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31 May 2015

CSIR NET Life Sciences Mock Test I

  Edubio       31 May 2015
Questions

1. The greatest distance among the three genes is between a and c. What does this mean?
a. Gene a is closest to b.
b. Genes are in the order: a–b–c.
c. Gene a is not recombining with c.
d. Gene a is between b and c.
2. Which of the following mutations is most likely to cause a phenotypic change?
a. a duplication of all or most introns
b. a large inversion whose ends are each in intergenic regions
c. a nucleotide substitution in an exon coding for a transmembrane domain
d. a single nucleotide deletion in an exon coding for an active site

3. Genes a and b are 20 cM apart. An a+ b+/a+ b+ individual was mated with an a b/a b individual. If the F1 was crossed to a b/a b individuals, what offspring would be expected, and in what proportions?
a. 40% a+ b+/a b, 40% a b/a b, 10% a+ b/a b, 10% a b+/a b
b. 50% a+ b+/a b, 50% a b/a b
c. 40% a+ b+/a b, 10% a b/a b, 40% a+ b/a b, 10% a b+/a b
d. 40% a+ b+/a b, 10% a b/a b, 10% a+ b/a b, 40% a b+/a b

4. 5-Bromuracil in keto form induces mutations because it
a. replaces a T and binds to G rather than A
b. replaces a G and binds to A rather than C
c. changes the binding affinity of G
d. changes the binding affinity of

5. A certain kind of snail can have a right-handed direction of shell coiling (D) or left-handed coiling (d). If direction of coiling is due to a protein deposited by the mother in the egg cytoplasm, then a Dd egg-producing snail and a dd sperm-producing snail will have offspring of which genotype(s) and phenotype(s)?
a. 1/2 Dd:1/2 dd; all right coiling
b. all Dd; all right coiling
c. 1/2 Dd:1/2 dd; half right and half left coiling
d. all Dd; all left coiling

6. A couple has a child with Down syndrome. The mother is 39 years old at the time of delivery. Which of the following is the most probable cause of the child's condition?
a. The woman inherited this tendency from her parents.
b. One member of the couple carried a translocation.
c. One member of the couple underwent nondisjunction in somatic cell production.
d. One member of the couple underwent nondisjunction in gamete production.

7. Suppose that a gene on human chromosome 18 can be imprinted in a given pattern in a female parent but not in a male parent. A couple in whom each maternal meiosis is followed by imprinting of this gene have children. What can we expect as a likely outcome?
a. All sons but no daughters will bear their mother's imprinting pattern.
b. All daughters but no sons will bear their mother's imprinting pattern.
c. All sons and daughters will have a 50% chance of receiving the mother's imprinting pattern.
d. All the children will bear their mother's imprinting pattern but only daughters will then pass it dow

8. A phenotypically normal prospective couple seeks genetic counseling because the man knows that he has a translocation of a portion of his chromosome 4 that has been exchanged with a portion of his chromosome 12.Although he is normal because his translocation is balanced, he and his wife want to know the probability that his sperm will be abnormal. What is your prognosis regarding his sperm?
a. 1/4 will be normal, 1/4 will have the translocation, and 1/2 will have duplications and deletions.
b. All will carry the same translocation as the father.
c. None will carry the translocation since abnormal sperm will die.
d. His sperm will be sterile and the couple might consider adoption.

9. In humans, male-pattern baldness is controlled by an autosomal gene that occurs in two allelic forms. Allele Hn determines nonbaldness, and allele Hb determines pattern baldness. In males, because of the presence of testosterone, allele Hb is dominant over Hn. If a man and woman both with genotype HnHb have a son, what is the chance that he will eventually be bald?
a. 0%
b. 25%
c. 75%
d. 50%

10. If nondisjunction occurs in meiosis II during gametogenesis, what will be the result at the completion of meiosis?
a. All the gametes will be diploid.
b. Half of the gametes will be n + 1, and half will be n - 1.
c. 1/4 of the gametes will be n + 1, 1/4 will be n - 1, and 1/2 will be n.
d. There will be three extra gametes
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