2002_Midterm_HMB265 - HMB265H1 MIDTERM Duration - 1 hours...

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Monday March 4 th , 2002 Page 1 of 10 HMB265H1 MIDTERM Duration - 1 hours Definitions 1. The general range of different phenotypes resulting from a genotype under different genetic and environmental backgrounds is … (a) discontinuous variation (b) continuous variation (c) norm of reaction (d) developmental noise (e) environmental determination 2. A one-to-many relationship of genes to phenotypes is … (a) phenocopy (b) pleiotropy (c) penetrance (d) polymorphism (e) pseudogene 3. A measure of the extent to which a given genotype is expressed at the phenotypic level is … (a) expressivity (b) epistasis (c) epitope (d) enhancer (e) equilibrium 4. The association of separate amino acid chains to constitute one active protein is called the … (a) primary structure (b) secondary structure (c) tertiary structure (d) quaternary structure (e) quinternary structure 5. A mutation that occurred in a plant petal would best be termed … (a) germinal (b) somatic (c) suppressor (d) dominant (e) recessive 6. A homozygous mutant allele results in no detectable activity; the mutation is best described as … (a) dominant (b) recessive (c) conditional (d) gain of function (e) null 7. A base change resulting in a codon specifying the same amino acid is a … (a) silent mutation (b) synonymous mutation (c) nonsense mutation (d) antisense mutation (e) missense mutation 8. A chromosome having the primary constriction located slightly nearer one end than the other is … (a) acrocentric (b) telocentric (c) metacentric (d) acentric (e) telomeric
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Monday March 4 th , 2002 Page 2 of 10 B. Pedigree Analysis 9. The following pedigree concerns the autosomal recessive disease phenylketonuria (PKU). The couple marked A and B are contemplating having a baby but are concerned about the baby’s having PKU. What is the probability of the first child having PKU? (Unless you have evidence to the contrary, assume that a person marrying into the pedigree is not a carrier; black indicates the idividual has the disease.) A B ? ½ chance B is P/p x 2/3 chance A is P/p (since we know A is NOT p/p; A is P/P or P/p) x ¼ chance their child is p/p = ½ x 2/3 x ¼ = 1/12 (a) 0 (b) 1/12 (c) ¼ (d) ¾ (e) 1/16 10. A couple are both heterozygous for the autosomal recessive allele for albinism. They have two children. What is the probability that both children will be phenotypically identical with regard to skin colour? Chance both children are albino is ¼ x ¼ = 1/16; chance both are non-albino = ¾ x ¾ = 9/16; therefore probability they are identical wrt skin colour is 1/16 + 9/16 = 10/16 = 5/8 OR chance that one child is albino and one is non-albino is 2 (since 2 birth orders) x ¼ x ¾ = 3/8; therefore probability that this is incorrect and that they are both the same phenotype is 1 – 3/8 = 5/8 (a) ¾ (b) 1/16 (c) ¼ (d) 9/16 (e) 5/8 11. Red-green colour blindness is X-linked in humans. If a male is red-green colour blind, and both parents have normal colour vision, which of the male’s grandparents is most likely to be
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This note was uploaded on 04/08/2010 for the course HMB 265 taught by Professor French during the Spring '09 term at University of Toronto- Toronto.

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2002_Midterm_HMB265 - HMB265H1 MIDTERM Duration - 1 hours...

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