Molecular Basis of Inheritance Test

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Molecular Basis of Inheritance Test - 89

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Question 1
1 / -0

The element required for the activation of DNA as well as DNA polymerase is

A
Ca$$^{++}$$

B
Mg$$^{++}$$

C
Cu$$^{++}$$

D
K$$^+$$
Solution
- A non-protein chemical that is necessary for the activity of protein/enzyme is known as a co-factor.
- Mg$$^{2+}$$ serves as co-factor for DNA polymerase. It activates the DNA polymerase that is used to synthesise the DNA during replication.
- Thus, the correct answer is B.
Question 2
1 / -0

The width of B-DNA is ______ and each turn of its helix has ______ nucleotides.

A
34 $$\mathring A$$; 11

B
20 $$\mathring A$$ ; 12

C
40 $$\mathring A$$ ; 8

D
20 $$\mathring A$$; 10

Solution

The Watson-Crick structure is the B form DNA, or B-DNA that is the most stable structure for a random-sequence DNA molecule under physiological conditions. The DNA is arranged in a right-handed double helix, but the helix is 20$$\mathring A$$ wider and the number of base pairs per helical turn is around 10.5. The plane of the base pairs in B-DNA is tilted about 6 degree with respect to the helix axis. Correct answer is D.
Question 3
1 / -0

The codon for anticodon 3' UUA 5' is

A
5' AAU 3'

B
3' AAU 5'

C
5' AAT 3'

D
3' AAG 5'

Solution

Transfer RNA has an anticodon loop with a three-base sequence that base-pairs with mRNA codons. The mRNA codon is read in 5'-->3 direction and the first base of the codon in mRNA pairs with the third base of the anticodon. Therefore, codon for anticodon 3' UUA 5' is 5' AAU 3'. Codon and anticodon are never read in the same direction; an anticodon with 3' UUA 5' can not have a codon with 3' UUA 5' which makes option B wrong. RNA has uracil (U) in place of thymine; option C is wrong. Adenine never pairs with guanine; option D is wrong. The correct answer is A.
Question 4
1 / -0

Which one of the following group of codons is called as degenerate codons?

A
UAA, UAG and UGA

B
GUA, GUG, GCA, GCG and GAA

C
UUC, UUG, CUU, CUC, CUA and CUG

D
AAC, AAG, GAC and CGG

Solution

Genetic codon show degeneracy which means that one amino acid can be encoded by more than one codons. Codons in option A "UAA, UAG and UGA" are the stop codons which cause termination of translation. Codons in option B are GUA and GUG (valine); GCA and GCG (alanine); and GAA (glutamine). Codons in option C are UUC and UUG (phenylalanine); CUU, CUC, CUA and CUG (leucine). Codons in option D are AAC (Asparagine); AAG (lysine); GAC (aspartic acid) and CGG (arginine). As we can see that all codons in option A signify codon redundancy and hence are example of codon degeneracy. Codons in option B and C have both multiple and single codon examples but option D has all single codons; each codon for a specific amino acid. Hence, option A is the correct answer.
Question 5
1 / -0

Consider the following statements and choose the correct option.
In eukaryotes
(i) RNA polymerase I transcribes rRNA.
(ii) RNA polymerase I transcribes snRNA.
(iii) RNA polymerases II transcribes tRNA.
(iv) RNA polymerase II transcribes hnRNA.

A
1 and 2 are correct.

B
1 and 3 are correct.

C
1, 2 and 4 are correct.

D
2 and 3 are correct.

E
1 and 4 are correct.

Solution

Eukaryotes have three RNA polymerases which are structurally distinct complexes, though share certain subunits in common, and have a specific function and specific promoter sequence. RNA polymerase I synthesize preribosomal RNA (pre-rRNA), which contains the precursor for the 18S, 5.8S, and 28S rRNAs.
RNA polymerase II transcribes the presursor of mRNA which is heterogenous RNA (hnRNA).
RNA polymerase III makes tRNAs, the 5S rRNA, and sNRNAs.
Therefore, the correct answer is option (E).
Question 6
1 / -0

Which of the following substances was used as a screening agent in replica plating experiment?

A
Vitamin B$$_{12}$$

B
Cofactors and streptomycin

C
Amino acid lysine

D
Antibiotic streptomycin

Solution

Replica plating technique rapidly transfers microorganism colonies to numerous petri plates and is used for detection of auxotrophs. It was given by Joshua Lederberg & Esther Lederberg. A pattern of colonies growing on the complete medium is imprinted on velveteen by inverting the petri plate. Plates with streptomycin were pressed on that master plate to transfer the impression. Cells carrying streptomycin resistance grow on the plate with streptomycin. The areas on the second plates where colonies fail to grow to correspond to wild-type colonies without streptomycin resistance on the original plate which is then used for isolation and multiplication of mutant colonies. The correct answer is D.
Question 7
1 / -0

Which one is wrong?

A
DNA can not produce its copies without DNA polymerase.

B
DNA cannot produce RNA.

C
RNA produces complementary DNA.

D
DNA helps in protein synthesis.

Solution

DNA polymerase is the polymerizing enzyme that adds deoxyribonucleotides to free 3’ hydroxyl group of primer using the parental DNA strand as the template. It is the essential component of DNA replication process which makes statement A correct. Transcription is the process of RNA synthesis by enzyme RNA polymerase using DNA template strand; the anticoding strand serves as the template in mRNA synthesis which makes statement B incorrect. The DNA synthesized using RNA template is known as cDNA or complementary DNA; RNA retroviruses have reverse transcriptase enzyme that synthesizes a DNA copy (cDNA) of viral RNA. Statement C is correct. Replication of DNA copies the genetic information present in it which is transcribed into RNA by RNA polymerase enzyme. The RNA sequence of ribonucleotides is translated into an amino acid sequence of polypeptides by the process of translation. Thus, flow of genetic information follows the path: DNA --> RNA --> protein; statement D is correct. Option B is the correct answer.
Question 8
1 / -0

Mode of DNA replication in Escherichia coli is

A
Conservative and unidirectional

B
Semiconservative and unidirectional

C
Conservative and bidirectional

D
Semiconservative and bidirectional

Solution

DNA replication is bidirectional mechanism in E. coli because two replication forks are formed at origin which moves in opposite directions, with both template strands being copied at each fork. If it was unidirectional, the replication fork would move in single direction only. This makes options A and B incorrect. In E. coli, the parental DNA strand serves as a template for the synthesis of a new strand and produces two new DNA molecules, each with one new strand and one parental strand; thus it is the semiconservative mechanism of DNA replication. In the conservative mode of DNA replication, new DNA molecules have both newly synthesized DNA strands while the parental DNA regains its duplex nature with both original strands. This makes options A and C incorrect. The correct answer is D. DNA replication is semi-conservative and bidirectional in E. coli.
Question 9
1 / -0

Instead of 3, if it was 2 bases code for an amino acid, the degeneracy of codons coding for the same amino acid would have

A
Increased

B
Decreased

C
Remained the same

D
Been uncertain

Solution

In genetic code table, each box is specified by the first and second positions (for example the AAX box, in which X is any of the four bases) therefore eight of the sixteen boxes contain just one amino acid per box. This means codon need only be read in the first two positions for these eight amino acids because the same amino acid will be represented regardless of the third base of the codon. That’s why multiple codons specify single amino acid and make the genetic code degenerate. According to the question, an amino acid that was encoded by three codons is now encoded by two codons. This means that degeneracy for that codon has been decreased. If the degeneracy was increased, a number of codons specifying the amino acid would have increased which makes option A wrong. If the degeneracy was same, the number of codons for the amino acid would remain same; this makes option C wrong. The correct answer is B.
Question 10
1 / -0

DNA molecule has small units called

A
Uracil

B
Muton

C
Cistron

D
Recon

E
Nucleotide

Solution

A cistron is defined as a unit of function that exhibits cis-trans test. A gene carries several cistrons. The smallest unit capable of undergoing recombination is called as recon. A recon consists of one to several mucons, which are the smallest unit of mutation. This makes option C incorrect. DNA is the polymer of deoxyribonucleotides; each of which is made up of sugar, the nitrogenous base and the phosphate moiety. DNA has four bases; two purines and two pyrimidines, in the nucleotide chains. There are two purines namely adenine and guanine and two pyrimidines namely cytosine and thymine.

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Molecular Basis of Inheritance Test - 89

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Molecular Basis of Inheritance Test - 89

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Question 1

Ca$$^{++}$$

Mg$$^{++}$$

Cu$$^{++}$$

K$$^+$$

Solution

Question 2

34 $$\mathring A$$; 11

20 $$\mathring A$$ ; 12

40 $$\mathring A$$ ; 8

20 $$\mathring A$$; 10

Solution

Question 3

5' AAU 3'

3' AAU 5'

5' AAT 3'

3' AAG 5'

Solution

Question 4

UAA, UAG and UGA

GUA, GUG, GCA, GCG and GAA

UUC, UUG, CUU, CUC, CUA and CUG

AAC, AAG, GAC and CGG

Solution

Question 5

1 and 2 are correct.

1 and 3 are correct.

1, 2 and 4 are correct.

2 and 3 are correct.

1 and 4 are correct.

Solution

Question 6

Vitamin B$$_{12}$$

Cofactors and streptomycin

Amino acid lysine

Antibiotic streptomycin

Solution

Question 7

DNA can not produce its copies without DNA polymerase.

DNA cannot produce RNA.

RNA produces complementary DNA.

DNA helps in protein synthesis.

Solution

Question 8

Conservative and unidirectional

Semiconservative and unidirectional

Conservative and bidirectional

Semiconservative and bidirectional

Solution

Question 9

Increased

Decreased

Remained the same

Been uncertain

Solution

Question 10

Uracil

Muton

Cistron

Recon

Nucleotide

Solution

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