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Unformatted text preview: Answers to Lecture Questions from 2‐3‐10 Lecture Plus Remaining Genetic Questions not Covered in Class (Question numbers below match those in the VNET document, “1362 CH 12‐16 Questions to be covered during lecture”) 12. Carnosinemia is a metabolic disorder and is an autosomal recessive trait. Parents Mike & Jane have a normal phenotype, but produced 2 children having carnosinemia. What is the probability that they will have another affected child? Designate C,c as alleles. An individual expressing the carnosinemia phenotype would only do so if he/she possessed 2 recessive alleles; i.e., cc (remember C dominant to c). The affected children have the cc genotype; as a result, both parents (normal phenotype) must be heterozygous (Cc) children inherited 1 recessive allele from each parent. The parental gametes are then C and c, now do the Punnett square: Probability of having another affected child, i.e., cc genotype, is ¼, or 25%. 13. Huntington’s disease is inherited as a dominant trait. Key: Dominant allele is always expressed whether genotype is homozygous or heterozygous; use alleles H, h for answers below. A. What is the genotype of an unaffected individual? hh B. What is the phenotype of a heterozygote? has Huntington’s C. What is the probability of having a normal child from a normal mother and a heterozygous father? Normal mother = hh, her gametes all have h allele; father = Hh, his gametes: ½ have H allele, ½ have h allele. Do Punnett square; 50% chance of having a normal child 18. In the pedigree to the right, is the trait expressed by the shaded individuals inherited as a dominant or recessive trait? (or are both possibilities?) Cannot be recessive; look at 2nd & 3rd generation on right side. 2nd generation parents are both affected; if it were a recessive trait, they would both be aa. These parents would not produce unaffected offspring as shown (i.e., the unaffected daughter in 3rd generation). It is a dominant trait. Genotypes: • A: Aa • B: aa • C: Aa • D: AA or Aa 19. White eyes are a recessive sex‐linked trait in fruit flies. The normal eye color is red. If a white‐eyed female fruit fly is mated to a red eyed male, their offspring would be…? Red eye color dominant to white eye color: white‐eyed female = XrXr red‐eyed male = XRY Offspring: all females red‐eyed, all males white‐eyed 20. Color blindness is a sex‐linked recessive trait. What are the genotypes of individuals A‐F in the following pedigree? Alleles B, b A. XBY; does not have the recessive allele (i.e. trait), or else it would be shaded (color blind, recessive phenotype). B. XBXb; must be a carrier, her son has the recessive trait. C. XBXBor XBXb; either genotype, when crossed with the XBY father, both have a chance of yielding unaffected (XBY sons) D. XbY; must have the recessive allele, since it is shaded (recessive phenotype). E. XBY; does not have the recessive allele, or else it would be shaded (indicating recessive phenotype). F. XBXb; must be a carrier; all daughters here inherit recessive allele (b) from father. Since they do not have the recessive phenotype (unshaded) they must be carriers. 21. A certain (hypothetical) organism is diploid, has either blue or orange wings as the consequence of one of its genes, and has either long or short antennae as the result of a second gene, as shown in the figure below. A female with a paternal set of one orange and one long gene chromosomes and a maternal set comprised of one blue and one short gene chromosome is expected to produce which of the following types of eggs after meiosis? A. Only blue gene eggs B. Only orange gene eggs C. ½ blue and ½ orange gene eggs D. ¾ blue and ¼ orange gene eggs E. An intermediate frequency of blue and orange gene eggs Answer: of the choices given, only “C” fits. 22. A mouse has the genotype aa for a specific trait. Which of the following statements in A‐D is incorrect regarding this individual? a) each parent of this mouse must have at least one copy of the a allele. b) all of the gametes produced by this mouse will have an a allele. c) there is a 100% probability that the offspring of this mouse will have the a allele. d) the aa phenotype would always represent the one that occurs at the lowest frequency in the mouse population. e) All statements in A‐D are correct; none are false. Answer: “D” 14. A dominant allele K is necessary for normal hearing. A dominant allele M on a different locus results in deafness no matter what other alleles are present (what do you call this?). If a kkMm individual is crossed with a Kkmm individual, what percentage of the offspring will be deaf? A) 0 B) 25 C) 50 D) 75 E) None of the above Answer: “D” Two genes, K & M; two alleles per gene: K gene: dominant K allele: hearing; recessive k allele: deafness. M gene: dominant M allele + any other allele (i.e., K or k) = deafness Thus, “hearing” genotypes = KKmm, Kkmm “deaf” genotypes = any genotype with M, i.e., _ _ M _ or kk _ _; Punnett square below: ...
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- Spring '05