Molecular Basis of Inheritance Test

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Molecular Basis of Inheritance Test - 92

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Weekly Quiz Competition

Question 1
1 / -0

The process of the formation of RNA from DNA is

A
Transduction

B
Transcription

C
Transversion

D
Transition

Solution

a) Transduction is the process of gene transfer wherein a foreign DNA, brought in a cell by viral particles, is incorporated into cell’s genome.
b) Transcription is the process of synthesis of RNA by RNA polymerase using DNA template strand.
c) Transversion is the substitution of purine base for a pyrimidine base or substitution of pyrimidine base for a purine base. Hence adenine is replaced with thymine/ uracil and guanine with cytosine.
d) Transition is the substitution of one type of base by other of its own type. Hence, purine is replaced with purine (adenine to guanine and vice versa) and pyrimidine is replaced with a pyrimidine (cytosine to thymine/uracil and vice versa) only.
So, the correct answer is option B.
Question 2
1 / -0

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

A
Both strands of DNA will be radioactive.

B
One strand of DNA will be radioactive.

C
Each strand of DNA will be half radioactive.

D
None of the strands of DNA will be radioactive.

Solution

Matthew Meselson and Franklin Stahl demonstrated the semi-conservative mode of replication in DNA and chromosomes in E.coli cells. After the incorporation of radioactive thymidine 3H, the cells were transferred to growing medium. Radioactive chromosomes (with labelled DNA having 3H) appeared in the form of scattered black dots of silver grains. After replication of DNA and constitution of chromosomes following facts were noticed: (a) It was found in first generation that radioactivity was uniformly distributed in both the chromosomes. In such cases original strand of DNA double helix was labelled with 3H and newly formed strand was non- labelled. (b) During second division only one of two chromosomes represented radioactivity by showing one strand radioactive (original) and other one non-radioactive (newly formed). Hence in E.coli, after the second division, only one strand of DNA will be radioactive.
Therefore, the correct answer is option B.
Question 3
1 / -0

Because most of the amino acids are represented by more than one codons, the genetic code is said to be

A
Deaminated

B
Comma less

C
Degenerate

D
Overlapping

Solution

Degeneracy of genetic codons means a specific amino acid can be coded by more than one triplet codon. For example, glutamic acid can be coded by GAA and GAG codons.
Leucine is coded by UUA, UUG, CUU, CUC, CUA, CUG codons.
Serine is coded by UCA, UCG, UCC, UCU, AGU, AGC.
So, the correct option is 'Degenerate'.
Question 4
1 / -0

The Human Genome Project (HGP) was initiated in

A
$$1988$$

B
$$1990$$

C
$$1992$$

D
$$1994$$

Solution

The Human Genome Project was started in 1990 by National Human Genome Research Institute (NHGRI) of the National Institutes of Health (NIH), the U.S. Department of Energy (DOE), with the international partners in UK, France, Germany, Japan and China.
Human genome project aims at determining the sequence of nucleotide that makes the whole human DNA. Genome sequencing then has its applications in identification of disease-causing genes and their elimination, to introduce new genes that code for desired protein products and to exploit polymorphism to compare DNA samples from two individuals. Thus, the correct answer is option B.
Question 5
1 / -0

Which of the following is correct pair of pyrimidine bases?

A
Adenine and Thymine

B
Adenine and Guanine

C
Thymine and Cytosine

D
Guanine and Cytosine
Solution
DNA molecule is made up of two types of nitrogen bases purine and pyrimidine.
Cytosine and Thymine (C, T) are examples of pyrimidine which is composed of a single ring 6 atom structure.
Adenine and Guanine (A, G) are the examples of purine which is composed of a double ring 9 atom structure.
A base pair is one of the pairs A-T or C-G.
The A-T pair forms two hydrogen bonds.
The C-G pair is formed by three hydrogen bonding between complementary bases that holds the two strands of DNA together.
So, the correct answer is 'Thymine and Cytosine'.
Question 6
1 / -0

Helical structure of DNA was proposed by

A
Chargaff

B
Watson and Crick

C
Griffith

D
All

Solution

James Watson and Francis Crick (1353) proposed the helical structure of DNA for which they were awarded the Nobel prize.
Question 7
1 / -0

How many sense codons code for $$20$$ known essential amino acids?

A
$$61$$

B
$$62$$

C
$$63$$

D
$$64$$

Solution

The essential feature of the genetic code is the strict one-to-one correspondence between codons and amino acids. The canonical code consists of three stop codons and 61 sense codons that encode 20% of the amino acid repertoire observed in nature
So, the correct answer is '61'
Question 8
1 / -0

Length of one turn of DNA is

A
3.4 $$\mathring{A}$$

B
34 $$\mathring{A}$$

C
20 $$\mathring{A}$$

D
3.04 $$\mathring{A}$$

Solution

Distance between two pained nucleotides is 3.4 $$\mathring{A}$$ ($$\mathring{A}$$ = Angstrom). Each turn of double helices contains ten pained nucleotides. DNA helix makes one fall turn every 34 $$\mathring{A}$$
Question 9
1 / -0

A $$340$$ $${^\circ}{A}$$ long segment of DNA molecule has $$20$$ thymine nitrogenous bases, what will be the number of guanine nitrogen bases in the same segment?

A
$$10$$

B
$$40$$

C
$$80$$

D
$$160$$

Solution

Length of one base pair = 3.4 $$\overset {\circ}{A}$$
If the segment of DNA has 20 thymine bases, then it will have 20 adenine bases (as A-T is formed in DNA).
So, the length covered by 20 thymine base = 20 X 3.4 = 68 $$\overset {\circ}{A}$$
The length covered by G-C will be 340 - 68 = 272 $$\overset {\circ}{A}$$
So, the number of guanine will be 272/3.4 = 80
Thus the correct answer is option C.
Question 10
1 / -0

The codon for anticodon 3'-UUUA-5' is

A
3'-AAU-5'

B
5'-UAAA-3'

C
5'-AAAU-3'

D
3'-UAAU-5'

Solution

Uracil (U) always pairs with adenine (A). Further, the mRNA codon is read in 5’ to 3’ direction. Alignment of the mRNA and tRNA is antiparallel, which means that 5’ end of mRNA pairs with 3’ end of tRNA. This makes options A and D incorrect. The first base of the codon in mRNA (read in the 5’ to 3’direction) pairs with the third base of the anticodon, thus the anticodon 3'-UUUA-5' would pair with 5'-AAAU-3'codon. Thus option C is correct.

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Molecular Basis of Inheritance Test - 92

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Question 1

Transduction

Transcription

Transversion

Transition

Solution

Question 2

Both strands of DNA will be radioactive.

One strand of DNA will be radioactive.

Each strand of DNA will be half radioactive.

None of the strands of DNA will be radioactive.

Solution

Question 3

Deaminated

Comma less

Degenerate

Overlapping

Solution

Question 4

$$1988$$

$$1990$$

$$1992$$

$$1994$$

Solution

Question 5

Adenine and Thymine

Adenine and Guanine

Thymine and Cytosine

Guanine and Cytosine

Solution

Question 6

Chargaff

Watson and Crick

Griffith

All

Solution

Question 7

$$61$$

$$62$$

$$63$$

$$64$$

Solution

Question 8

3.4 $$\mathring{A}$$

34 $$\mathring{A}$$

20 $$\mathring{A}$$

3.04 $$\mathring{A}$$

Solution

Question 9

$$10$$

$$40$$

$$80$$

$$160$$

Solution

Question 10

3'-AAU-5'

5'-UAAA-3'

5'-AAAU-3'

3'-UAAU-5'

Solution

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