Question

3. (2 points) In the nematode roundworm, Caenorhabditis elegans, the LIN-14 protein controls the timing of certain cell divisions during development. LIN-14 protein levels are normally high in early development but decrease in the later stages. In a lin-4 mutant, the level of LIN-14 protein stays high throughout development, changing the pattern of cell divisions in the animal and producing defects in the shape of the animal. The lin-4 gene encodes a microRNA that binds to a sequence in the 3' UTR of the lin-14 mRNA a) How does the lin-4 microRNA likely regulate LIN-14 protein levels? Explain why the lin-4 mutant has high levels of LIN-14 throughout development. b) Mutations in the 3' UTR of lin-14 have been identified that alter the sequence to which lin-4 normally binds. What effect would these mutations be expected to have on the expression of LIN-14 protein in the animal? 4. (2 points) In the bacterium, Martian coli, it was discovered that the lac operon is positively regulated. Lactose, when present, binds to the regulatory protein A. The protein A-lactose complex then binds to the operator domain O and induces synthesis of lacZ+ and lacY+ MRNAS. What would be the most common phenotype of O mutants in terms of lactose utilization? What would be the most common phenotype of A mutants? Explain your answers. 5. (2 points) Eukaryotic genes can be introduced into bacteria by recombinant DNA techniques. If the introduced gene encodes a protein that is also found in bacteria-for example, a universally used glycolysis enzyme-then, expression of the eukaryotic gene may produce a protein that functions in the bacterial cell. The mouse gene for a glycolysis enzyme is introduced into an E. coli cell that has a mutant gene for the bacterial version of the same enzyme. Even though the mouse enzyme should function in the bacterial cell and restore the cell's ability to perform glycolysis, it does not. Provide two possible reasons why this experiment does not work and propose a solution to overcome one of the problems you suggest.

Answer 1

3) a. As lin-4 is a microRNA and it binds to the 3' UTR of LIN-14 mRNA, it inhibits translation of this mRNA. So, when lin-4 mutant is present, it can not bind the 3' UTR of LIN-14 mRNA and LIN-14 expression is higher throughout development.

b. When mutations in the 3' UTR of lin-14 have been identified that alter the sequence to which lin-4 normally binds, expression of lin-14 will be higher in the animal, as lin-4 does not inhibit it's expression.

4) In O- mutant, lacZ+ and lacY+ mRNAs will not be synthesized, as lactose-protein A complex does can not bind it. In A- mutant also lacZ+ and lacY+ mRNAs will not be synthesized, as without regulatory protein A, lactose can not bind operator alone.

5) One possible reason could be that genetic transfer failed to occur. If it happens, one can use different mode of genetic transfer. Transformation and transduction are some effective ways of it. Otherwise, recombination may not happened if bacteria is RecA mutant. In that case, one should depend on RecA-independent recombination, which is efficient but limited by exonucleases.