Could someone please help me the following questions?
- What is the role of reverse transcriptase in "Gene Cloning"?
- How can a mutation in a single base be as damaging as a mutation in a sequence of bases?
- Explain how mutation can be harmful or beneficial
- Explain why non-disjunction in meiosis 1 results in a higher proportion of faulty gametes than non-disjunction in meiosis II
Big thanks to y'all
1. Reverse transcriptase, as the name suggests, is an enzyme which catalyses the production of complementary DNA (cDNA) from an RNA template strand. In gene cloning, by inserting a gene of interest in the form of mRNA into bacteria, along with reverse transcriptase, the gene of interest can then be inserted into the bacteria's genome (transforming it). So reverse transcriptase would convert mRNA into complementary cDNA, which can then be integrated into the bacteria's genome. As the bacteria undergoes binary fission, the gene which the original mRNA codes for is cloned.
2. As mutations are random, you can't really determine that unless you're given an example. So a point mutation which results in the same amino acid (as DNA is redundant; each amino acid can be by more than one DNA triplet) is "harmless", compared to another point mutation which results in a stop codon (which is serious). A mutation in several bases may or may not be more harmful than a point mutation. But I'd say the probability of it harbouring more harm is greater.
3. Mutations which result in alleles which contribute to a phenotype that is selected for or favourable in an organism's environment is beneficial. But as mentioned, they can also be harmful - and usually are (e.g. stop codons).
4. Okay, so think of the reductive division of meiosis; homologous chromosomes are separated from each other into two different daughter cells. Say we've got 2 chromosome-20s and two chromosome-21s. Non-disjunction would cause one daughter cell to have, say, two chromosome-20s and one chromosome-21 (extra one). The other daughter cell would then have only one chromosome-21 (less one). So both daughter cells would be faulty. Then meiosis II occurs which results in 4 gametes: two gametes would have chromosome-21, 2 x chromosome-20s (n+1) and the other two gametes would only have chromosome-21 but no 22 (n-1). So they would all be faulty.
If non-disjunction occurred in meiosis II, then only two cells would be abnormal (one is n+1 the other is n-1).
It's best to explain these types of questions with diagrams haha
Hope that helps
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