AIPMT DISCUSSION QUESTIONS ON GENETIC CODE

Q1. Which of the following is a characteristic of genetic code?

a)      The genetic code is non overlapping

b)      The genetic code is unambiguous

c)       The genetic code is degenerative

d)      All of the above

Explanation:

The correct answer is‘d’. The genetic code is non overlapping means that the successive triplets are read in order.

The genetic code is unambiguous means each codon or triplet specifies a particular amino acid and only one amino acid.

The genetic code is degenerative means each amino acid can be specified by more than one triplet codon.

Q2. Which of the following is a nonsense codon?

a)      UAG

b)      UGA

c)       UAA

d)      All of the above

Explanation:

The correct answer is‘d’. Out of 64 triplet codons, three codons are called nonsense codons because they terminate the process of translation. UAG is amber, UGA is opal and UAA is ochre.

Q3. Which of the following poly bases could not form any amino acid in Nirenberg’s experiment?

a)      Poly U

b)      Poly A

c)       Poly G

d)      Poly C

Explanation:

The correct answer is ‘c’ i.e., Poly G because it formed triple stranded structure and thus no translation occurred.

Q4. Hargobind Khorana chemically synthesized amino acids:

a)      Cysteine/valine 50:50 percent

b)      Proline/Cysteine 50:50 percent

c)       Phenyalanine/ Proline 50:50 percent

d)      All of the above

Explanation:

The correct answer is ‘a’ i.e., cysteine/valine. He used UGUGUGUGU poly nucleotide bases to synthesize the same.

Q5. Which one represents serine?

a)      CUU, CUC, CUA, and CUG

b)      UAU, UAC, UGU, and UGC

c)       UCU, UCC, UCA, and UCG

d)      UGU, UGC, UGA, and UAG

Explanation:

The correct answer is ‘c’. In order to remember which triplet codon codes for which amino acid, one needs to see the codon chart. If we reference Proline which falls on the second row and second column of the codon chart, we can remember the codons. However, a practice is needed for the same.

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AIPMT DISCUSSION QUESTIONS ON TRANSCRIPTION IN PROKARYOTES

Q1: What is the correct direction of transcription?

a)      5’à 3’

b)      3’à 5’

c)       5’à 5’

d)      3’à 3’

Explanation:

The correct answer is ‘a’ i.e., transcription always occurs in the 5’ to 3’ direction. 5’ is called the head and 3’ is called the tail of a nucleotide chain. Replication also occurs in the same direction.

 Q2. A transcription unit comprises of:

a)      A promotor, structural gene, and a terminator

b)      A promotor and structural genes

c)       Structural genes and a terminator

d)      A promotor and a terminator

Explanation: 

The correct answer is ‘a’ i.e., a transcription unit comprises of a promoter, the structural gene and a terminator. Promoter is responsible for binding the enzyme RNA polymerase, strcutural genes are expressed to form proteins and a terminator sequence is required to end the process of transcription and translation. 

Q3. Which of the following statement is correct of a Lac operon?

a)      There are three promotor regions present for each gene Z, Y, and A

b)      There is a single promotor region for each gene Z, Y, and A

c)       There could be more than one promotors for the three genes Z, Y, and A

d)      No promotor is required for the three genes Z, Y, and A

Explanation:

The correct answer is ‘b’ i.e., for the Lac operon, only one promotor is required to express the three genes Z, Y, and A. On the other hand, in eukaryotes, each gene has its own promotor region.

Q4. Coding strand in DNA is the one:

a)      Which does not take part in transcription, but its bases will be the same as a new transcript with T replaced by U

b)      Which takes part in transcription, and its bases will the same as a new trascript with T replaced by U

c)       Which does not take part in transcription, but its bases will be the opposite of the new transcript bases

d)      None of the above

Explanation:

The correct answer is ‘a’ because, a coding strand is one that does not take part in transcription. It can also be called as a non template strand. However, since RNA strand does not contain Thymine, it is replaced by Uracil. 

Q5. The transcription start site i.e., +1 region contains:

a)      Mostly Purine bases

b)      Mostly Pyrimidine bases

c)       50 percent purine and 50 percent pyrimidine

d)      75 percent purine and 25 percent pyrmidine

Explanation:

The correct answer is ‘a’. In 90 percent of the cases, the +1 transcription start site contains purine bases especially Adenine.

Q6. Which of the following conserved sequence is called Pribnow box?

a)      TATAAT

b)      TATATA

c)       TTGACG

d)      GCAGAT

Explanation:

The correct answer is ‘a’ i.e., TATAAT a conserved seequence, which is present -10 seequence upstream on the DNA. This conserved sequence is essential for replication as RNA polymerase recognises this conserved sequence to start replication at the transcription start site.

Q7. Which of the following is the function of a sigma factor in transcription?

a)      Sigma factor is responsible for termination

b)      Sigma factor is responsible for elongation of RNA strand

c)       Sigma factor is involved in transcription initiation

d)      Sigma factor is involved in inhibiting the transcription process

Explanation:

The correct answer is ‘c’ i.e., sigma factor is responsible for transcription initiation. Sigma factor when attaches itself to the core enzyme, the combination is called Holoenzyme i.e., RNA polymerase. Sigma factor helps recognize the RNA polymerase to the correct place on the template strand to initiate transcription. Once the transcription process starts, sigma factor is released and the core enzyme gets involved in the elongation of RNA strand.

Q8. What mechanisms are involved in the termination of transcription process?

a)      Rho dependent and rho independent mechanism

b)      Rho dependent mechanism only

c)       Rho independent mechanism only

d)      None

Explanation:

The correct answer is ‘a’ i.e., rho dependent and rho independent mechanisms. The former is called protein based and the latter is called RNA based. 

Q9. Which of the following is true of rho dependent mechanism of termination?

a)      Near the end of the gene, RNA polymerase encounters a run of ‘A’ nucleotides

b)      Near the end of the gene, RNA polymerase encounnters a run of ‘G’ nucleotides

c)       Near the end of the gene, RNA polymerase encounters a run of ‘C’ nucleotides

d)      Near the end of the gene, RNA polymerase encounters a run of ‘U’ nucleotides

Explanation:

The correct answer is ‘b’ i.e., at the end of the gene, RNA polymerase encounters a run of G nucleotides. As a result, the rho protein collides with the polymerase. The interaction with the rho protein releases the mRNA from the termination bubble.

Q10. In RNA based termination, the polymerase enzyme encounters a region rich in:

a)      C-G nucleotides

b)      A-G nucleotides

c)       A-T nucleotides

d)      A-U nucleotides

Explanation:

The correct answer is ‘a’ i.e., the C-G nucleotides. The mRNA folds back on itself and the complementary C-G nucleotides bind together. The result is a hairpin that causes the polymerase to stall as soon as it begins to transcribe a region rich in A-T nucleotide.

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TRANSCRIPTION IN PROKARYOTES

Transcription is considered as the very first step in gene expression.  In other words, it is an act of making a copy. In genetics, however, it means copying DNA into an RNA sequence. Unlike DNA replication, transcription involves a single strand of DNA.

TRANSCRIPTION UNIT

A transcription unit in DNA consists of the following regions:

1)      A Promoter

2)      The Structural Gene

3)      A Terminator

Note: Like DNA replication, transcription also occurs in 5’ à 3’ direction.

Note: The RNA strand is similar to the coding strand; the only difference is that in place of T, U is added in the RNA strand (T is replaced by U).

Transcriptionin Bacteria

There are three basic steps of bacterial transcription

i)                    Initiation

ii)                   Elongation

iii)                 Termination

Promoters: 

• ·         These are DNA sequences that promote gene expression.
• ·         They promote the exact location for the initiation of transcription
• ·         They are located upstream of the site where transcription of a gene actually begins
• ·         The basees in a promoter are numbered in reference to the transcription start site

The conventional numbering system of promoters

• ·         Most of the promoter regions are labelled with negative numbers
• ·         There is no base numbered zero
• ·         Bases to the right are numbered in a positive direction
• Initiation of Bacterial Transcription
• RNA polymerase is the enzyme that catalyzes the synthesis of RNA i.e., a single RNA polymerase is responsible for catalyzing all three RNAs (rRNA, mRNA, and tRNA).
• In E.coli, the RNA polymerase holoenzyme is composed of
• ·         Core enzyme (5-subunit) = α₂ββ’ώ
• ·         Sigma factor (1-subunit) = σ
• Holoenzyme = Core enzyme + Sigma factor

                       = α2ββ’ώ σ =4, 50,000 mol wt

Stepsof Transcription in Bacteria:

• ·         The RNA polymerase holoenzyme binds loosely to the DNA away from -35 region
• ·         It then scans along the DNA until it finds both -35 and -10 upstream regions.
• ·         A region within the sigma factor that contains a helix-turn-helix structure is involved in a tighter binding to the DNA
• ·         The binding of the RNA polymerase to the promoter forms a closed complex
• ·         The open complex is formed whenTATAAT box in the -10 region is unwound
• ·         A short RNA strand is made within the open complex (the sigma factor is released at this point). And this marks the end of initiation.

 

Elongation:

·         The core enzyme now slides down the DNA to synthesize an RNA strand. This is known as elongation phase.

Termination

·         It is the end of RNA synthesis. It occurs when the short DNA-RNA hybrid of the open complex is forced to separate. This releases the newly made RNA as well as the RNA polymerase.

E.coli has two different mechanisms of termination.

a) rho-dependent termination = protein based

b) rho-independent termination = RNA based

rho-dependent termination:

• ·         It is controlled by rho-protein, which tracks along behind the polymerase on the growing mRNA chain.
• ·         Near the end of the gene, the polymerase encounters a run of G-nucleotides on the DNA template and it stalls. As a consequence, the rho-protein collides with the polymerase.
• ·         The interaction with the rho-protein releases the mRNA from the termination bubble.

Rho-independenttermination

• ·         It is controlled by specific sequence in the DNA template strand.
• ·         As the RNA polymerase reaches near the end of the gene being transcribed, it encounters a region rich in C-G nucleotides.
• ·         The mRNA folds back on itself and the complementary C-G nucleotides bind together. The result is a stable hairpin that causes the polymerase to stall as soon as it begins to transcribe a region rich in A-T nucleotides.
• ·         The complementary U-A region of the mRNA transcript forms only a weak interaction with the template DNA. This coupled with the stalled polymerase induces enough instability for the core enzyme to break away the new mRNA transcript.
• ·         Upon termination, the process of transcription is complete.

Note: The bacterial RNA does not require any processing to become active.

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