C785 Biochemistry Module 1 Quiz 2020 – Western Governors University | C 785 Biochemistry Module 1 Quiz 2020

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C785 Biochemistry Module 1 Quiz 2020 – Western Governors University You are designing a piece of DNA for a project and must select the appropriate nitrogenous bases. Which combination of nitrogen... ous bases would you find most suitable for your work? Select one: a. Adenine, Guanine, Thymine, Cytosine ! These nitrogenous bases, in combination with deoxyribose sugar and phosphate, form the nucleotides that are bonded together to form a long polymer of DNA - the carrier of genetic information. b. Adenine Guanine, Uracil and Thymine c. Thymine, Uracil, Adenine, Cytosine d. Adenine, Thymine, Cytosine, Uracil The DNA sequence 3’-ATGATCCTT-5’ is part of the template strand during DNA replication. Which of the following would be the sequence for the newly made DNA? Select one: a. 5’-ATGATCCTT-3’ b. 5’-GCAGCTTCC-3’ c. 5’-TACTAGGAA-3’ d. 5’-UACUAGGAA-3’ In. In this option U has been used instead of T. Although this is how an mRNA molecule would be created from the DNA, it is not the appropriate complementary base when replicating the DNA. Recall that during replication, the bases on the newly made daughter strand are complementary to the bases on the template strand using the following base pairing rules: A with T and G with C. Which of the following statements is TRUE about RNA? Select one: a. There are several types of RNA in the cell, including mRNA, rRNA, and tRNA. b. RNA nucleotides include the bases A, U, C, and G. c. RNA is single stranded and contains a ribose sugar in the sugar-phosphate backbone. d. All of these ! All of these statements about RNA are true. The cell contains mRNA, rRNA, and tRNA molecules, RNA is single stranded, RNA contains a ribose sugar in the sugar-phosphate backbone, and RNA nucleotides include the bases A, U, C, and G. What is the complementary DNA strand for the DNA sequence 5'-ATT CGG GCT CCC -3'? Select one: a. 5' - CCC TCG GGC TTA -3 b. 5'- UAA GCC CGA GGG -3' c. 5'- ATT CGG GCT CCC -3' d. 5'- GGG AGC CCG AAT -3 ! The DNA complementary strand would be 5'- GGG AGC CCG AAT -3', when written in the conventional way starting with the 5' end of the strand. A research scientist is trying to create a new strain of bacteria that can produce a protein needed for her research. Which of the following molecules will act as a template in creating the protein? Select one: a. Ribozyme In. Ribozymes are certain RNA molecules which are able to catalyze certain reactions. They do not provide the template sequence for the creation of the Protein. DNA carries the genetic information in the form of specific DNA sequences. The stretches of DNA which contain information for specific proteins are referred to as genes. In order to make a protein, the gene must first be transcribed into messenger RNA (mRNA) which contains the complementary sequence to one of the DNA strands of that specific gene. This message within the mRNA is then translated by the ribosome to produce a protein. b. transfer RNA (tRNA) c. messenger RNA (mRNA) d. Ribosomal RNA(rRNA) What is one direct effect of inhibiting translation? Select one: a. The level of proteins would remain the same. b. Transcription would cease. In. Transcription is the copying of DNA into mRNA, and it happends before translation. While mRNA is used during translation, inhibiting translation would not stop transcription because the RNA polymerase that makes the mRNA would still be functional. Translation is the production of proteins, so inhibiting (stopping) translation would stop protein production. c. DNA would no longer be replicated. d. Production of proteins would stop. How many amino acids, and which ones, would be encoded by the following mRNA sequence: 5’-AUGACUCCCGGG-3’? (The codon table below can assist you.) Select one: a. Three amino acids - Met, Ile, and Arg b. Three amino acids - Met, Thr, and Pro c. Four amino acids - Met, Thr, Pro and Gly ! Amino acids are coded by 3-letter codons, and the sequence contains 12 nucleotides, which means is codes for four amino acids. The first is AUG for Met, the second is ACU for Thr, the third is CCC for Pro, and the fourth is GGG for Gly. d. Four amino acids - Met, Leu, Pro, Val Which of the following is NOT part of mRNA processing prior to translation? Select one: a. Poly(A) tail addition In. Poly(A) addition is an mRNA processing step that occurs soon after transcription of the mRNA ends. 5’ cap addition is the first mRNA processing step, which occurs soon after transcription of the mRNA begins. The final step of mRNA processing is splicing - removing introns and connecting the exons of a gene - prior to the exit of mRNA from the nucleus for translation. Promoters are DNA sequences near the start of a gene that are recognized by transcription factors which recruit RNA polymerase to that gene for expression. This is an important aspect of transcription, but is not part of mRNA processing. b. Splicing c. 5’ cap addition d. Promoters DNA replication is ___________, which allows each of the two strands to serve as a _______ for the new strands. Select one: a. semiconservative, template ! DNA replication is semiconservative, meaning that each new duplex has one original (parent) strand and one new strand. Because the two parent strands are separated during replication and the base pairing is predictable, each parent strand can serve as a template for the new strand synthesis. b. semiconservative, replacement c. semicontinuous, replacement d. unreliable, template Recall that the DNA polymerase must bind a double-stranded nucleotide polymer before it can start making its own DNA polymer. What helps the DNA polymerase to overcome this problem in DNA replication? Select one: a. RNaseH b. Ligase c. RNA polymerase d. RNA primers ! RNA primers are made at the start of DNA replication to allow the DNA polymerase to bind and begin making the new DNA strand. Several components of cigarette smoke, including benzopyrene, insert themselves (intercalate) into the DNA and lead to several types of mutations such as frameshift mutations, including both insertions and deletion. Which of the following repair pathways would be used to repair this type of damage? Select one: a. Base excision repair b. Nucleotide Excision Repair ! Nucleotide excision repair is used to repair deletions, insertions, and helix-distorting lesions, such as thymine dimers. c. Homologous Recombination d. Mismatch Repair Maternal smoking during pregnancy is hazardous yet common in many places. Many studies have associated prenatal smoking to unhealthy physical and psychological outcomes for the baby. Researchers know that maternal smoking affects are epigenetic in nature. Which of the following events can be considered epigenetic in nature? Select one: a. Changes in chromatin structure b. Double stranded breaks c. Frameshift mutations In! Frame shift mutations are a kind of mutations which result from addition of deletion of a nucleotide base resulting in an altered reading frame and ultimately a different protein, than the one the gene originally encoded. Frameshift mutations are genetic changes because they alter the DNA sequence, whereas epigenetic changes do not alter the DNA sequence. Epigenetic changes are modifications to genomic structure (not sequence) that are caused by the external environment. These environmental factors affect the overall chromatin structure to allow more or less “access” to the DNA by gene expression machinery to turn the genes “on” or “off”. In other words, epigenetics can alter gene expression without changing the underlying DNA sequences. The changes may or may not be heritable, depending on the location and circumstances. d. Replication Blood type is an example of what type of inheritance? Select one: a. Sex-linked inheritance b. Incomplete dominance In. Incomplete dominance means that the heterozygote will have a distinct phenotype from either homozygote. Hair shape is an example of this, but blood type is not. The genes that produce the A and B antigen proteins can both be expressed independently, and a heterozygote (someone with both genes) will be produce both A and B proteins - neither will dominate the other. This is an example of codominance c. Complete dominance d. Codominance What is the expected probability that a child will have an autosomal dominant disease if their father is heterozygous for the allele and their mother is homozygous for the normal allele? Select one: a. 25% b. 0% In. Because the disease is autosomal dominant, the presence of the disease allele in either parent makes this probability non-zero. If D is the disease-conferring dominant allele and d is the normal allele, the father has the genotype Dd and the mother's genotype is dd. Each child can only inherit a d allele from their mother, and they have a 50% chance of inheriting the D allele from their father. As a result, the expected probability that their child will inherit the disease is 50%. c. 50% d. 100% The physical trait of lip protrusion exhibits a characteristic type of inheritance, as shown by the pedigree above. What type of inheritance best describes this inheritance pattern? Select one: a. Incomplete dominance b. Sex-linked inheritance In. The answer is incomplete dominance. The blending of the large and small lip protrusion into an intermediate, medium lip protrusion, as well as the presence of all three variations in the offspring, demonstrate a clear example of incomplete dominance. c. Codominance d. Complete dominance Which of the following is the order for the steps in the Polymerase Chain Reaction (PCR) process? Select one: a. Denaturation, annealing, and elongation b. Annealing, elongation, and denaturation In. Annealing of the primers takes place after the denaturation of the DNA double strands. Primer elongation follows the annealing of the PCR primers step. The answer is denaturation, annealing, and elongation. First, the DNA double helix is denatured. The annealing of the DNA primers follows this step. Finally, the primers are extended by DNA polymerase in a process referred to as elongation. c. Elongation, annealing, and denaturation d. Elongation, denaturation, and annealing Which of the following is NOT a required “ingredient” in a PCR reaction? Select one: a. RNA primers ! RNA primers are used in DNA replication inside the cell, but the quick degradation of RNA makes it less useful for PCR reactions. Instead, PCR reactions contain primers made of DNA to anneal to the region of DNA that will be amplified and serve as a starting point for DNA polymerase. b. DNA polymerase c. DNA nucleotides d. DNA primers Which of the following changes can NOT be detected using PCR? Select one: a. Insertions b. Deletions c. Epigenetic changes ! Epigenetic changes are not detectable via PCR because they don’t affect the DNA sequence. d. Differences in DNA sequence A small segment of Kevin’s green opsin gene is shown below. What would be the resulting mRNA sequence? Kevin’s opsin gene at nucleotide positions 936 to 941. 5’-G.C.C.T.A.G-3’ (coding strand) 3’-C.G.G.A.T.C-5’ (template strand) Select one: a. 5’-GCCUAG-3’ b. 5’-GCCTAG-3’ c. 5’-CGGATC-3’ In! This sequence matches that of the template sequence. Recall that the mRNA is complementary to the template and matches the coding strand. Additionally, RNA does not contain any thymine (T) bases. d. 5’-CGGAUC-3’ The gene for blue opsin protein is located on chromosome 7, a non-sex chromosome. What kind of inheritance pattern would you expect to see with color blindness due to a mutated blue opsin? Select one: a. More males than females with blue color blindness. b. An equal distribution of blue color blindness between males and females. ! The blue opsin gene is located on a somatic (non-sex) chromosome, and therefore will be inherited with the same frequency for both sexes. c. More females than males with blue color blindness In Kevin’s case, a specific kind of mutation in the opsin gene resulted in the premature termination of the translation process. This resulted in a shorter opsin protein than usual. What specific mutation could have caused this? Select one: a. Nonsense mutations ! a nonsense mutation is a kind of point mutation that changes a single base pair in a codon to a stop codon resulting in termination of the translational signal. This results in a truncated protein. b. Missense mutation c. Frame shift mutation [Show More]

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