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- Problem 3: Nitrous acid causes all possible transition mutations, C T and G A. If the normal sequence of a protein contains a valine residue (val) at a particular place, what new amino acid(s) would most often appear in place of the valine residue as a result of nitrous acid mutagenesis? Use a codon chart. O Isoleucine, methionine, alanine O Aspartic acid, glutamic acid, alanine O Only termination codons O Alanine or threonine O AlanineEXERCISE 1 MUTATION Work in a small group or alone to complete this exercise. In Lab 2, Exercise 8, you determined the amino acid sequence for the following strand of DNA: AGCAATC CGTCTTGG TCGTTAGGCAGAACC That strand has mutated. It is now AGCAACCCGTCTTGG TCGTTGGGCAGA ACC Use your knowledge of mutation and protein synthesis to answer the following questions. 1. What mutation has occurred? 2. Will this mutation have a real effect? Why or why not? (Hint: You may want to try Exercise 10 from Lab 2 again, using the mutated DNA strand to make the mutated protein.)* REQUIRED 34. The diagram below represents one of a number of different types of mutations that can occur in DNA ....CACTAGCAG.... DNA Sequence mutation ....CACTAACAG.... DNA Sequence This mutation can best be described as the the insertion of an adenine (A) base into both strands of the DNA molecule pairing of an adenine (A) base with thymine (T) the substitution of an adenine (A) base for guanine (G) deletion of an adenine (A) base from the DNA molecule 0 1
- Problem #2 DNA TACT TAA A AAT G A (TS) DNA (CS) MRNA Amino Acids: Please continue to the next page. Activity 2- Reinforcing Protein Synthesis Use your codon chart to determine the amino acid sequence. Read through the strand and ONLY start on AUG and STOP when it tells you to stop. Follow example below: Example: DNA > AGA CGG TAC CTC CGG TGG GTG CTT GTC TGT ATC CTT CTC AGT ATC MRNA > UCU GCC AUG GAG GC ACC CAC GAA CAG ACA UAG GAA GAG UCA UAG protein > start - glu - ala -thre - hist – asp -glu – threo - stop 1. DNA → CCT CTT TAC ACA CGG AGG GTA CGC TAT TCT ATG ATT ACA CGG TTG CGA TCCATA ATC MRNA protein > 2 DNA → AGA ACA TAA TAC CTC TTA ACA CTC TAA AGA CCA GCA CTC CGA TGA ACT GGA GCA mRNA > pratein > 3. DNA > TAC CTT GGG GAA TAT ACA CGC TGG CTT CGA TGA ATC CGT ACG GTA CTCGCC ATC MRNA > protein → 4. Fill in the diagram below. DNA MRNA ERNA Amino Acids Please continue to the next page. Universal Genete Code Chart Mensenger RNA Codans and the Amina Asd for Which They Code SECONDBASE…Drug 1-Ivacaftor (VX-770): Ivacaftor is a potentiator that increases CFTR channel opening time. We know from cell culture studies that this increases chloride transport by as much as 50% from baseline and restores it closer to what would be expected in wild type CFTR. Basically, the drug increases CFTR activity by unlocking the gate that allows for the normal flow of salt and fluids. For which class of mutations do you think Ivacaftor will be most effective? Explain your choice.Question: Genetically modified animal that might be approved for human consumption is a super “muscly” pig made by the inactivation of the myostatin gene. During normal development, the myostatin protein prevents the overgrowth of muscles. How would such a pig be achieved using CRISPR? Why would it not considered the GMO?
- Problem 3: Proteins that bend DNA Integration Host Factor, IHF, is an architectural protein in E. coli that helps to package the DNA inside the cell as well as to organize the DNA for higher-order nucleoprotein complexes. IHF can bend some DNA segments are 35 base-pairs long into a U-turn as shown in the figure to the right. The Young's modulus of DNA is 300 MPa, and its cross-sectional radius is 1 nm. Model DNA as a cylindrical rod. A. What is the bending coefficient of DNA in units of J. m? B. How long is a 35 base-pair long segment of DNA? (Express your answer in nm given that the separation between base pairs in double-stranded DNA is 0.34 nm.) C. By what angle (in radians) is the DNA bent in the IHF-bound complex? D. Determine the radius of curvature (in nm) for the bent DNA in this IHF-bound complex.Question:- What biological rationale can explain why there are so few variants observed at position 65 of the heme distal ligand and position 94 of the heme proximal ligand of myoglobin? Why does the number of variants differ between the two sites?Question: What is a loss of function of a certain protein that prevents nucleotide metabolism and prevents DNA synthesis - this gene is expressed in all tissues but is expressed at the highest levels in bone marrow?
- Question: What is mutation? Distinguish between a gene mutation and a chromosomal mutation?The folding and unfolding rate constants for a myoglobin mutant have been determined. The unfolding rate constant ke-u = 3.62 x 10-55 and the folding rate constant ku-p = 255 s1, where Fis the folded protein and U is the unfolded (denatured) protein. For wild-type myoglobin, AG;u = +37.4 kJ/mol. Which myoglobin is more thermodynamically stable, the mutant or the wild-type?The distal Histidine (His 64) in myoglobin is subjected to three different mutations, this is one of them: H64N. (Histidine to Aparagine) For the mutation, draw a theoretical binding curve and CO relative to the O2 and CO binding curves for wild-type Mb (see example below). Provide a clear rationale for the binding curve.