Molecular Basis of Inheritance Test

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Molecular Basis of Inheritance Test - 78

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Weekly Quiz Competition

Question 1
1 / -0

Hypothesis of ''African origin'' of man is based on the study of____________

A
Nuclear DNA

B
Mitochondrail DNA

C
rRNA

D
X-chromosome
Solution
Hypothesis of ''African origin'' of man is based on the study of Mitochondrial DNA.
Mitochondrial DNA possesses several favorable characteristics, including large quantity in the cell, small genome size, haploid, maternal inheritance and extremely low probability of paternal leakage , higher mutation rate than nuclear DNA, and change mainly through mutation rather than recombination.
All these features make mt-DNA a useful and one of the most frequently used markers in molecular systematics and has been widely employed to address questions of genetic diversity, population structure, phylogeography and population evolution.
So, the correct option is 'Mitochondrial DNA'.
Question 2
1 / -0

Calculate the number of nitrogen bases in E.coli, if the length of E.coli DNA is 1.36 mm.

A
$$4 \times 10^6$$

B
$$8 \times 10^6$$

C
$$2 \times 10^6$$

D
$$16 \times 10^6$$

Solution

The distance between two base pairs is 0.34 X 10$$^{-9}$$ m. The total length of DNA in E. coli is 1.36 mm.
Total length of DNA = the total number of base pairs x distance between two consecutive base pairs
Hence, total number of base pairs = total length of DNA / distance between two consecutive base pairs
= 1.36 / 0.34 X 10$$^{-9}$$ = 4 X 10$$^6$$ base pairs.
Thus, the correct answer is '4 x 10$$^6$$.'
Question 3
1 / -0

1.38 milimeter DNA is present in E. coli. How many base pair will be present in it?

A
$$4 \times 10^6 bp$$

B
$$4 \times 10^7 bp$$

C
$$4 \times 10^8 bp$$

D
$$4 \times 10^5 bp$$

Solution

The distance between two base pairs is 0.34 X 10$$^{-9}$$ m. The total length of DNA in E. coli is 1.38 millimeter.
Total length of DNA = the total number of base pairs x distance between two consecutive base pairs
Hence, total number of base pairs = total length of DNA / distance between two consecutive base pairs
= 1.38 / 0.34 X 10$$^{-9}$$ = 4 X 10$$^6$$ base pairs.
Thus, the correct answer is '4 x 10$$^6$$ bp'.
Question 4
1 / -0

Name the enzyme which removes the non coding split genes

A
Ribosome

B
Lysozyme

C
Ribozyme

D
Helicase
Solution
- Ribozymes (ribonucleic acid enzymes) are RNA molecules that are capable of catalyzing specific biochemical reactions, similar to the action of protein enzymes.
- They also participate in a variety of RNA processing reactions, including RNA splicing, viral replication, and transfer RNA biosynthesis.
- Hence. Ribozyme removes the non-coding split genes.
- So, the correct answer is 'Ribozyme'.
Question 5
1 / -0

Maximum possible transversion, which may take place in genetic code

A
$$366$$

B
$$183$$

C
$$549$$

D
In each coden $$9$$ only

Solution

Transversion is a type of gene mutation in which single purine is changed into pyrimidine. The process may be transversion. It explains the distributation of gene coding.
So, the correct option is '366'.
Question 6
1 / -0

Transfer of information from molecule of DNA to mRNA is called as

A
Transcription

B
Translation

C
Replication

D
None of the above

Solution

Transfer of information from molecule of DNA to mRNA is called as transcription. Translation of mRNA into amino acid sequence of protein is called as translation. The process of producing two identical copies of DNA from one original double stranded DNA molecule is called as replication.
Thus, the correct answer is 'Transcription.'
Question 7
1 / -0

A virus containing $$ssRNA$$ act as a template for $$DNA$$ synthesis is called as ...........

A
Polio virus

B
Retro virus

C
Adenovirus

D
Pox virus

Solution

A retrovirus is a group of viruses that belong to the family 'Retroviridae' and they characteristically carry genetic material in the form of ssRNA. Retrovirus is named for an enzyme 'Reverse transcriptase'. Retrovirus contains ssRNA which acts as a template for the synthesis of DNA.
So, the correct answer is 'Retrovirus'.
Question 8
1 / -0

Theta replication of DNA not occur in

A
E.coli

B
Mitochondrial

C
Chloroplast

D
Nucleus
Solution
Replication of DNA occurs in circular DNA molecules is called Theta replication of DNA. It is found in prokaryotes, mitochondria, and chloroplast.
It does not occur in eukaryotic chromosomes present in the nucleus, because in this, linear dsDNA is present.
So, the correct option is 'Nucleus'.
Question 9
1 / -0

Purine base of DNA molecule has

A
A and C

B
C and T

C
A and G

D
None of the above

Solution

Purine base of DNA molecule has Adenine(A) and Guanine(G) which bond with their complimentary pyrimidines, Thymine(T) and Cytosine(C) respectively.
A purine is a heterocyclic aromatic organic compound, consisting of a pyrimidine ring fused to an imidazole ring.
So, the correct answer is 'A and G'.
Question 10
1 / -0

All living organisms of the present,past and future are

A
Not linked to any extent

B
Linked to one another in their cell membrane

C
Linked to one another by the sharing of the common genetic materials ,but to varying degrees

D
Linked to another by the sharing of the common genetic material to $$100\%$$ extent

Solution

All living organisms—present, past and future are linked to one another by the sharing of the common genetic material, but to varying degrees.
The genetic material of living organisms undergoes mutations and reshuffling of genes. This causes variations in the genetic material. Variations are so abundant that no two individuals of the same race are exactly similar.
So the answer is 'Linked to one another by the sharing of the common genetic materials, but to varying degrees'.

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Molecular Basis of Inheritance Test - 78

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Molecular Basis of Inheritance Test - 78

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Question 1

Nuclear DNA

Mitochondrail DNA

rRNA

X-chromosome

Solution

Question 2

$$4 \times 10^6$$

$$8 \times 10^6$$

$$2 \times 10^6$$

$$16 \times 10^6$$

Solution

Question 3

$$4 \times 10^6 bp$$

$$4 \times 10^7 bp$$

$$4 \times 10^8 bp$$

$$4 \times 10^5 bp$$

Solution

Question 4

Ribosome

Lysozyme

Ribozyme

Helicase

Solution

Question 5

$$366$$

$$183$$

$$549$$

In each coden $$9$$ only

Solution

Question 6

Transcription

Translation

Replication

None of the above

Solution

Question 7

Polio virus

Retro virus

Adenovirus

Pox virus

Solution

Question 8

E.coli

Mitochondrial

Chloroplast

Nucleus

Solution

Question 9

A and C

C and T

A and G

None of the above

Solution

Question 10

Not linked to any extent

Linked to one another in their cell membrane

Linked to one another by the sharing of the common genetic materials ,but to varying degrees

Linked to another by the sharing of the common genetic material to $$100\%$$ extent

Solution

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