The Hemagglutinin protein in Influenza virus contains a remarkably long a- helix (7.95 nm) with 53 residues. How many turns does this helix have?
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A:
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- Question 14 Why are long reads the preferred method to sequence repetitive elements? Select the statement that is FALSE. ◇ If a repetitive region is longer than 300 basepairs (the length of a short read) it's impossible to determine the total length. O Long reads are more likely to span repetitive regions on either side, making it easier to map the repetitive regions. O It's hard to determine where short reads should be placed, as they could map to multiple repetitive regions. O Long reads are less error prone and thus better for sequencing overall. Question 15 Which of the following statements is NOT true about GWAS? When identifying alleles associated with a disease, we compare the frequency of an allele in a control group versus a cash groupQuestion 6 1 pts You are assembling the genome sequence of a newly discovered bacterium by aligning a set of BAC clones. Correct assembly requires O spacer regions O repeated sequences O overlapping sequencesQuestion 5 Two different types of mastermixes were used to idetify the GMO food. What was the purpose of using plant mastermix? a. to show that both samples are of plant source and PCR worked for both mastermixes b. None of them C. Both of them d. to show that both samples are not of plant source and PCR worked for both mastermixes
- Remaining Time: 1 hour, 07 minutes. 53 seconds. v Question Completion Status: A Moving to another question will save this response. Question 2 Multiple Choice Question (suggested time - between 1 and 2 minutes): The Wobble Hypothesis proposed by Francis Crick postulates that... O A. Base pairing between nucleotides in the 1st codon and the 3rd anticodon positions does not strictly follow complementarity rules O B. The two ribosomal subunits are loosely associated and wobble during translation. Oc Codorns and anticodons can pair when aligned in the same 5'-to-3' direction O D. Nucleotides in the 3rd codon and the 1st anticodon positions do not interact with each other. DEG can form a base pair with A. OF G can form a base pair with U A Moving to another question will save this response. IypQuestion one parts A though E a. True / False: Guanine to cytosine intermolecular interactions (hydrogen bonding) is stronger than adenine to thymine. b. What peptide would be made from the following DNA sequence? 5'ATCCCGGGTACTCACTCCCAT3' Start-Gly-Pro-Stop Start-Gly-Ser-Pro-Val-Arg-Val Start-Gly-Gly-Thr-lle-Arg Start-Arg-Arg-Gly-Gly c. Starting from the MRNA strand below - what peptide would be produced? Remember your start and stop codons...5'CCAUGCGGCAUACCAAAUUACUAAACUAGC3' Start-Arg-His-Thr-Lys-Leu-Leu-Asn-Stop Start-Asn-Leu-Leu-Lys-Thr-His-Arg-Stop Start-Arg-Lys-Leu-Asn-Stop Pro-Met-Arg-His-Leu-Leu-Asn d. Which type of RNA comprises over 80% of total cellular RNA? ribosomal RNA Messenger RNA Transfer RNA e. True / False - All of the DNA nucleotides are attached to the deoxyribose in the BETA configuration (at the anomeric carbon of the sugar). B (Ctrl) -Question 25 If one DNA segment has the following base composition, 5-CAGTTAGTCA-3', which of the following sequences is complementary? A 3'-CAGTTAGTCA-5' 3'-TGACTAACTG-5' 3-TGACTAAСTG-5 D 5-TGACTAACTG-3'
- Question 5 If an mRNA has the sequence 5'--AUGGUGUUA--3' what is the sequence of the coding DNA strand? Please input just the letters with no other marks. ATGGTGTTA 1 ptsQUESTION 21 Suppose the following are sequences at the 5' splice junction and the 3' splice junction, with intervening intron sequences shown as dashes and the branchpoint A shown, not surprisingly, as 'A'. What is the sequence formed after splicing? 5'-CUCAAUGGUACA- -----CGAUACGAGCACUGACC-3' O A. 5' CUCAAUGCACUGACC 3' O B. 5' CUCAAUGACC 3' O C. 5' CUCAAUGGGCACUGACC 3' O D. 5' CUCAAUGGUGACC 3' O E. 5' CUCAAUGGUACACGAUACGAGCACUGACC 3'QUESTION 8 Consider the pathway for the synthesis of the amino acid arginine in Neurospora: ARG-E → citrulline ARG-F ARG-H ornithine argininosuccinate arginine Mutant strains of Neurospora may carry one or more mutations. Neurospora mutant strain b is grown on minimal media plus supplements as shown. Growth is shown by (+) and no growth is shown by (o). Supplements ornithine mutant nothing citrulline arginino- succinate arginine strain What can you conclude from these data? O Strain a has only one mutation and it is in ARG-E. O Strain b has only one mutation and it is in ARG-H. O Strain a has a mutation in ARG-F and strain b has a mutation in ARG-E. O Strain b has mutations in ARG-E, ARG-F, and ARG-H. + +
- Question 18 The tRNA having a cloverleaf structure suggests the following features EXCEPT some segments can hydrogen bond because of complementary A bases it is a single stranded linear structure with protein coding B sequences © it has segments that can form stem-loops modified bases are usually found at the loops in a stem-loop D segmentQuestion 1 options: The specificity pocket of the serine protease chymotrypsin, which interacts with Tyr and Phe-containing peptide sequences, contains a Ser residue. A research group is trying to modify chymotrypsin such that it has a low KM with Trp-containing peptides. Enter the name or abbreviation of an amino acid that the Ser could be mutated to that would likely have the desired effect. (Hint: look at the diagrams of the specificity pockets shown in the course slides, and consider how the Ser would need to change to account for the difference between Tyr/Phe and Trp.)Question 7: Peptide nucleic acids (PNAS) exhibit greater stability as heteroduplexes with DNA (ie PNA-DNA duplex) than does double-stranded DNA (i.e. DNA-DNA duplex). However, peptide nucleic acids lack charged groups, making them largely insoluble under near-physiological conditions in aqueous buffer. Provide (1) an explanation for the increased stability of PNA-DNA duplexes (hint: consider intermolecular forces). (2) Additionally, propose modification(s) of the PNA scaffold DNA that could increase solubility without drastically reducing duplex stability. PNA R1 HN но. Base Base o=P-O NH Base Base 8 o=P-O- NH 8. Base Base НО NH