Q. No.Peptide bond formation between adjacent amino acids during translation is because of:
1. Peptidyl transferase activity of the large subunit.
2. Proteins in the large and small subunits of the ribosome.
3. Ribozyme activity of tRNA.
4. The charging effect of aminoacyl tRNA synthetase.
1. Peptidyl transferase activity of the large subunit.
2. Proteins in the large and small subunits of the ribosome.
3. Ribozyme activity of tRNA.
4. The charging effect of aminoacyl tRNA synthetase.
The amino acid sequence for a eukaryotic gene is placed in a plasmid just downstream of a promoter. Which of the following can be the most likely result?
| 1. | The cell would die because of the antigenicity of the eukaryotic proteins. |
| 2. | The prokaryotic cell will demonstrate the eukaryotic processes such as splicing. |
| 3. | An mRNA would form, but would it would be much longer than that formed directly from the template in the original eukaryote. |
| 4. | If the ribosome could bind the mRNA, a polypeptide that is longer than the one from the original eukaryote would be created. |
The number of genes in human genome is found to be much less than what was initially expected as humans are capable of making large number of diverse proteins. This paradox can be best explained by:
1. Gene editing
2. Alternative splicing
3. Intron excision
4. Post transcriptional modification
Wobble" pairing, a property of tRNA, put forth by Francis Crick is largely responsible for:
| 1. | the redundancy of the genetic code. |
| 2. | the conservation of the genetic message. |
| 3. | the large variety of amino acids encoded by the genetic code. |
| 4. | the excess of different kinds of tRNA that are required for all the amino acids. |
hnRNA from the nucleus of a eukaryotic cell is hybridized with mature mRNA produced from it. Some regions loop out and represent:
1. excised introns
2. exons in the pre-mRNA
3. the poly-A tail in the pre-mRNA
4. introns to be excised later
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Let us presume that it is possible to view enzymes and organelles in a cell, including those involved with gene expression. A prokaryotic cell can be differentiated from a eukaryotic cell because it can be seen that:
1. the nucleotides in the promoter are different in prokaryotes and eukaryotes.
2. the actual genes are more widely-spaced apart in eukaryotes than prokaryotes.
3. ribosomes have started translating mRNA that was still being synthesized by RNA polymerase.
4. prokaryotic mRNA does not contain introns.
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RNA polymerase is able to identify the template strand from the coding strand because of:
| 1. | promoter | 2. | initiator |
| 3. | transcription factor | 4. | start codon |
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It is believed that all organisms have arisen from a single distant ancestor. The strongest evidence for this would be:
| 1. | DNA stores genetic information in all organisms. |
| 2. | The genetic code is universally a triplet code. |
| 3. | The amino acids specified by particular triplets are almost always identical between any two organisms. |
| 4. | The genetic code is degenerate but unambiguous. |
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The addition of deoxyribonucleotides by DNA polymerase during DNA synthesis is an endergonic process. The source of energy used is:
| 1. | ATP |
| 2. | Energy from photons of light |
| 3. | Entropy created from protein digestion |
| 4. | The nucleotides themselves |
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Heat-killed S form Streptococcus extract treated with proteases is mixed with live R form Streptococcus cells and injected into living mice. What is most likely to happen?
| 1. | The mice would live. |
| 2. | The mice would die. |
| 3. | The mice's blood would have dead R cells. |
| 4. | A new strain of Streptococcus will be present in mice. |
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