This set of Cytogenetics Multiple Choice Questions & Answers (MCQs) focuses on “Chromosomal Abnormalities : Duplication”.
1. Which of the following case of duplication involves more than one chromosome?
a) Tandem duplication
b) Reverse tandem duplication
c) Displaced
d) Transposed
View Answer
Explanation: In transposed the segment could be passed to another homologous chromosome or extra chromosome with a centromere. However, tandem, revere tandem and displaced occurs within the same chromosome.
2. Tandem duplication creates ____________ and reverse tandem creates_____________
a) Buckling, looping
b) Hairpin, buckling
c) Buckling, Hairpin
d) Loop with turn, hairpin
View Answer
Explanation: In case of tandem repeat the same sequence in same order is repeated. As parent chromosome has only one copy of that sequence the daughter bulges out. This, causes buckling. In reverse tandem the duplication is inverted in sequence which leads to formation of hairpin.
3. The Streisinger’s model explains__________
a) Frame shifts
b) Transposition
c) Inversion
d) Transition
View Answer
Explanation: Streisinger’s model explains that the Frame shift mutations are more frequent in repetitive sequences. This causes duplicative insertion and also at times deletion.
4. If there are two sequences CGCGCGCG and ATATATAT, which will have a higher probability of being duplicated?
a) ATATATATAT
b) CGCGCGCGCG
c) Both will have equal probability
d) Both are unlikely to be duplicated
View Answer
Explanation: Streisinger model works on repeated sequences, it is not base specific. In this case both the AT rich and CG rich sequences should have an equal probability of getting duplicated.
5. Which of the following will have the highest number of eye facete?
a) Ultra bar female Drosophila
b) Heterozygous Bar male Drosophila
c) Heterozygous Bar female Drosophila
d) Homozygous bar Female
View Answer
Explanation: The bar gene is present in the X chromosome, thus, males can’t have two copies for being homozygous or heterozygous. Ultra bar condition is most severe, then is the homozygous bar and then follows bar.
6. Duplication of Vermillion eye colour gene leads to the expression of the mutant phenotype. Which of the following can lead to normal phenotype with this duplication present?
a) Reverse tandem duplication
b) Duplication of normal eye colour gene
c) Deletion of both the Vermillion gene
d) Translocation of Vermillion gene to another chromosome
View Answer
Explanation: Duplication in normal eye colour gene would balance the mutant gene product and lead to expression of wild type phenotype. Reverse tandem duplication will only create more copies of mutant making phenotype worse, also deletion of both genes will cause imbalance. Translocation makes no difference in expression unless it’s beside a heterochromatin.
7. Which human chromosomes are involved in Down’s syndrome?
a) 6
b) 14 and 21
c) 8 and 12
d) X and Y
View Answer
Explanation: In Down’s syndrome a part of the 21st chromosome’s long arm duplicates and transfers to chromosome 14’s long arm.
8. Which of the following could be due to duplication?
a) Co-dominance
b) Dominance
c) Incomplete dominance
d) Pleiotropy
View Answer
Explanation: In case of Pleiotropy one gene effects several character; this may be due to mutations of the initial duplicated sequences.
9. If a gene undergoes duplication and one of the duplicated copy is mutated to render unexpressed protein. What will be the effect on phenotype?
a) Mutant phenotype
b) Normal phenotype
c) Expression of the mutant gene is less than that would occur if both copy active
d) The gene not expressed at all
View Answer
Explanation: In this case as the product of the 2nd copy is mutated and not affecting the phenotype, only one copy is producing the required product in normal quantity. This is similar to the duplication being silenced.
10. During Strisinger’s model, what happens?
a) The Ribosome slips which causes duplication
b) Parent strand slippage polymerase
c) Daughter strand slippage polymerase
d) The polymerase back tracks and resynthesizes
View Answer
Explanation: Daughter strand slippage will cause looping which is stabilized by repetitive sequences. This leads to an increase in a number of repeats in the daughter.
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