Genetics Test 3 Ch. 4 Flashcards
(46 cards)
1) Which mode of inheritance produces heterozygotes with phenotypes that differ from either homozygote but typically more closely resembles one homozygous phenotype than the other?<br></br>A) complete dominance<br></br> <br></br>B) incomplete dominance<br></br>C) codominance<br></br>D) epistasis<br></br>E) incomplete penetrance
Answer: B
2) Which mode of inheritance results in the phenotype of a heterozygote being indistinguishable from that of an organism homozygous for the dominant allele?<br></br>A) complete dominance<br></br>B) incomplete dominance<br></br>C) codominance<br></br>D) epistasis<br></br>E) incomplete penetrance
Answer: A
3) Which mode of inheritance results in both alleles being detected equally in the heterozygous phenotype?<br></br>A) complete dominance<br></br>B) incomplete dominance<br></br>C) codominance<br></br>D) epistasis<br></br>E) incomplete penetrance
Answer: C
4) A mutation results in an enzyme that is partially active compared to the wild- type allele. This type of “leaky” mutation is classified as ________.<br></br>A) null/ amorphic<br></br>B) hypomorphic<br></br>C) hypermorphic<br></br>D) neomorphic<br></br>E) dominant negative
Answer: B
5) A mutation resulting in an inactive gene product is classified as ________. <br></br>A) null/ amorphic<br></br>B) hypomorphic<br></br>C) hypermorphic<br></br>D) neomorphic<br></br>E) dominant negative
Answer: A
6) Two proteins interact to form a multimeric complex. When one of the proteins is mutated, there is a substantial loss of functional activity in the multimeric protein. This type of mutation is classified as ________.<br></br>A) null/ amorphic<br></br>B) hypomorphic<br></br>C) hypermorphic<br></br>D) neomorphic<br></br>E) dominant negative
Answer: E
7) Many oncogenes result from mutations that cause excessive expression of a protein in cells where it is normally not expressed or is expressed at inappropriate times during development. This type of mutation can be described as ________.<br></br>A) hypomorphic<br></br>B) dominant negative<br></br>C) amorphic<br></br>D) hypermorphic <br></br>E) neomorphic
Answer: D
8) A mutation results in a gene product with a novel function that is not normally found in wild- type organisms. This type of mutation is known as ________.<br></br>A) dominant negative<br></br>B) hypermorphic <br></br>C) hypomorphic<br></br>D) amorphic<br></br>E) neomorphic
Answer: E
9) Which of the following is correct regarding individuals who are blood type AB?<br></br>A) Their blood cells clump when they receive blood from an individual with the genotype IAi.<br></br>B) They express B- transferase that adds N- acetylgalactosamine to the H antigen.<br></br>C) They do not carry the A or B antigen.<br></br>D) They are the universal recipients.<br></br>E) They carry both the anti- A and anti- B antibodies.
Answer: D
10) A man with blood type A (whose mother was blood type O) has children with a woman that has blood type AB. The man and the woman are also heterozygous for the H antigen. What is the probability that they will have a child with blood type A?<br></br> A) 0 B) 1/8 C) 3/4 D) 1/2 E) 3/8
Answer: E
12) What type of allele is often detected as a distortion in segregation ratios, where one class of expected progeny is missing?<br></br>A) incompletely penetrant allele <br></br>B) partially dominant allele<br></br>C) dominant negative allele<br></br>D) temperature- sensitive allele<br></br>E) lethal allele
Answer: E
13) You discover a new allele of a gene important for tail formation in mice. WT mice have long tails, but mice heterozygous for the allele have short tails. When you cross two heterozygous mice together, you obtain a 2:1 ratio of short- tailed mice to long- tailed mice. None of the short- tailed progeny are homozygous. What type of allele results in short tails?<br></br>A) temperature- sensitive allele<br></br>B) lethal allele<br></br>C) incompletely penetrant allele<br></br>D) partially dominant allele<br></br>E) dominant negative allele
Answer: B
14) The amount of enzyme activity in a cell that is homozygous for a mutant allele is 400 units. The amount of enzyme activity in a cell homozygous for the WT allele is 200 units. The amount of enzyme activity in a heterozygote is 300 units. What type of allele is the mutant allele?<br></br>A) hypermorphic<br></br>B) hypomorphic <br></br>C) neomorphic<br></br>D) null/ amorphic<br></br>E) dominant negative
Answer: A
15) The amount of enzyme activity in a cell that is homozygous for a mutant allele is 400 units. The amount of enzyme activity in a cell homozygous for the WT allele is 200 units. The amount of enzyme activity in a heterozygote is 300 units. What is the dominance relationship between the WT and mutant allele for the phenotype of amount of enzyme per cell? <br></br>A) The WT and mutant alleles show incomplete dominance.<br></br>B) The mutant allele is dominant.<br></br>C) The WT allele is dominant.<br></br>D) The WT and mutant alleles are codominant.<br></br>E) It is impossible to determine from the information given.
Answer: A
16) The amount of enzyme activity in a cell that is homozygous for a mutant allele is 0 units. The amount of enzyme activity in a cell homozygous for the WT allele is 200 units. The amount of enzyme activity in a heterozygote is 0 units. What type of allele is the mutant allele?<br></br>A) neomorphic<br></br>B) hypermorphic<br></br>C) hypomorphic<br></br>D) null/ amorphic<br></br>E) dominant negative
Answer: E
17) Which phenomenon explains differences in the inheritance patterns of the appearance of a chin beard between males and females of certain species of goats, even when their genotypes are the same?<br></br>A) sex- limited trait<br></br>B) lethal allele<br></br>C) sex- influenced trait<br></br>D) incomplete penetrance<br></br>E) variable expressivity
Answer: C
18) In certain goat breeds, appearance of a chin beard is a sex- influenced trait. Recall that bearding is inherited as an autosomal trait determined by two alleles, B1 and B2, and females must be homozygous for the bearded allele, B2, to have a beard. What genotypes must a bearded billy goat (male) and a beardless female goat have if they have a bearded female offspring?<br></br>A) The bearded billy goat must be heterozygous for the bearded allele, while the beardless female must be homozygous for the bearded allele.<br></br>B) The bearded billy goat could be heterozygous or homozygous for the bearded allele, while the beardless female must be homozygous for the bearded allele.<br></br>C) Both the bearded billy goat and beardless female must be heterozygous for the bearded allele.<br></br>D) The bearded billy goat could be heterozygous or homozygous for the bearded allele, while the beardless female must be heterozygous for the bearded allele.<br></br>E) Both the bearded billy goat and beardless female must be homozygous for the bearded allele.
Answer: D
21) King George III of England and other members of the royal family were afflicted with a series of strange, seemingly unrelated symptoms including abdominal pain, rapid pulse, convulsions, and insanity. It has been determined that he likely suffered from porphyria, caused by a mutation in a single allele. What is the genetic term describing the alteration of multiple, distinct traits of an organism by a mutation in a single gene?<br></br>A) incomplete dominance<br></br>B) pleiotropy<br></br>C) epistasis<br></br>D) codominance<br></br>E) incomplete penetrance
Answer: B
22) Gene interactions in which an allele of one gene modifies or prevents expression of alleles of another gene is known as ________.<br></br>A) incomplete dominance<br></br>B) codominance<br></br>C) epistasis<br></br>D) incomplete penetrance <br></br>E) pleiotropy
Answer: C
23) Bateson and Punnett crossed two white- flowered lines and saw all purple flowers in the F1 generation. They concluded this was an example of complementary gene interactions because a cross of the F1 plants yielded what ratio in the F2 generation?<br></br>A) 16 purple to 0 white <br></br>B) 7 purple to 9 white<br></br>C) 0 purple to 16 white<br></br>D) 9 purple to 7 white<br></br>E) 8 purple to 8 white
Answer: D
24) The 9:6:1 ratio seen in the dihybrid cross of summer squash indicates what genetic relationship between the two genes controlling fruit shape?<br></br>A) recessive epistasis<br></br>B) dominant suppression<br></br>C) complementary gene interaction<br></br>D) dominant epistasis<br></br>E) dominant gene interaction
Answer: E
27) Wild- type bacteria can grow on minimal medium. Four mutants that cannot grow on minimal medi but can grow on minimal medium supplemented with the nutrient “H” are isolated. It is suspected th metabolites T, P, and A are in the biochemical pathway for synthesis of H, so each mutant is tested fo ability to grow on minimal medium supplemented with these metabolites: <br></br>A. Mutant 1: can grow on minimal medium supplemented with T, but not P or A<br></br>B. Mutant 2: is unable to grow on minimal medium supplemented with T, P, or A<br></br>C. Mutant 3: is able to grow on minimal medium supplemented with A or T, but not P<br></br> D. Mutant 4: can grow on minimal medium supplemented with T, P, or A.<br></br>Which of the following is NOT consistent with this information?<br></br>A) Mutant 1 will cause a build up of metabolite A<br></br>B) Mutant 4 blocks a step before all the metabolites<br></br>C) Mutant 3 blocks the conversion of metabolite P metabolite A<br></br>D) Mutant 3 will cause a build up of metabolite T<br></br>E) Mutant 2 blocks a step right before metabolite H
Answer: D
28) Wild- type bacteria can grow on minimal medium. Four mutants that cannot grow on minimal medi but can grow on minimal medium supplemented with the nutrient “H” are isolated. It is suspected the metabolites T, P, and A are in the biochemical pathway for synthesis of H, so each mutant is tested for<br></br> <br></br>ability to grow on minimal medium supplemented with these metabolites: A<br></br>A Mutant 1: can grow on minimal medium supplemented with T, but not P or A<br></br>B. Mutant 2: is unable to grow on minimal medium supplemented with T, P, or A <br></br>C Mutant 3: is able to grow on minimal medium supplemented with A or T, but not P<br></br> D. Mutant 4: can grow on minimal medium supplemented with T, P, or A.<br></br>Which of the following is NOT consistent with this information?<br></br>A) Mutant 3 blocks the conversion of metabolite P metabolite A <br></br>B) Mutant 4 blocks a step before all the metabolites <br></br>C) Mutant 1 will cause a build up of metabolite A<br></br>D) Mutant 1 blocks the conversion of metabolite T metabolite H <br></br>E) Mutant 2 blocks a step right before metabolite H
Answer: D
29) Independent assortment predicts a 9:3:3:1 ratio with four different phenotypes in the F2 progeny. If the alleles are epistatic, what would you predict?<br></br>A) heterozygotes with a novel phenotype between the dominant and recessive homozygotes<br></br>B) 1/3 of the progeny with the dominant phenotypes and 2/3 recessive<br></br>C) more than four phenotypes<br></br>D) fewer than four phenotypes<br></br>E) no change in the 9:3:3:1 ratio
Answer: D
A) The genes are part of two distinct biosynthetic pathways.
B) The allele is pleiotropic.
C) The allele exhibits incomplete dominance.
D) More than one gene is involved in determining the phenotype.
E) The two lines exhibit different mutations in the same gene.
A) 15:1 B) 12:3:1 C) 9:3:4 D) 9:6:1 E) 13:3
A) The white sheep could be A_B_ or aabb.
B) The white sheep must all be A_B_.
C) The white sheep could be A_B_ or A_bb.
D) The white sheep must all be aabb.
E) The white sheep could be A_B_ or aaB_.
Which the following is NOT consistent with this information?
A) Blue- flowering plants are either A_bb or aaB_.
B) The pure- breeding parental parents are homozygous recessive for mutations in two different genes.
C) Dominant gene interaction appears to result in a 9:6:1 ratio.
D) Analysis of the F1 and F2 progeny phenotypes suggests epistasis.
E) Red- flowering plants are homozygous recessive for both genes.
A) The mutations are codominant to the normal allele.
B) The parents have the same mutated protein involved in inner ear development.
C) The parents have mutations in different genes.
D) The mutations are incompletely dominant to the normal allele.
E) The parents have mutations in the same gene.
Which of the following can you definitively conclude?
A) The haploid strains have identical mutations in the same genes.
B) The haploid strains must belong to the complementation group encoding the first enzyme in the biosynthetic pathway.
C) The haploid strains have identical mutations in different genes.
D) The haploid strains have mutations in the same gene.
E) The haploid strains have mutations in different genes.