Cytogenetics Questions and Answers – Linkage and Crossing Over – 2

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This set of Cytogenetics Multiple Choice Questions & Answers focuses on “Linkage and Crossing Over – 2”.

1. Bar is a gene often used in recombination study. Which of the following is true for the Bar gene?
a) It is normally present in two copies
b) Three copies of bar are called ultra bar
c) Heterozygous bar females have a less severe effect than homozygous bar female
d) Homozygous bar are of common occurrence
View Answer

Answer: b
Explanation: Bar is present in single copy in normal genes. Recombination results in two bar genes which in homozygous condition can produce a severe reduction in number of eye facets. Recombination can also result in 3 copies of bar gene in a chromosome which reduces the number to 45, known as the ultra bar.
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2. Consider that recombination occurs in 2 strand stage of meiosis. If a Drosophila is heterozygous for the bar locus what will be the result of recombination?
a) The recombination will make them homozygous
b) It will produce ultra bar phenotype
c) It will be heterozygous
d) No exchange
View Answer

Answer: c
Explanation: The exchange in two strand stage will lead to the initial strand having 2 copies of Bar becomes normal and the normal chromosome has two copies. But the resultant will still be heterozygous. Only exchange in 4 strand stage will give ultra-bar.

3. Chiasmatype theory states recombination occurs before chiasma formation. State whether it is true or false.
a) True
b) False
View Answer

Answer: a
Explanation: It is stated in the chiasmatype theory, that strand breakage and reunion occurs prior to chiasma formation. Both sides of a chiasma represent a reduction division of chromatids. These recombination sites lead to the appearance of O shaped structures during anaphase.
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4. How many rounds of replication in BrdU will be needed to visualize Harlequin chromosome?
a) Single
b) Two
c) Three
d) Four
View Answer

Answer: b
Explanation: Due to the semiconservative nature of DNA replication in the 1st round the chromosome will have only one parental strand and one strand with BrdU. When two rounds are complete one chromosome had only BrdU strand and other has parental type. The BrdU strand is lighter so distinguishes the appearance.

5. If two genes are unlinked the recombination frequency will be?
a) 25%
b) 50%
c) 75%
d) 100%
View Answer

Answer: b
Explanation: When two genes are unlinked they assort independently. So there is an equal chance of production of parental and recombinant phenotype. Thus recombination frequency is 50%.
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6. Is two crossovers occur between same two strands, this will result in__________
a) All recombinant strands
b) Three recombinant strands with different loci exchanged
c) Exchange of middle portion between two strands
d) Exchange of ends
View Answer

Answer: c
Explanation: Double cross over event occurring between two strands only result in the production of two strands that exchanged a portion between them in the middle. The other two strands remain unaffected.

7. Double cross over involving ________ strands result in 100% recombinant strands.
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: d
Explanation: Double cross over event between for strands will result in all the strands exchanging their genes. Thus, all the resultant will be recombinant so recombination frequency is 100%.
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8. If a recombination event of three points crossing produces 6 DCO, 142 SCO and 352 NCO. What will be the percentage cross over between the terminal genes.
a) 10%
b) 20.8%
c) 14.8%
d) 30.8%
View Answer

Answer: d
Explanation: Recombination frequency between terminal genes will be as {(Single cross over frequency)+2(double cross over frequency)}/100. This will result in 30.8%.

9. If single cross over frequency between two genes r and q and q and s in 3 point mapping is 0.6 and 0.2 respectively. What will be the expected double cross over frequency?
a) 0.012
b) 0.010
c) 0.022
d) 0.024
View Answer

Answer: c
Explanation: Expected double cross over frequency by probability rule is product of the two single crossover frequencies in three point mapping. Here 20.8X10.0/10000 results on 0.022.

10. In an experiment you calculate the expected DCO frequency to be 0.022 but in reality you observe that only 0.012 recombination frequency. What is the phenomenon resulting in this?
a) Coincidence
b) Interference
c) Penitence
d) Expressivity
View Answer

Answer: b
Explanation: A single cross over in nearby region depreciates that probability of a DCO occurring in nearby regions. This phenomenon is known as interference.
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11. Which of the following will result in parental phenotype?
a) DCO in between two genes considering four strands
b) Even number of DCO between two genes
c) Odd number of DCO between two genes
d) Single cross over at 4 strand stage between two genes
View Answer

Answer: b
Explanation: Even number of DCO between two genes leads to exchange of small parts between the two genes but the resultant is the same as the parental as the terminal genes remain unchanged.

12. What will be the nature of curve with map distances plotted in X axis and percentage recombination in Y axis?
a) Parallel to X axis
b) Increasing in a straight line
c) Parabolic
d) Exponential
View Answer

Answer: d
Explanation: The observed curve is seen to be exponential which reaches a saturation point when the recombination frequency is about 50%.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter