Biochemistry Readiness Check II

Biochemistry Readiness Check II 5' UAC CGU 3' 3' UAC CGU 5' 3' TAC CGT 5' 5' TAC CGT 3' Correct! First, note that we are asked for a template/non-coding strand of DNA. DNA uses the base T where RNA uses the base U. Therefore, we can eliminate the two answers containing U because they must be RNA and we want DNA. Then, recall that any strands of nucleic acid that bind to each other (base pairing; A-T/U or C-G) must be each be running in opposite directions (antiparallel; a strand running 5'-->3 is bound to a strand running 3'-->5'). If any two strands of nucleic acid (for example 2 strands of DNA in a double helix) are complementary (A-T/U or C-G), the two strands must run IN OPPOSITE directions, 5'--->3' and 3'--->5'. 5' TTC GCC ATG CAT 3' 3' TTC GCC ATG CAT 5' Correct! All complementary base pairing must be antiparallel. The strand that is complementary to 5’ AAG CGG TAC GTA 3’ is 3' TTC GCC ATG CAT 5'. If we simply 'flip' the sequence, we get 5' TAC GTA CCG CTT 3'. Thus, both of these are the correct answers. 3' TTC GCC ATG CAT 5' 3' TTC GCC ATG CAT 5' 5’ AAG CGG TAC GTA 3’ 3' AAG CGG TAC GTA 5' is similar to the original sequence (5’ AAG CGG TAC GTA 3’), but its 5' and 3' ends are swapped. 5' TAC GTA CCG CTT 3' Correct! All complementary base pairing must be antiparallel. The strand that is complementary to 5’ AAG CGG TAC GTA 3’ is 3' TTC GCC ATG CAT 5'. If we simply 'flip' the sequence, we get 5' TAC GTA CCG CTT 3'. Thus, both of these are the correct answers. 3' TTC GCC ATG CAT 5' 3' TTC GCC ATG CAT 5' 5’ AAG CGG TAC GTA 3’ 3' AAG CGG TAC GTA 5' is similar to the original sequence (5’ AAG CGG TAC GTA 3’), but its 5' and 3' ends are swapped. 3' AAG CGG TAC GTA 5' 5’ ATG TAC GGC GAA 3' 3' ATG GCG TTC GAA 5' 5' AUG GCG UUC CUU 3' 5' AUG GCG UUC GAA 3' Correct! The coding strand and the mRNA both go in the same direction, but do not both contain Ts. 5' UTG GCG TTC GUU 3' Amino Acids RNA nucleotides DNA polymerase Correct! In PCR, a DNA sample is separated into single strands and incubated with DNA polymerase, deoxynucleotides (dNTPs), and two short DNA primers whose sequences flank the DNA segment of interest and thus define the region to be amplified. Recall from the section on DNA replication that DNA polymerase needs a primer to begin DNA synthesis. This requirement means the primers will direct the DNA polymerase to only synthesize complementary strands of the target DNA. DNA nucleotides Correct! In PCR, a DNA sample is separated into single strands and incubated with DNA polymerase, deoxynucleotides (dNTPs), and two short DNA primers whose sequences flank the DNA segment of interest and thus define the region to be amplified. Recall from the section on DNA replication that DNA polymerase needs a primer to begin DNA synthesis. This requirement means the primers will direct the DNA polymerase to only synthesize complementary strands of the target DNA. Template DNA Correct! In PCR, a DNA sample is separated into single strands and incubated with DNA polymerase, deoxynucleotides (dNTPs), and two short DNA primers whose sequences flank the DNA segment of interest and thus define the region to be amplified. Recall from the section on DNA replication that DNA polymerase needs a primer to begin DNA synthesis. This requirement means the primers will direct the DNA polymerase to only synthesize complementary strands of the target DNA. DNA Primers Correct! In PCR, a DNA sample is separated into single strands and incubated with DNA polymerase, deoxynucleotides (dNTPs), and two short DNA primers whose sequences flank the DNA segment of interest and thus define the region to be amplified. Recall from the section on DNA replication that DNA polymerase needs a primer to begin DNA synthesis. This requirement means the primers will direct the DNA polymerase to only synthesize complementary strands of the target DNA. Ribosomes RNA Primers Elongation Annealing Cycling Denaturation Correct! Denaturation separated the DNA strands. The DNA strands must be separated to expose the bases to which the primers will bind. 6 Which statement accurately describes how a frameshift mutation affects a gene? (0/1 Points) It will change a single nucleotide in a mutant gene compared to the normal gene. Examples are missense, nonsense, and silent mutations. It will change the number of nucleotides in a mutant gene compared to the normal gene. Examples are missense, nonsense, and silent mutations. It will change a single nucleotide in a mutant gene compared to the normal gene. Examples are insertion and deletion mutations Incorrect. - If you chose ‘change a single nucleotide in a mutant gene compared to the normal gene. Examples are missense, nonsense, and silent mutations”, this is incorrect. This is the correct definition of Point Mutations. Background: Errors that change the number of nucleotides in a gene (insertion/deletion) are said to ‘shift’ the ‘reading frame’ from the correct groups of three basses to new groups. To visualize what a frameshift is, imagine you are given a string of letters and told to start at the beginning and read every group of three letters: CATDOGRATPIGAPE. You would get "cat dog rat pig ape". But if we add an extra letter, say CATDFOGRATPIGAPE, following our "read every group of three" rule, we would get "cat dfo gra tpi gap e". Shifting our 'reading frame' by adding one more letter completely changes what we read. This is similar to what happens with a frameshift mutation. It will change the number of nucleotides in a mutant gene compared to the normal gene. Examples are insertion and deletion mutations. 7 If the coding DNA sequence of a normal gene is 5' GTC GCA TGG TGA 3', what kind of mutation would create the mutant gene sequence 5' GTC GAC ATG GTG A 3'? (1/1 Points) Silent mutation Nonsense mutation Insertion mutation Correct! Background. An insertion mutation adds nucleotides to a gene and is a type of frameshift mutation, which changes the way the mRNA codons are read, changing the amino acid sequence. See Figure