This set of Cytogenetics Questions and Answers for Aptitude test focuses on “Gene Mapping in Bacteria by Transformation”.
1. In an experiment you have a U-Tube with two strains of bacteria in two arms. In one arm the bacteria are Gal+ and His-, in other the strain is Gal- and His+. If the two arms are joined by a membrane that doesn’t allow bacterial cells to pass to opposite site what will you observe when you plate each arm bacterial strain on minimal media?
a) None will grow
b) Some colonies seen in right arm
c) Some colonies seen in left arm
d) Some colonies seen in both arms
Explanation: Here the transfer of gene is mainly due to transformation. Although the bacterial cells couldn’t come in physical contact yet their nucleic acids could flow with the liquid to other arm and it can be incorporated by the bacterial cells in the other arm.
2. State whether it is true or false- Broad scale mapping can be done using transformation?
Explanation: In broad scale mapping we generally use concepts like- A gene enter the bacterial cell 10 minutes after B gene entry, so A and B are 10 unit apart. This is mainly done using conjugation. Transformation mapping is much finer.
3. If the genes order in bacteria is His gal met. Which of the following genes have the least probability of getting incorporated alone?
a) All have equal probability
Explanation: Gal can be incorporated alone only by the event of double cross over. Double crossover at such short stretch is of least occurrence so the probability of it being incorporated alone is least.
4. Which of the following species has natural competence?
a) Saccharomyces cerevisiae
b) Staphylococcus Aeruginosa
c) Gelidium corneum
d) Streptococcus pneumoniae
Explanation: Only 1% of all the bacterial strains are naturally competent. One among these is Streptococcus pneumonia, which can take up the chromosomes from medium without prior treatment.
5. Competence can be generated in some strains of non-competent bacteria by __________
a) Treating then with Chitinase
b) By treating them with penicillin
c) By electroporation
d) By pentrifugation
Explanation: Electroporation is a technique where a short pulse f electricity is used to induce artificial competence in some cells. Treating with chitinase or penicillin will severely affect the cell wall and the cells will die.
6. If a bacterium is heterozygous for a gene then will both the genes have equal probability of incorporation?
Explanation: Incorporation depends on which strand the gene is in. One of the strands gets degraded while entering the bacteria. This is via a specific 5’ or 3’ exonuclease.
7. What is the minimum length of DNA fragment in exogenote required for incorporation of the gene in the endogenote?
Explanation: Although very small fragments of DNA are taken up by the cell, a minimum of 500bp is necessary for recombinational machinery to act on, thus for incorporation.
8. If the probability of incorporation of a gene A from a bacterial cell that has lysed in the medium is a. And the probability of incorporation of another gene B from the same bacteria that has lysed is b. What is the probability of one recipient receiving both?
c) a X b
Explanation: For both genes to be incorporated together we must multiply the probability of incorporation of either genes. Thus, the probability value overall decreases.
9. Probability of incorporation of a gene P is 0.2 and the probability of incorporation of another gene is 0.1. The probability of both being incorporated is seen experimentally to be equal to 0.1. What is our inference?
a) Both the genes are incorporated randomly
b) The genes are linked
c) The bacteria has preference for the two gene
d) It was just an artifact
Explanation: The probability of incorporation of both should be theoretically 0.2X0.1= 0.02. But experimentally it is found to be 0.1 which is same as that of 2nd gene. Thus these genes are completely linked i.e. whenever b is incorporated a is incorporated as well.
10. If co-incorporation probabilities are given as p,o 0.1; o,r 0.02 ; p,r 0.15. What is the gene order
a) o r p
b) o p r
c) r o p
d) Can’t be determined due to insufficient data
Explanation: This is because o and p have an appreciable degree of co-incorporation, as does r and p but o and r have much lower degree of co-incorporation. This shows that p must be at the middle of o and r.
Sanfoundry Global Education & Learning Series – Cytogenetics.
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