us to observe at a different level. b. Benefit of carrier detection is to warn of elevated inherited cancer risk. 17. Both are recessive, so the chance a child would have either is not elevated above the general population risk. A SECOND LOOK 1. A gene can vary in many ways because it is made up of a long sequence of nucleotides. 2. Variable expressivity 3. Alkaptonuria is pleiotropic because all tissues that have melanin may be affected.