Molecular Basis of Inheritance Test

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Molecular Basis of Inheritance Test - 90

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Weekly Quiz Competition

Question 1
1 / -0

Degeneration of genetic code is due to

A
First member of codons

B
Second member of codons

C
Third member of codons

D
Entire codons

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 the 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. There is the strong base pairing between first two bases of an mRNA codon and corresponding bases of the tRNA anticodon and first base of the anticodon (5’-3’direction) pairs with the third base of the codon; this makes options A and B wrong. Thus, the codon base of lesser importance is the number-three base, whereas it is the number-one base in the anticodon. The correct answer is C.
Question 2
1 / -0

RNA code for DNA codon ATG will be

A
ATG

B
AUG

C
UAC

D
TAC

Solution

RNA is synthesized by transcription of DNA template strand; therefore nucleotide sequence of newly formed RNA is complementary to its DNA template strand. DNA and RNA both have four bases; two purines and two pyrimidines, in the nucleotide chains. There are two purines namely adenine and guanine and three pyrimidines namely cytosine, thymine and uracil. The purines adenine and guanine and the pyrimidine cytosine are common to both DNA and RNA. DNA has the pyrimidine thymine; RNA has the pyrimidine uracil. A purine base pairs with pyrimidine and vice versa; adenine and guanine pair with thymine and cytosine; this makes option B wrong. Since RNA has uracil in place of thymine; adenine pairs with uracil in RNA. A DNA codon ATG will code for RNA codon “UAC”. Option C is correct. RNA has uracil in place of thymine; this makes options A and D wrong. The correct answer is C.
Question 3
1 / -0

Find out the correct matching.
a Helicase i Joining of nucleotides
b Gyrase ii Opening of DNA
c Primase iii Unwinding of DNA
d DNA polymerase III iv RNA priming

A
a- ii, b- iii, c- iv, d- i

B
a- i, b- ii, c- iii, d- iv

C
a- iv, b- iii, c- i, d- ii

D
a- ii, b- iv, c- iii, d- i

Solution

DNA gyrase is topoisomerase II that can introduce negative supercoils using the energy of ATP. It cleaves both strands of a DNA molecule and passes another duplex through the break thereby unwinding them.
Helicases move along the double-stranded DNA and separate/open the strands by breaking hydrogen bonds between base pairs. ATP hydrolysis provides the required energy for breaking of hydrogen bonds.
Primase synthesise RNA/DNA primer which is a stranded segment (complementary to the template) and has a free 3’ hydroxyl group to facilitate the addition of nucleotides by polymerase enzymes.
DNA polymerase III is main DNA polymerising enzyme that has 5’ to 3’ polymerization and 3’ to 5’ exonuclease proofreading activities.
So, the correct answer is option A.
Question 4
1 / -0

The two polynucleotide chains of DNA are

A
Discontinuous

B
Antiparallel

C
Parallel

D
Conservative
Solution
Correct Option: B
Explanation:
The two DNA strands are antiparallel, and have opposite polarity, means that 5’ end of one strand faces 3’ end of other strand and vice versa. This means that 5’ phosphate of one strand faces 3’ hydroxyl group of other strand and that the 5’ phosphate groups of two strands are present in opposite position.
DNA replication is semiconservative in which a 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.
Synthesis of lagging DNA strand is discontinuous; DNA strands are continuous. Thus, the correct answer is option B.
Question 5
1 / -0

Which enzyme catalyses the synthesis of a new strand for a DNA molecule by linking nucleotides to the developing strand?

A
RNA polymerase

B
DNA ligase

C
DNA polymerase

D
Topoisomerase
Solution
- RNA polymerase helps in DNA dependent RNA synthesis. This process is known as transcription. DNA strands are used as a template to synthesize RNA molecules.
- DNA ligase is used to join DNA segments during DNA repair or during replication (to join Okazaki fragments of the lagging strands).
- DNA polymerase works in the pair to synthesis new strand for DNA molecule from deoxyribonucleotides. In other words, it helps in DNA dependent DNA synthesis.
- Topoisomerase enzyme participates in overwinding or unwinding of DNA. During DNA replication or transcription, DNA becomes overwound ahead of the replication fork. It binds to DNA and cut the phosphate backbone of either one or both the DNA strands.
So, the correct option is 'DNA polymerase'.
Question 6
1 / -0

Okazaki fragments during DNA replication

A
Polymerise in 3' $$\rightarrow$$ 5' direction and form replication fork

B
Prove semiconservative DNA replication

C
Short, newly synthesized DNA fragments that are formed on the lagging template strand

D
Cause transcription process

Solution

DNA synthesis always occurs in the 5'-->3' direction because chain growth results from the formation of a phosphodiester bond between the 3' oxygen of a growing strand and $$\alpha$$ phosphate of a dNTP; option A is incorrect.
To facilitate the growth of the lagging strand in the 5'-->3' direction, copying of its template strand must somehow occur in the opposite direction from the movement of the replication fork, i.e., in 3’ to 5’ direction; option C is correct. This is done by synthesizing a new primer every few hundred bases on the second parental strand. Each of these primers is elongated in the 5'-->3' direction and form discontinuous segments called Okazaki fragments which are later joined together by DNA ligase.
The semiconservative mechanism of DNA replication in which a 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, is followed by both DNA strands. Option B is incorrect. Transcription is the process of RNA synthesis using DNA template strand which is carried out by RNA polymerase; any of two DNA strands can serve as the template, option D is incorrect. The correct answer is C.
Question 7
1 / -0

E.coli about to replicate was placed in a medium containing radioactive thymidine for five minutes. Then it was made to replicate in a normal medium. Which of the following observation will be correct?

A
Each strand is half radioactive

B
Both the strands of DNA will be radioactive

C
One strand will be radioactive

D
None of the above

Solution

A.Correct option -C

B.Explanation for correct option -C
●DNA replication is semi conservative in which a 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.
●Culture of E. coli bacteria in radioactive thymidine incorporates tritium in place of the hydrogen atom in thymidine.
●DNA replication in tritiated thymidine medium synthesizes daughter DNA strand with tritiated thymidine and the new DNA duplex is hybrid of daughter strand with tritiated thymidine and parental DNA strand with normal thymidine.
Question 8
1 / -0

Which amino acid is specified by genetic codes ACU, ACC, ACA, ACG showing degeneracy?

A
Leucine

B
Methionine

C
Glycine

D
Threonine

Solution

Genetic code shows degeneracy which means that multiple codons specify particular amino acid. Amino acid leucine is encoded by 6 codons; UUA, UUG, CUU, CUC, CUA and CUG, thus it shows degeneracy but is not encoded by codons given in the question which makes option A wrong. Methionine is the exception to code degeneracy as it is encoded by single codon “AUG” which makes option B wrong. Glycine is encoded by GGU, GGC, GGA and GGG; thus shows code degeneracy but is not specified by codons mentioned in question which makes option C incorrect. Threonine is encoded by multiple codons ACU, ACC, ACA and ACG; option D is the correct answer.
Question 9
1 / -0

Complete turns in 45,000 bp DNA would be

A
45

B
450

C
4,500

D
45,000

Solution

Since one complete turn of the double helix is 3.4 nm long and each of which has 10 base pairs, 45000 base pairs DNA will have 45000/10 = 4500 turns. Correct answer is option C.
Question 10
1 / -0

For the structure of nucleic acid, which of the following statements is wrong?

A
DNA can be single stranded in some viruses.

B
RNA can be double stranded occasionally.

C
There are as many as 12 bases per turn in Z DNA.

D
The length of one helix is 45$$\mathring A$$ in B-DNA and in Z-DNA.

Solution

In a DNA molecule, the two strands are not parallel, but intertwined with each other. Each strand looks like a helix. The two strands form a "double helix" structure, which was first discovered by James D. Watson and Francis Crick in 1953. In this structure, also known as the B form, the helix makes a turn every 3.4 nm, and the distance between two neighboring base pairs is 0.34 nm. Hence, there are about 10 pairs per turn. The intertwined strands make two grooves of different widths, referred to as the major groove and the minor groove, which may facilitate binding with specific proteins.
In a solution with higher salt concentrations or with alcohol added, the DNA structure may change to an A form, which is still right-handed, but every 2.3 nm makes a turn and there are 11 base pairs per turn.
Another DNA structure is called the Z form, because its bases seem to zigzag. Z DNA is left-handed. One turn spans 4.6 nm, comprising 12 base pairs. The DNA molecule with alternating G-C sequences in alcohol or high salt solution tends to have such structure.
In a Z-DNA, one complete tun of the helix is 45$$\mathring A$$ long, while in B DNA it is 34 $$\mathring A$$ long.

Genetic information is carried in viruses by either RNA or DNA. The genome of a virus can be either single stranded RNA (ssRNA), double stranded RNA (dsRNA), single stranded DNA (ssDNA), double stranded DNA ds (DNA).
Thus, the correct answer is option D.

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a	Helicase	i	Joining of nucleotides
b	Gyrase	ii	Opening of DNA
c	Primase	iii	Unwinding of DNA
d	DNA polymerase III	iv	RNA priming

Molecular Basis of Inheritance Test - 90

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Molecular Basis of Inheritance Test - 90

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Question 1

First member of codons

Second member of codons

Third member of codons

Entire codons

Solution

Question 2

ATG

AUG

UAC

TAC

Solution

Question 3

a- ii, b- iii, c- iv, d- i

a- i, b- ii, c- iii, d- iv

a- iv, b- iii, c- i, d- ii

a- ii, b- iv, c- iii, d- i

Solution

Question 4

Discontinuous

Antiparallel

Parallel

Conservative

Solution

Question 5

RNA polymerase

DNA ligase

DNA polymerase

Topoisomerase

Solution

Question 6

Polymerise in 3' $$\rightarrow$$ 5' direction and form replication fork

Prove semiconservative DNA replication

Short, newly synthesized DNA fragments that are formed on the lagging template strand

Cause transcription process

Solution

Question 7

Each strand is half radioactive

Both the strands of DNA will be radioactive

One strand will be radioactive

None of the above

Solution

Question 8

Leucine

Methionine

Glycine

Threonine

Solution

Question 9

45

450

4,500

45,000

Solution

Question 10

DNA can be single stranded in some viruses.

RNA can be double stranded occasionally.

There are as many as 12 bases per turn in Z DNA.

The length of one helix is 45$$\mathring A$$ in B-DNA and in Z-DNA.

Solution

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