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BIOCHEM C785 2 OA readiness check QUESTIONS & ANSWERS BEST EXAM SOLUTION 2024 RATED A+
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The correct answer is 3' TAT TCG CAT 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’ ATG CGA ATA 5’ (original sequence) 5’ TAC GCT TAT 3’ (complementary sequence) But we asked for it in the 3’ direction, so mirror answer to give correct answer: 3’ TAT TCG CAT 5’ If you chose 3' TAC GCT TAT 5', this is incorrect. Although the nucleotides are in the correct order, the 5' and 3' are in the wrong place. DNA strands that are complementary run in opposite directions. This means that if one strand starts with a 3', the the complementary strand must start with a 5'. If none of the answer choices give the correct answer in 5' ---> 3', then you need to write its mirror, which will run from 3' ---> 5'. 3’ GAT AGC ATA 5’ 5’ ATA AGC GTA 3’ 2
Tyr Val Tyr Ile Gln Ile Ile Asp Val The correct answer is Ile Asp Val. 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’ ATG CAG ATA 5’ coding 3’ AUG CAG AUA 5’ mRNAMirror by changing orientation: 5’ AUA GAC GUA 3’ Read chart Ile Asp Val (chart is in direction of 5' ---> 3') Leu His Lys Tyr Tyr Arg Ile Met Ala Ser Val Ile The correct answer is Ser Val Ile because 5’ TAT TAC CGA 3’ template is complementary and antiparallel so 3’ AUA AUG GCU 5’ but it is in the wrong orientation, so mirror 5’ UCG GUA AUA 3’ and read the chart Ser Val Ile Ser His Gln dNTPs, Primer, RNA Polymerase, template RNA
codes for the same amino acid. The coding sequence 5' TAC 3' corresponds to the mRNA sequence 5' UAC 3' (Tyr), and the coding sequence 5' TAT 3' corresponds to the mRNA sequence 5' UAU 3' (Tyr). Since the C changed to at T, this is a point mutation. If the point mutation results in the same amino acid in the new sequence as in the original sequence, the point mutation is a silent mutation. Missens e Nonsens e Insertion Option 1
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. Option 2 Option 3 Option 4 Autosomal dominant Autosomal recessive X-linked Dominant X- Linked recessive 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.
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. The homologous chromosome is used to replace the incorrectly added base with the correct one.
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. Thymine dimers occur. Distortion of the double helix occurs and is repaired by RNA Polymerase. Removal of a single damaged nucleotide Damage to a few or several nucleotides are identified, then many nucleotides are removed and all are replaced to repair the DNA segment The correct answer is "Damage to a few or several nucleotides are identified, then many nucleotides are removed and all are replaced to repair the DNA segment." In nucleotide excision repair, several nucleotides are removed whereas, in BER (base excision repair), a single nucleotide is removed. Required when there are breaks in the double stranded DNA strand which causes discontinuity in both strands Insertion of a thymine dimer
Option 4 Hydrophobic interactions Hydrogen bonds Disulfide bonds The correct answer is covalent (disulfide) bonds. Hydrophobic interactions are disrupted by heat. Hydrogen bonds and ionic bonds are both disrupted by pH and by salt. Ionic bonds 57
When oxygen is present, pyruvate is converted lactate during fermentation When oxygen is not present, pyruvate enters the mitochondrial matrix and is converted to lactate When oxygen is present, pyruvate is converted to acetyl-CoA in the mitochondrial matrix The correct answer is, "When oxygen is present, pyruvate is converted to acetyl-CoA in the mitochondrial matrix." If it has oxygen, it undergoes aerobic metabolism. If it does not, it is converted into lactic acid in the cytosol in fermentation. When oxygen is present, pyruvate produces glucose to reverse the cycle The starchy foods lead to an increase in acetyl-CoA, which is necessary for fatty acid synthesis. The correct answer is, "The starchy foods lead to an increase in acetyl-CoA, which is necessary for fatty acid synthesis." Glucose and fatty acids do not share similar structures, so glucose cannot be stored as fatty acids.Glucose is stored as glycogen, and most of our energy is stored as triglycerides. Glucose does result in glycosylation, although most of the material is used in fatty acid synthesis. Glucose is processed to acetyl-CoA through glycolysis, and then pyruvate is converted, in the mitochondrion, to acetyl-CoA. If there are high levels of acetyl-CoA present, acetyl-CoA is used to produce fatty acids.
Insulin lowers blood glucose levels by allowing glucose to enter into cells through Glut4 transporters when blood sugar is high The correct answer is, "Insulin lowers blood glucose levels by allowing glucose to enter into cells through GluT4 transporters when blood sugar is high." Once the glucose enters the cells, it will stimulate the cells to break down the glucose through glycolysis, and to store the extra as glycogen. Insulin raises blood glucose levels by allowing glucose to leave cells through the use of Glut4. Insulin has no effect on blood glucose Insulin has no relationship with Glut4, it just signals the blood Phospholipid Triglyceride The correct answer is, "Triglyceride." A triglyceride consists of three fatty acids that are linked to one glycerol molecule. Triglycerides are the primary means of fat storage in the adipose tissue. Trans fatty acid Saturated fatty acid
Option 1 Option 2 Option 3 Option 4 The correct answer is Option 4. Option 4 has the structure of a long-chain fatty acid that has omega-3 and an omega-6 (and also an omega-9) bonds, which are the characteristics of an essential fatty acid. Cis The correct answer is a cis bond. The hydrogens are on the same side of the double bond. Trans Essenti al Omega- 9
Option 1 The correct answer is Option 1, arachidonic acid, which is made from essential fatty acids. Arachidonic acid is best recognized by its characteristic "hairpin" and polyunsaturated structure. Eicosanoids have twenty carbon atoms. Option 2 Option 3 Option 4 Long-chain unsaturated The correct answer is, "Long chain unsaturated." This fatty acid has a single double bond, and it has more than 12 carbons. Long-chain saturated Medium-chain unsaturated Medium- chain saturated Short- chain unsaturated
Short-chain saturated Liqui d Solid The correct answer is, "Solid." It is a long-chain saturated fatty acid. Long-chain fatty acids have more than 12 carbon atoms, and they have no double bonds. Since the chain is so long, it is a solid at room temperature and stacks well. Bilayer Phospholipid During fasting, these individuals will produce excessive amounts of ketone bodies. They should avoid intake of carbohydrates and instead focus on a high-fat diet. They should ensure that they eat frequently so that their body has adequate supplies of glucose. The correct answer is, "They should ensure that they eat frequently so that their body has adequate supplies of glucose." Individuals that are not able to breakdown lipids should limit their intake of lipids and consume more protein and carbohydrates instead. They will break down CH3(CH2)8CH=CH(CH2)10COOH without a problem.
Hydrophobic interaction Hydrogen bond Message for respondents who select this answerThe correct answer is "hydrogen bond." Each of the amino acids shown is a polar amino acid, and polar amino acids interact through hydrogen bonds. The NH combined with the C=O in glutamine places this amino acid side chain into the polar category. The OH bond in serine places this amino acid side chain into the polar category. Disulfide bond Activation energy Enzymes Dehydration The correct answer is a dehydration reaction. When two amino acids join together, they lose water. One loses an O from a carboxyl group, and the other loses two Hs from its amino group, to give a loss of water. See Study Guide Step 7 for review. Hydrolysis Primary Secondar y Tertiary The correct answer is "tertiary." Myoglobin is made from one polypeptide chain,
or is a "single subunit" protein. Myoglobin's tertiary structure gives it its functional 3D shape. A hyperbolic curve indicates a "single subunit" protein. A sigmoidal curve indicates a quaternary (multi-subunit) protein. Quaternary Primary Seconda ry The secondary level has alpha helices and beta pleated sheets. Primary structure is formed by joining together amino acids through peptide bonds. Secondary structure is formed by hydrogen bonds between backbone amino acids.Tertiary structure is formed form interactions between amino acid side chains. Quaternary structure is formed from interactions between amino acid side chains. Proteins with quaternary structure have two or more subunits. Tertiary Quaternary