Cytogenetics Multiple Choice Questions on “Gene Interactions”.
1. The difference in epistasis and gene interaction is ___________________
A. Gene interaction involves two genes
B. Epistasis modifies the normal Mendelian ratio
C. Epistasis involves two genes
D. Gene interaction produces a novel phenotype
Answer: D
Clarification: In case of gene interaction the product of two genes affects the same phenotype and the resultant product might express a novel phenotype. In epistasis, no new phenotypic expression is seen.
2. Gene interaction always modifies the normal Mendalian ratio.
A. True
B. False
Answer: B
Clarification: In case of gene interactions the ratio of the normal dihybrid cross may be modified due to the appearance of the novel phenotype, but in some cases, the ratio remains the same.
3. In case of comb shape of a chicken, which of the following when inbred will breed true?
A. Rose
B. Pea
C. Walnut
D. Single
Answer: D
Clarification: This is the case when the chicken is homozygous recessive in both the R and P locus. Thus, they will always breed true.
4. You cross a true breeding pea comb chicken with a single comb chicken. What will you observe in the F2 generation?
A. Ratio 1:2:1
B. Ratio 3:1
C. Ratio 9:3:3:1
D. All pea combed
Answer: D
Clarification: Pea comb has genotype P/P r/r and single has genotype p/p r/r. Thus the cross will give F1 P/p r/r. This is homozygous recessive at R locus so F2 will only differ in P locus giving monohybrid ratio 3:1.
5. If you cross two walnuts, which of the following is impossible?
A. Progeny to be a walnut
B. Progeny to be a single
C. Progeny to be a homozygous Rose
D. Progeny to have a phenotype other than rose, single, pea or walnut
Answer: D
Clarification: The gene interactions in case of chicken comb can only produce these four phenotypes in the ratio 9:3:3:1.
6. If a true breeding rose combed chicken is crossed to walnut combed chicken, what are the only possibilities of the F1 progeny?
A. Walnut
B. Rose
C. Walnut or rose
D. Walnut or rose or pea
Answer: C
Clarification: Considering the most variable of the condition when the walnut is heterozygous in both the loci it can result in walnut or rose in the ratio of 3:1. In other cases, it can mostly result in walnut while parent heterozygous in P locus could also give a rose.
7. In summer squash if you cross different varieties of true breeding sphere shaped squash, you get a disk shaped progeny. What phenomenon is seen here?
A. Gene interaction
B. Epistasis
C. Multiple allele
D. Codominance
Answer: A
Clarification: As the cross lead to the production of a novel phenotype, it is not the case of epistasis but rather a case of gene interaction.
8. Which one of the following summer squash would be true breeding?
A. Sphere
B. Disc
C. Long
D. Short
Answer: C
Clarification: The long phenotype is produced because of the recessive genes in both the locus. Thus it will be true breeding.
9. If you cross a true breeding long with a true breeding disc shaped squash, what will you get in the F1?
A. Disc
B. Sphere
C. Long
D. Disc and long in 3:1
Answer: A
Clarification: Considering a true breeding disc to be AA BB and a long is aa bb. The cross will yield a genotype Aa Bb which is also disc.
10. If you cross two sphere shaped squash, you always get a disc.
A. True
B. False
Answer: B
Clarification: Only cross between two different strains of sphere shared squash with genotypes A/- bb and aa B/- will give A/- B/- necessary for disc shaped.
11. You cross two heterozygous disc shaped squash. What will be the ratio obtained?
A. 1:2:1
B. 3:1
C. 9:3:3:1
D. 9:6:1
Answer: D
Clarification: Here you expect to see a deviation from the normal dihybrid ratio as two genotypes yield the same phenotype. Thus the A/- bb and aa B/- give spheres in 6/16th of the progeny.
12. The cross between a true breeding sphere shaped and a long shaped squash gives 30 sphere squash in F2. What is the total number of expected progeny?
A. 30
B. 40
C. 50
D. 60
Answer: B
Clarification: In F the sphere and the long progenies should be produced in a 3:1 ratio. Thus sphere being 30, there must be 10 long. Thus the expected total number is 30+10= 40.
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