Please Help me. Explain your answers 11. a. In studying mutant variants of a par
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Question
Please Help me. Explain your answers
11. a. In studying mutant variants of a particular enzyme, a researcher finds that one mutation results in a protein that is 245 amino acids long, whereas the normal protein is 227 amino acids long. In the mutant protein, there is a block of 18 extra amino acids at the end of the polypeptide sequence. The inserted sequences do not correspond in any way to the normal protein sequence. What are some possible explanations (types of changes to the DNA) for this mutant variant.
b. For another mutant variant, the researcher finds that one mutation results in a protein that is 312 amino acids long. In this mutant protein, there is a block of 85 extra amino acids in the middle of the polypeptide sequence. The inserted sequences do not correspond in any way to the normal sequence. What are some possible explanations (types of changes to the DNA) for this mutant variant.
12. The lin-4 microRNA influences the activity of lin-14 gene across developmental time in the nematode C. elegans. Below are developmental timecourse data from normal animals, with lin-4 northern blot (RNA) in the top panel, and LIN-14 western blot (protein) in the bottom two panels. Shown are only the results from wild type animals. Fill in the expected bands in the western blot of LIN-14 protein for animals homozygous for a null (no function) allele of lin-4 (the lin-4(-)/lin-4(-) panel).
13. MicroRNAs are small RNA molecules that bind to the 3’ end of mRNAs and suppress translation of the RNA. How miRNAs suppress translation is still being investigated. Some eukaryotic mRNAs have internal ribosome binding sites downstream of the 5’ cap, where ribosomes normally bind. In one investigation, miRNAs did not suppress the translation of ribosomes that attach to the internal ribosome-binding sites (Pillai et al, 2005, Science 309: 1573-1576). What does this finding suggest about how miRNAs suppress translation?
14. In the nematode C. elegans, sup-5 is a Trp tRNA gene. Mutant alleles of sup-5 function as an amber (tRNA) suppressor.
a. What is the anticipated sequence change to the DNA sequence of the sup-5 gene that would be associated with the suppressor (mutant) allele? Indicate the sequence of the double stranded DNA, as well as which strand is the template strand for transcription (or, the direction of transcription of the sup-5 gene).
b. Animals homozygous for the suppressor allele of sup-5 are viable. What does this suggest about the genome of C. elegans?
15. a. For the C. elegans sup-5 suppressor allele (of #14), do you expect this allele to be dominant or recessive to the wild type allele? Explain your reasoning.
b. The unc-22(s32) allele of the C. elegans unc-22 gene has a premature stop codon that is amber suppressible. Animals homozygous for unc-22(s32) have muscle defects, and exhibit a “twitching” phenotype. unc-22(s32) is recessive to unc-22(+). Let sup-5(s) = the suppressor allele, and assume that unc-22 and sup-5 are unlinked (on separate chromosomes). Animals of unc-22(s32)/unc- 22(+); sup-5(s)/sup-5(+) genotype are crossed with each other. Among the offspring, what proportion of normal-moving, and what proportion of “twitching” animals do you predict?
Practice problem 8.pdf-Adobe Acrobat Reader DC File Edit View Window Help Home Tools Practice problems 8... x 115% lin-4 RNA developmental time (in hours) genotype 8 10 12 14 16 wild type LIN-14 protein developmental time (in hours) 10 12 14 16 wild type lin-4 )/lin-4(-) O Ask me anything xB sign in Export PDF a Adobe Export PDF Convert PDF Files to Word or Excel Online Select PDF File Practice problems 8.pdf X Convert to Microsoft Word docx Document Language: English (US) Change Convert LB Create PDF Store and share files in the Document Cloud Learn More 9-09 PM 3/6/2017Explanation / Answer
11. Mutations are spontaneous. Mutations occur due to a change in single base of a nucleotide of DNA (point mutations) or due to chromosomal aberrations. Chromosomal aberrations include missing or extra chromosomal segment could be found within normal chromosome. The inserted sequences do not correspond in any way to the normal protein sequence shows that the attachment of the new segment has been from a non homologous chromosome. The inserted block of segment at the end shows that the type of chromosomal aberration involved is translocation. It is a simple translocation where the translocated segment could be observed at the end of a non homologous chromosome. If the inserted block of segment has been found at the middle shows that the type of chromosomal aberration involved is duplication from a non homologous chromosome. The duplication is inter-chromosomal duplication if it involves non homologous chromosomes. Bar eye of drosophila explains about this theory of duplication. Interstitial transposition also involves where a middle sequence which has been separated from a chromosome has been gets inserted interstitially at the middle of a non homologous chromosome.