A DNA sequence consists of a string of elements called nucleotides, in a defined order. Suppose the DNA sequence of a virus is 20 nucleotides long. If each nucleotide can be either a G, T, C, or A, how many different sequences are possible?
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- One common type of non-homologous end joining, to repair a double stranded break, occurs when enzymes cut out a few nucleotides around the break, and then fuse the ends back together true or false?You have isolated a fragment of viral DNA that supposedly encodes for two proteins, 196 and 220 amino acids long respectively. The isolated DNA fragment is found to be 1100 base pairs in length. Argue why is the isolated DNA cannot code the two proteins identified earlier?What is the conformation of the glycosidic bond (i.e. orientation of the nucleobase relative to the the sugar) in the B-form of DNA? syn anti alpha hetero
- Write the base sequence and label the 3' and 5' ends of the complementary strand for a segment of DNA with the following base sequences: 5'CGGAC3'Please explain why the order is 3–1-4-2 and not another way, thank you so much. Question is in the image.If you analyze a double-stranded DNA molecule and find that 15% of all the nucleotide bases are Adenines, you know that there must also be [ Select ] Thymines, [ Select ] Guanines, [ Select ] v Cytosines and [ Select ] Uracils. (Count each of the bases in any double stranded DNA molecule and calculate their percentages to find the simple key for this if you haven't seen it yet.)
- In Figure 1-8b, can you tell if the number of hydrogenbonds between adenine and thymine is the same as thatbetween cytosine and guanine? Do you think that aDNA molecule with a high content of A + T would bemore stable than one with high content of G + C?As you should recall, DNA, when not being actively transcribed, has a double helical structure. This portion of the DNA has had the two strands separated in preparation of transcribing for a needed protein. The following is one of the two complimentary strands of DNA: 3' - AACCAGTGGTATGGTGCGATGATCGATTCGAGGCTAAAATACGGATTCGTACGTAGGCACT - 5' Q: Based on written convention, i.e. the 3'-5' orientation, is this the coding strand or the template strand? ______________________________ Q: Assuming this strand extends from base #1 to #61 (going left to right), interpret the correctly transcribed mRNA and translated polypeptide for bases 24 - 47: mRNA: ___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___-___- polypeptide chain: ________--________--________--________--________--________--________--________The sequence of a polypeptide is determined by the order of codons that specify the amino acids in the polypeptide. How many different sequences of codons can specify the polypeptide sequence methionine-histidine-lysine? (Use the table to find the number of possibilities.) SECOND BASE UAU UACFTyrosine (Tyr) UAA -Stop codon UAG -Stop codon UUUL UGU Cysteine (Cys) UCU uc UCA FSerine (Ser) uca Uuc Phenylalanine (Phe) UUAL Leucine (Leu) CAU CAC CAA Glutamine (Gin) CAGF UGA -Stop codon uaa -Tryptophan (Trp) CGU сос CGA FArginine (Arg) CU CU Histidine (His) CuA FLeucine (Leu) Cua) Proline (Pro) CCA cca AAU Asparagine (Asn) AGU Serine (Ser) AGC AUU ACU ACC Threonine (Thr) AACF AAA AAGLysine (Lys) AUC Fisoleucine (lle) AUA Methionine (Met) AUG - Start codon ACA ACG AGA AGGFArginine (Arg) GU GACAspartic acid (Asp) GGA GAA Glutamic acid (Glu) Gaa) GcU -Valine (Val) G GUA GCA FAlanine (Ala) Glycine (Gly) 8. 1 4 THIRD BASE 2. FIRST BASE
- DNA structure depends on base pairing of its four nucleotides, A, C, T, and G. Nucleotide A pairs with T, and nucleotide C pairs with G. This forms a four-letter DNA “alphabet." Because DNA codes for amino acids in sets of three nucleotides, there are 4 cubed (4'), or 64, possible combinations, coding for 20 different amino acids. What is the best explanation for why there is no selective advantage for DNA to have five nucleotides (e.g., A, C, T, G, and E) with C pairing with either G or functionally equivalent E? It would be impossible to form the DNA molecule, because it must have an equal number of Cs and Gs. Because G and E have the same role, there would still be four functional letters of the alphabet. Replication would be inaccurate because sometimes C would bond with G and sometimes C would bond with E. There would be a five-letter alphabet with 125 combinations, which is too numerous. It is impossible because there are not five known nucleotides in the cell.Cystic fibrosis (CF) is an inherited disorder caused by different types of mutations, many of which prevent ions from moving across cell membranes. Normally there are channel proteins that allow passage of the ions, but in patients with one kind of CF these proteins seem odd. Closer examination shows that these proteins display the correct amino acid sequence. However, they fail to do their job. A) Given that the primary structure of the protein is correct, what can you infer about the DNA sequence for the gene coding this protein on this patient, is there a mutation? Explain. B) Why is the primary structure insufficient to guarantee the proper function of the protein?For a closed-circular DNA molecule of 6,825 base pairs in the fully relaxed form, the linking number (Lk) is: 65 6825 650 700