Molecular Basis of Inheritance Test

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Molecular Basis of Inheritance Test - 94

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Question 1
1 / -0

E. coli bacteria grew in $$^{15}NH_4Cl$$ medium for several generation are allowed to grow in $$^{14}NH_4Cl$$ medium. After 2 generations, the bacteria are isolated from the medium and DNA of bacteria centrifuged in CsCl. The result of the density gradient consists of DNA that is

A
Only hybrid DNA.

B
Both hybrid and heavy DNA.

C
Both heavy and light DNA.

D
Both hybrid and light DNA.

Solution

Meselson-Stahl showed the semi-conservative mechanism of DNA replication. They cultured E. coli cells initially in a medium containing ammonium salts prepared with heavy nitrogen $$^{15}$$N to label, using all cellular DNA. Then cells were transferred to a medium containing the normal light isotope $$^{14}$$N. DNA was analyzed by equilibrium density-gradient centrifugation in the periodically collected sample to separate heavy-heavy (H-H), light-light (L-L), and heavy-light (H-L) duplexes into distinct bands. The sample was placed on a preformed gradient of cesium chloride. Centrifugation separates the macromolecules in different bands of density gradient which can be collected as a pure fraction. After the first generation, the density gradient contained only one band having H-L nitrogen. After the second generation, the density gradient contained 2 bands- one will have hybrid DNA (H-L) and the other will have light DNA (L-L).
Thus, the correct answer is option D.
Question 2
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The codons UUU and UUC codes for phenylalanine only. This feature of genetic code is called

A
Commaless

B
Non- overlapping

C
Degenerate

D
Non-ambiguous
Solution
(A) Correct option (C)
(B) Explanation of correct answer:
The codons UUU and UUC code for phenylalanine only. This feature of the genetic code is called degenerate.
When multiple codons code for a single amino acid it is known as the degeneracy of code.
Question 3
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A section of DNA appears as shown above.
When replication occurs, if this molecule is being unzipped in from left to right, which strand will be the leading strand, and why?

A
The top strand, because DNA polymerase reads the template strand in a $$5\rightarrow 3$$ direction forming the new strand in a $$3\rightarrow 5$$ direction.

B
The bottom strand, because DNA polymerase reads the template strand in a $$3\rightarrow 5$$ direction making the new strand in segments running $$3\rightarrow 5$$.

C
The top strand, because DNA polymerase replicates the template strand in a continuous $$5\rightarrow 3$$ direction.

D
The bottom strand, because the leading strand is always replicated in a $$5\rightarrow 3$$ direction and the lagging strand is always replicated $$3\rightarrow 5$$.

Solution

The enzyme for the DNA replication is the DNA polymerase that works in the direction 5'-3'. So, the strand that has the opposite poarity will form the leading starnd and will form contiiuous DNA and the other will be known as the lagging strand and DNA frangments will be formed on that.

So, the correct otpyion is 'The top strand, because DNA polymerase reads the template strand in a 5→3 direction forming the new strand in a 3→5 direction.'
Question 4
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In nucleoside, the nitrogen base is attached to the first carbon of sugar by ............ bond.

A
Glycosidic

B
Hydrogen

C
Phosphodiester

D
Phosphate

Solution

The nitrogen base is attached to the sugar molecule to form the nucleoside. The phosphate groups are attached to the structure to form the nucleotide. There is N- glycosidic bond in the nucleoside which connects the carbon number 1 of sugar with purine or pyrimidine base. The chemical bonding is known as N-glycosidic type of bond.
Thus, the correct answer is option A.
Question 5
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Wobble hypothesis deals with

A
Phenomena of multiple codons coding for the code for a single amino acid

B
Phenomena of multiple codons coding for the code for a different amino acid

C
Phenomena of anticodon coding for the code for a different amino acid

D
Phenomena of anticodon coding for the code for more than one amino acid
Solution
The wobble hypothesis deals with the phenomenon of degeneracy that is seen in the genetic code through tRNA recognition of more than one codon.
Each tRNA contains a triplet anti-codon that is complementary to a codon in mRNA. The first two pairs should be complementary for binding to occur, but the third pair need not always be complementary. This is the wobble in the genetic code.
Each mRNA is made up of four different nucleotide bases (A, G, U, and C) and codons are made up of three of these bases,
Thus, 64 possible codon configurations exist. So it would be expected that there would be 64 tRNA and 64 associated amino acids.
However, only 21 amino acids exist. This means that some tRNA must bind multiple codons and provide the same amino acid for different codon configurations.
Thus, the correct answer is option A.
Question 6
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Which of the following loop in the given diagram has ribosome recognition site?

A
D-loop

B
T loop

C
Anticodon Loop

D
None of the above

Solution

The T$$_Ψ$$C pseudouridine arm (T loop) consists of a stem of 5 base pairs and a loop of 7 unpaired bases. The T$$_Ψ$$C loop consists of a T$$_Ψ$$C sequence at 5′ - 3′ direction. The T$$_Ψ$$C arm has a ribosome recognition site and binds the tRNA molecules to the ribosome.
Question 7
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tRNA was discovered by ____ and _____.

A
Brenner and Jacob

B
Jacob and Monod

C
Hoagland and Zamecnik

D
Payen and Persoz
Question 8
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Which option is correct for the amino acid and the total number of their genetic codons?

A
Val = 6, Pro = 6

B
Arg = 6, His = 6

C
Pro = 4, Thr = 4

D
Thr = 4, Arg= 4

Solution

Pro (Proline) has 4 codons CCA, CCG,CCC and CCU
Thr (Threonine) also has 4 codons ACA, ACU, ACG, ACC.
Question 9
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Which of the following amino acids is specified by 4 codons?

A
Tyrosine (Tyr)

B
Proline (Pro)

C
Lysine (Lys)

D
Methionine (Met)

Solution

Proline (Pro) amino acid is the amino acid which is specified by 4 mRNA codons; CCU, CCC, CCA, CCG. While methionine is specified by just one codon, the amino acids Lysine and Tyrosine are specified by two codons each. So, the correct answer is option B.
Question 10
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First triplet pair deciphered by Khorana was _________________.

A
Serine and Leucine

B
Threonine and Histidine

C
Cysteine and Valine

D
Phenylalanine and Isoleucine
Solution
- Khorana generated a poly-UC mRNA and it was translated into polypeptides with an alternating pattern of serine and leucine amino acids.
So, the correct answer is option B.