The median size protein in humans is 8600 amino acids. How does this compare to if our genome was random (21 amino acids long) Explain the difference.
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The median size protein in humans is 8600 amino acids. How does this compare to if our genome was random (21 amino acids long) Explain the difference.
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- Below is a sequence of 540 bases from a genome. What information would you use to find the beginnings and ends of open reading frames? How many open reading frames can you find in this sequence? Which open reading frame is likely to represent a protein- coding sequence, and why? Which are probably not functioning protein-coding sequences, and why? Note: for simplicitys sake, analyze only this one strand of the DNA double helix, reading from left to right, so you will only be analyzing three of the six reading frames shown in Figure 19.4.The genetic information contained in DNA consists of a linear sequence of coding units known as codons. Each codon consists of three adjacent DNA nucleotides that corresponto a single amino acid in a protien. The E.coli DNA molecule contains 4.70 x 10^6 base pairs. Determine the number of codons that can be present. Assuming that the average protein in E.coli consists of a chain of 400 amino acids, calculate the maximum number of protiens that can be coded by an E.coli DNA molecule.Suppose that there is an unknown protein that underwent Edman sequencing method. From N- terminal determination, a biochemist found out that there are two N-terminal amino acid residues, V and G. What is the original sequence of the protein given the following peptide fragments: after digestion with Chymotrypsin: G-L-S-R-G-M-w V-A-L-F Q-L-Y L-R-V-W G-M-V-E-A-D-I-P K-S-P-E-M-T-W R-M-A-S-E-K-P-G-H after digestion with Trypsin: P-G-H V-W-G-M-V-E-A-D-I-P M-A-S-E-K G-M-W-Q-L-Y-L-R S-P-E-M-T-W-R G-L-S-R V-A-L-F-K after digestion with Cyanogen Bromide: T-W-R-M W-Q-L-Y-L-R-V-W-G-M V-E-A-D-I-P A-S-E-K-P-G-H V-A-L-F-K-S-P-E-M G-L-S-R-G-M
- E. How many nucleotides would be required to generate a polypeptide that is 15 amino acids long? This requires knowing how many nucleotides of DNA code for one amino acid. F. Assuming that there are between 20,000-25,000 genes in the human genome, do you think there are 1) fewer, 2) approximately the same number, or 3) more proteins in the human genome? Explain your answer.If an average amino acid is 110Da, how many amino acids would you predict to have in a polypeptide that is 65,000 Daltons in size? How many nucleotides would this be if you were looking for the gene size?Pinker mentions that only 1% of the human genome codes for proteins (the rest included introns, regulatory sequences, and repetitive DNA, some of it parasitic—we will talk about this later). Given that the human genome contains 3 x 109 base-pairs of DNA, there are about 20,000 human genes, and three base pairs code for each amino acid in a protein, how many amino acids are in the average human protein? [Hint: start with what fraction of base pairs in the human genome code for proteins.]
- In not more than 200 words, explain how the human genome of 3.4 Gb would be in 2.3 meters long when stretched linearly if this size is fit into a cell of minuscule proportions.A compact disc (CD) stores about 4.8 × 109 bits of information in a 96 cm2 area. This information is stored as a binary code—that is, every bit is either a 0 or a 1. how many bits would it take to specify each nucleotide pair in a DNA sequence? how many CDs would it take to store the information contained in the human genome?The human genome (3.4Gb) would be 2.3 metres long if stretched linearly. In not more than 200 words, explain how a genome of this size is fit into a cell if minuscule proportions
- Below is a sequence of DNA. 5'-ttaccgataattctctctcccctcttccatgattctgattaaagaaggcgagaacgaaactatttgttaatacc-3' Using the one letter code for Amino Acids, what is the predicted AA sequence of the shortest ORF (from N to C-terminal end)? Using the one letter code for Amino Acids, what is the predicted AA sequence of the longest ORF (from N to C-terminal end)?What will be the order of amino acids derived from the following DNA sequence 5’-TGATCGCACAAT-3’? Explain briefly. (1.5) If the base G (denoted by an asterisk) in the sequence 5’-TGATCG*CACAAT-3’ is replaced by C due to a mutation, the new sequence will be 5’-TGATCCCACAAT-3’ what will be the new amino acid sequence? Explain briefly. (1.5) If the anticodon sequence of a tRNA is 5’-GCG-3’, what amino acid will it carry? Explain briefly. (1.5) What would be the effect of mutation if the C is changed to A in the anticodon? Explain briefly. (1.5)The mass of RNA that has copied a segment of DNA is 63,000 units. The RNA mass that is placed in the ribosome represents 60% of it. Determine the mass of the protein encoded by the gene, if the mass of the aminoacid is 110 and the mass of the nucleotide is 300.
