Q. No.Although gene expression in eukaryotes can take place at multiple levels. The most common level of gene expression would be:
1.
transcriptional level (formation of the primary transcript)
2.
processing level (regulation of splicing)
3.
transport of mRNA from the nucleus to the cytoplasm
4.
translational level
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| 1. | The Human Genome Project was a 13-year project coordinated by the U.S. Department of Energy and the National Institute of Health. |
| 2. | Many non-human model organisms, such as bacteria, yeast, Caenorhabditis Elegans (a free-living non-pathogenic nematode), Drosophila (the fruit fly), plants (rice and Arabidopsis), etc., have also been sequenced. |
| 3. | The sequence of chromosome 1 was completed only in May 2016 (this was the last of the 24 human chromosomes – 22 autosomes and X and Y – to be sequenced). |
| 4. | The fragments were sequenced using automated DNA sequencers that worked on the principle of a method developed by Frederick Sanger. |
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| A: | Expressed Sequence Tags | Fragments of mRNA sequences derived through single sequencing reactions performed on randomly selected clones from cDNA libraries. |
| B: | Sequence Annotations | The process of marking specific features in a DNA, RNA, or protein sequence with descriptive information about structure or function. |
| 1. | Only A | 2. | Only B |
| 3. | Both A and B | 4. | Neither A nor B |
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| 1. | Polymorphism (variation at the genetic level) arises due to mutations. |
| 2. | Allelic sequence variation has traditionally been described as a DNA polymorphism if more than one variant (allele) at a locus occurs in the human population with a frequency greater than 1.00. |
| 3. | The probability of polymorphic variation to be observed in non-coding DNA sequence would be higher as mutations in these sequences may not have any immediate effect/impact on an individual’s reproductive ability. |
| 4. | For evolution and speciation, polymorphisms play a very important role. |
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| (i) | isolation of DNA |
| (ii) | separation of DNA fragments by electrophoresis |
| (iii) | digestion of DNA by restriction endonucleases |
| (iv) | hybridisation using labelled VNTR probe |
| (v) | detection of hybridised DNA fragments by autoradiography |
| (vi) | transferring (blotting) of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon |
| 1. | (i) – (ii) – (iii) – (vi) – (iv) – (v) |
| 2. | (i) – (iii) – (ii) – (vi) – (iv) – (v) |
| 3. | (ii) – (iii) – (i) – (vi) – (v) – (iv) |
| 4. | (iii) – (i) – (ii) – (v) – (iv) – (vi) |
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The core RNA polymerase is capable of catalyzing which steps of transcription?
| 1. | Initiation only | 2. | Elongation only |
| 3. | Termination only | 4. | All of these |
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During DNA replication, Okazaki fragments are used to elongate:
| 1. | The lagging strand towards the replication fork. |
| 2. | The leading strand away from the replication fork. |
| 3. | The lagging strand away from the replication fork. |
| 4. | The leading strand towards the replication fork. |
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What would happen if in a gene encoding a polypeptide of 50 amino acids, 25th codon (UAU) is mutated to UAA?
| 1. | A polypeptide of 25 amino acids will be formed |
| 2. | Two polypeptides of 24 and 25 amino acids will be formed |
| 3. | A polypeptide of 49 amino acids will be formed |
| 4. | A polypeptide of 25 amino acids will be formed |
Identify the correct statement:
| 1. | The coding strand in a transcription unit is copied to an mRNA. |
| 2. | Split gene arrangement is characteristic of prokaryotes. |
| 3. | In capping, methylguanosine triphosphate is added to the 3' end of hnRNA. |
| 4. | RNA polymerase binds with the Rho factor to terminate the process of transcription in bacteria. |
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In the human genome, the chromosomes with the most and the fewest genes located on them respectively are:
| 1. | 1 and Y | 2. | X and Y |
| 3. | 21 and X | 4. | 11 and Y |
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