Molecular Basis of Inheritance Test

Result

Molecular Basis of Inheritance Test - 48

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The part of a DNA molecule that varies among DNA molecules is its

A
Glycerol attachments

B
Nitrogenous bases

C
Sugars

D
Phosphates
Question 2
1 / -0

mRNA that carries information for complete polypeptide synthesis is

A
Muton

B
Codon

C
Operon

D
Reading frame

Solution

An operon consists of the gene cluster, promoter and other regulatory sequences that function together in regulation of transcription; option A is incorrect. Codon is a set of three nucleotides that code for an amino acid; it does not code for the whole polypeptide. mRNA nucleotide sequence is read in codons which in turn determine the type and sequence of amino acids in synthesizing polypeptides. Mucons are the smallest unit of mutation of a gene; a recon consists of one to several muton. A stretch of mRNA that carries information for polypeptide synthesis constitutes reading frame. Any given mRNA sequence can have three possible reading frames each of which gives a different sequence of codons. A specific first codon in the sequence determines the reading frame, in which a new codon begins every three nucleotide residues. For example, a DNA sequence “AAAGGGTTTCCC” can have three possible reading frames:
1. “AAA’GGG’TTT’CCC”,
2. “A’AAG’GGT’TTC’CC” and
3. “AA’AGG’GTT’TCC’C”.
Therefore, the correct answer is option D.
Question 3
1 / -0

Which of the following is important for transcription?

A
CAAT box

B
Promoter

C
RNA polymerase

D
All of the above

Solution

CAAT is a promoter sequence that lies between -70 and -80 base pairs and is essential for transcription initiation. The sequence is GGT/ACAATCT. The promoter is a DNA segment that serves to initiate transcription of a particular gene. They are present upstream near the transcription start sites of genes and provide the binding site for RNA polymerase. Transcription is the first step of gene expression, in which a particular segment of DNA is copied into RNA (mRNA) by the enzyme RNA polymerase. Thus, the correct answer is option D.
Question 4
1 / -0

Diameter of DNA double helix is

A
34 $$A^0$$

B
20 $$A^0$$

C
3.4 $$A^0$$

D
340 $$A^0$$

Solution

The B-DNA is arranged in a right-handed double helix; the helix is 20 angstrom wider and the number of base pairs per helical turn is 10.5. The plane of the base pairs in B-DNA is tilted about 6 degree with respect to the helix axis. One complete turn of the double helix is 3.4 nm long and each of which has 10.5 base pairs, the distance between two consecutive base pairs 0.34 nm. Correct answer is B.
Question 5
1 / -0

Which is true for DNA helicases?

A
Separate DNA strands and establish replication forks

B
ATP requiring unwinding enzymes

C
Hydrolyse ATP

D
All of the above

Solution

During DNA replication, two hexamers of helicases are loaded onto each DNA strand. Helicases move along the double-stranded DNA and separate the strands by breaking hydrogen bonds between base pairs. ATP hydrolysis provides the required energy for breaking hydrogen bonds; options B and C are correct. Thereby, DNA is unwound bidirectionally and two potential replication forks are created. The point at which two DNA strands are separated to facilitate the replication is called replication fork; option A is correct.
Therefore, the correct answer is D.
Question 6
1 / -0

Okazaki fragments are

A
RNA primers

B
Short DNA fragments on leading strand

C
Short DNA fragments on lagging strand

D
DNA fragments from dimerization
Solution
Correct Option: C
Explanation:

RNA primer is a stranded segment (complementary to the template) and has a free 3’ hydroxyl group to facilitate the addition of nucleotides by polymerase enzymes; option A is incorrect.
The leading strand synthesis proceeds continuously from a single RNA primer in the 5'-->3' direction and there are no short fragments formed on it during replication which makes option B wrong.
To facilitate the growth of the lagging strand in the 5'-->3' direction, a new primer is added 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; option C is correct.
Ultraviolet light causes DNA dimerization by introducing pyrimidine dimers which are then repaired by DNA repair mechanisms; option D is incorrect.
Question 7
1 / -0

How many types of RNA polymerases are operative in eukaryotes?

A
Four

B
Three

C
Two

D
One

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 is synthesized mRNAs and some specialized RNAs. RNA polymerase III makes tRNAs, the 5S rRNA, and some other small specialized RNAs. Option B is correct.
Question 8
1 / -0

A double stranded DNA having 100,000 bp will have a length of

A
10,000 nm

B
$$10^5$$ nm

C
$$3.4\, \times\,10^4$$ nm

D
200,000 nm
Solution
- As we know that DNA is a double helix; one complete turn of the double helix is 3.4 nm long, each of which has 10 base pairs thus the distance between two consecutive base pairs is $$3.4/10 = 0.34$$ nm.
- Therefore, double-stranded DNA having 100,000 bp will have a length of $$0.34$$ $$\times$$ $$10^5$$ = 3.4 x $$10^4$$ nm.
Thus, the correct answer is C.
Question 9
1 / -0

Adenine is 30%, what would be the percentage of guanine?

A
10%

B
20%

C
30%

D
40%

Solution

According to Chargaff’s rule, in all cellular DNAs, regardless of the species, number of adenosine residues is equal to the number of thymidine residues which means that A=T; and the number of guanosine residues is equal to the number of cytidine residues; G = C. Hence, the sum of the purine residues equals the sum of the pyrimidine residues; i.e. A + G = T + C. Here, [A] = 30%, therefore, % of [T] is also 30%. Thus, [G]+[C] = 100 - 60 = 40%. [G] = 20% and [C] = 20%. The correct answer is B.
Question 10
1 / -0

Exon segments are reunited after splicing by

A
RNA primase

B
DNA polymerase

C
RNA polymerase

D
RNA ligase

Solution

RNA primase synthesizes RNA 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. RNA polymerase synthesizes RNA using DNA template strand by adding the ribonucleotides to a primer in 5' to 3' direction via phosphodiester bonds. Splicing is the removal of introns from the primary transcript and ligation of exons to form a continuous sequence specifying a functional polypeptide. RNA ligase enzyme joins two exons together via phosphodiester bond between 3’ hydroxyl at the end of one exon and a 5’ phosphate at the end of another exon. The correct answer is D.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Molecular Basis of Inheritance Test - 48

Result

Molecular Basis of Inheritance Test - 48

Score

-

Rank

-

SHARING IS CARING

Question 1

Glycerol attachments

Nitrogenous bases

Sugars

Phosphates

Question 2

Muton

Codon

Operon

Reading frame

Solution

Question 3

CAAT box

Promoter

RNA polymerase

All of the above

Solution

Question 4

34 $$A^0$$

20 $$A^0$$

3.4 $$A^0$$

340 $$A^0$$

Solution

Question 5

Separate DNA strands and establish replication forks

ATP requiring unwinding enzymes

Hydrolyse ATP

All of the above

Solution

Question 6

RNA primers

Short DNA fragments on leading strand

Short DNA fragments on lagging strand

DNA fragments from dimerization

Solution

Question 7

Four

Three

Two

One

Solution

Question 8

10,000 nm

$$10^5$$ nm

$$3.4\, \times\,10^4$$ nm

200,000 nm

Solution

Question 9

10%

20%

30%

40%

Solution

Question 10

RNA primase

DNA polymerase

RNA polymerase

RNA ligase

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.