BIOCHEM C785 pre-assessment #3

5' TTG TTA CTG 3 3’ GTC ATT GTT 5’ 3’ GTT ATT GTC 5’ 3' TTG TTA CTG 5 The correct answer is 3' TTG TTA CTG 5'. Remember complementary means “Matching or Pairing” You have to remember to pay attention to your numbers as well as your letters (A-T, G- C, 5'-3'). The correct answer is 3’ CAG TAA CAA 5’ (original sequence) 5’ GTC ATT GTT 3’ (complementary sequence) But we asked for it in the 3’ direction, so mirror answer to give correct answer: 3’ TTG TTA CTG 5’ Correct answers: 3' TTG TTA CTG 5 Gln Leu His The correct answer is Gln Leu His. We are starting at the coding strand, and have to remember the relationship between coding DNA and mRNA. These two strands are non-complementary and parallel. So we copy the coding strand , change T ---> U, and then write the mRNA sequence: 3’ TAC GTT AAC 5’ coding 3’ UAC GUU AAC 5’ mRNA Mirror by changing orientation: 5’ CAA UUG CAU 3’ Read chart Gln Leu His (chart is in direction of 5' ---> 3') If you chose Met Gln Leu, this answer is incorrect because this is the amino acid sequence that results from the mRNA 5' AUG CAA UUG 3' which would have been complementary to the given coding strand. Coding DNA is non-complementary and parallel to mRNA. Met Gln Leu Ile Asp Val Gln Ile Met Correct answers: Gln Leu His 33, Which amino acid sequence would be made from this template DNA strand? 5’ AGT TAT AAC 3’?Required to answer. Single choice. (0/1 Point) Met Leu Thr Val Ile Thr Ser Tyr Asn The correct answer is Val Ile Thr because 5’ AGT TAT AAC 3’ template is complementary and antiparallel so 3’ UCA AUA UUG 5’ but it is in the wrong orientation, so mirror 5’ GUU AUA ACU 3’ and read the chart Val Ile Thr Gln Tyr Your answer to question 3 is wrong. Correct answers: Val Ile Thr 44, 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 Correct answers: DNA polymerase,DNA nucleotides,Template DNA,DNA Primers 16 8 32 The correct answer is 32. 2x2x2x2x2=32 10 Correct answers: 32 5’ UAG AGC CAU 3’ 5' UAC AGC CAU 3' 5’ UAC AGC CAC 3’ Correct! The correct answer is 5' UAC AGC CAC 3', which also gives Tyr Ser His 5’ UAC AGC CAA 3’ Correct answers: 5’ UAC AGC CAC 3’ Silent Missense The correct answer is missense mutation. The nucleotide sequence changes, and one amino acid is changed. Since the third codon is changed from GGA to CGA, this is a point mutation. If the point mutation results in an amino acid change compared to the original sequence, the point mutation is a missense mutation. Nonsense Insertion Correct answers: Missense True The correct answer is Option 1 because an autosomal dominant disorder would be inherited on numbered chromosomes, not sex chromosomes X or Y. Also, at least one dominant allele (yellow box) needs to be present for the individual to have the dominant disease. False Correct answers: True True The correct answer is X- linked recessive because parents (carriers) do not have it (II-5-6) but a child does (III-5). You will get the same result if you consider parents (carriers) (I-1-2), who do not have the trait, but a child does (II-3). A third option that gives the same result (X-linked recessive) is by considering parents (carriers) who do not have the trait (III-1-2), and their child does (IV-1). The pattern is recessive because the selected parents are carriers, and it is X-linked because only males have the trait. False Correct answers: True True The correct answer is 50%. Homozygous recessive: aa Heterozygous: Aa When you see "percentage" or "probability," think Punnett square. 50% of the children would be expected to be Aa, and 50% of the children would be expected to be aa. False Correct answers: True Breastfeeding increases the methylation of the promoter of the LEP gene, decreasing the spacing between nucleosomes. Breastfeeding decreases the methylation of the promoter of the LEP gene, increasing the spacing between nucleosomes. Correct! When histones wrap up DNA, transcription will be repressed. Breastfeeding decreases the methylation of the promoter of the LEP gene, decreasing the spacing between nucleosomes. Breastfeeding increases the methylation of the promoter of the LEP gene, and the tight packing of the DNA alters binding of RNA polymerase. Correct answers: Breastfeeding decreases the methylation of the promoter of the LEP gene, increasing the spacing between nucleosomes. A DNA-binding protein blocks RNA Polymerase from binding to the promoter sequence, facilitating the transcription of the MECP2 gene. Transcription factors are unable to bind to the transcription start site of the MECP2 gene because nucleosomes are tightly packed together. The answer is "Transcription factors are unable to bind to the transcription start site of the MECP2 gene because nucleosomes are tightly packed together." Think "increased space gives increased access and increased expression." Gene expression is increased when nucleosomes are widely spaced and transcription factors and RNA Polymerase are able to bind to the transcription start site of the gene. In this question, decreased expression is resulting from decreased space between the nucleosomes, so the RNA Polymerase and transcription factors have decreased access to the transcription start site of the gene. Transcription activators cause nucleosomes to separate, exposing the MECP2 gene. RNA Polymerase binds to the MECP2 gene and begins translation. Correct answers: Transcription factors are unable to bind to the transcription start site of the MECP2 gene because nucleosomes are tightly packed together. The homologous chromosome is used to replace the incorrectly added base with the correct one. Thymine dimers occur. DNA Polymerase removes the incorrect base and adds in the correct base. The correct answer is "DNA Polymerase removes the incorrect base and adds in the correct base." DNA Polymerase repairs mismatch errors that occur during DNA replication. Distortion of the double helix occurs and is repaired by RNA Polymerase. Correct answers: DNA Polymerase removes the incorrect base and adds in the correct base. The error in one strand of DNA is removed as well as several nucleotides on either side of the error. The gap that was removed is filled in by DNA polymerase Correct! The correct answer is that nucleotide excision repair replaces several damaged nucleotides plus additional nucleotides and uses the DNA on the opposite strand as a template to fill in the gap. See Figure 1-22 in the Module 1 text for more information. Only one nucleotide is removed and DNA polymerase replaces the one abnormality The entire homologous chromosome is used to repair the double stranded DNA error The affected DNA strand and its complementary strand is discarded and a new double stranded DNA is created Correct answers: The error in one strand of DNA is removed as well as several nucleotides on either side of the error. The gap that was removed is filled in by DNA polymerase 1515,