WGU Biochemistry, Questions Bank. Final Review, Top Exam Questions and answers, 100% Accurate, rated A+

Document Content and Description Below

WGU Biochemistry, Questions Bank. Final Review, Top Exam Questions and answers, 100% Accurate, rated A+ 1. The polymerase chain reaction is a tool used to study protein structure. True False - �... ��✔-False Feedback PCR is a tool used to amplify a specific segment of DNA. What color is the primer in the following diagram? Red Purple Blue - ✔✔-Red Feedback 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. (Note: In DNA replication, the primers are RNA primers, while PCR generally uses DNA primers because they are more stable.) dNTPs are DNA nucleotides used in PCR. True False - ✔✔-True Feedback dNTPs stands for deoxynucleotide triphosphates, which are the nucleotides used in DNA synthesis. A thermocycler is a machine used for PCR that varies the temperature of a sample. True False - ✔✔-True Feedback For PCR, in each cycle, the two strands of the duplex DNA are separated by heating, then the reaction mixture is cooled to allow the primers to anneal (or pair) to their complementary segments on the DNA. Next, the DNA polymerase directs the synthesis of the complementary strands. The use of a heat-stable DNA polymerase eliminates the need to add fresh enzyme after each round of heating (heat inactivates most enzymes). Hence, in the presence of sufficient quantities of primers and dNTPs, PCR is carried out simply by cycling through the different temperatures for strand separation, primer annealing, and DNA synthesis. The thermocycler is the machine that is used to vary the temperature of the samples. DNA polymerase is used in DNA replication and in PCR. True False - ✔✔-True Feedback Both PCR and DNA replication require DNA polymerase to make new copies of DNA. Which of the following changes can NOT be detected using PCR? Differences in DNA sequence Epigenetic changes Deletions Insertions - ✔✔-Epigenetic changes Feedback Epigenetic changes do not affect the sequence of the DNA. PCR is used to look at the DNA sequence. Which of the following components is NOT used in PCR? DNA template DNA nucleotides RNA polymerase Primers - ✔✔-RNA polymerase Feedback PCR makes a DNA copy, so DNA polymerase is used. Assuming there is one copy of the target DNA sequence before PCR, how many copies of DNA are there after 5 PCR cycles? 256 6 65,536 16 32 - ✔✔-32 Feedback Each cycles double the numbers of DNA copy. Round 1: 1->2. Round 2: 2 ->4. Round 3: 4 -> 8. Round 4: 8 ->16. Round 5: 16 ->32. DNA polymerase can synthesize new DNA strands in which direction? 3' to 5' 5' to 3' any direction N-terminus to C-terminus - ✔✔-5' to 3' Feedback DNA needs a free 3' end to bind to and initiate synthesis of a DNA. It synthesizes in a 5' to 3' direction. The following are steps involved in a polymerase chain reaction. Which is the correct order: Elongation, denaturation, and annealing Denaturation, annealing, and elongation Annealing, elongation, and denaturation Elongation, annealing, and denaturation - ✔✔-Denaturation, annealing, and elongation Feedback PCR uses repeated cycles of temperature to amplify particular DNA segments. In the first step, the reaction mixture is heated to separate the DNA strands (denaturation). The reaction is then cooled to allow the DNA primers, which define the sequence to be amplified, to anneal (base pair) with the template DNA. In the third step, DNA polymerase extends the DNA primers to create a copy of the target DNA sequence. Heating the reaction to stop polymerization and separate the DNA strands starts the cycle over again Sickle Cell Anemia is inherited in an autosomal recessive pattern. Choose the set of chromosomes of a person that has inherited the disease. Green boxes represent normal alleles and yellow boxes represent mutant alleles. Captionless Image A. Chromosome 12 (Blue) vs. Chromosome 11 (Red) B. Chromosome X (Blue) vs. Chromosome Y (Red, Short) C. Chromosome 11 (Blue) vs. Chromosome 11 (Red w/green stripe) D. Chromosome 11 (Blue) vs. Chromosome 11 (Red w/yellow stripe) - ✔✔-D. Chromosome 11 (Blue) vs. Chromosome 11 (Red w/yellow stripe) Feedback This question depicts chromosomes and asks which pair represents an Autosomal Recessive inheritance pattern. Each of the chromosomes are pictured carrying an allele from each parent, which is depicted by a yellow or green box. The green box represents a normal or dominant allele, while the yellow box represents a mutant or recessive allele. An Autosomal trait will be carried on a numbered chromosome and both chromosomes should have the same number. An X-linked trait will be carried on an X-chromosome. This allows us to rule out answer choices A and B. To inherit a recessive trait, we need to inherit two recessive (or mutant alleles), which would be shown as two yellow boxes. A dominant trait can be inherited when one or two dominant alleles are present. Answer choice C depicts an Autosomal Dominant inheritance, while answer choice D is the correct answer because it depicts an Autosomal Recessive inheritance pattern. Hemophilia is an X-linked recessive condition. This means that: 1. A person with at least one normal X chromosome without the hemophilia gene will not show signs of hemophilia 2. The disease is more common in females 3. Affected men can pass the condition on to sons but not to daughters 4. Women can only pass the condition on to daughters - ✔✔-1. A person with at least one normal X chromosome without the hemophilia gene will not show signs of hemophilia Feedback An X-linked recessive condition is inherited when a female has a recessive allele on each of her X chromosomes. Males inherit an X-linked recessive condition if they inherit a recessive allele on their one and only X chromosome. Therefore if a female has a dominant (or normal) allele on one of her X chromosomes, she will not have the disease. Because men only have one X-chromosome, they tend to inherit X-linked conditions more easily than females. Men with X-linked recessive conditions do not pass the disease on to their sons, because they pass a Y chromosome on to their sons. Men will pass an X chromosome to their daughters. Women are able to pass an X chromosome on to their son or their daughters. A woman is homozygous for an abnormal allele on Chromosome 2 that codes for an autosomal dominant disease. This means that she: Will have the disease Will not have the disease Has one normal gene and one abnormal gene Can pass on two genes to any daughters and one gene to any son - ✔✔-Will have the disease Feedback Homozygous means that the woman will have two copies of the abnormal allele. An abnormal allele is the allele that can lead to disease. Since the disease is dominant, then the abnormal allele is dominant in this case. With two copies of a dominant allele, an individual will inherit a dominant disease. Mutations in the FANCA gene (located on chromosome 16) can lead to Fanconi Anemia. A healthy individual inherited one mutant and one normal allele of the FANCA gene. Which of the following describes the inheritance pattern of Fanconi Anemia? Autosomal Dominant Autosomal Recessive X-linked Dominant X-linked Recessive - ✔✔-Autosomal Recessive Feedback Chromosome 16 is an autosome because it is a numbered chromosome. For a healthy individual to have a mutant allele, the mutant allele must be recessive. A black female mouse mates with a white male mouse and produces a litter of all gray mice. Which inheritance pattern can be used to describe this situation? Complete Dominance Recessive Incomplete Dominance Codominance - ✔✔-Incomplete Dominance Feedback Incomplete dominance produces a blended phenotype. Both pedigrees show the inheritance of von Willebrand disease, a bleeding disorder in which platelets fail to clot properly. There are different types of von Willebrand disease that exhibit different inheritance patterns. Type I is inherited in an autosomal dominant fashion, while type III is autosomal recessive. Which of the following statements best describes the families depicted in the pedigrees below? Family 1 has type III and Family 2 is unlikely to have either of these two types. Family 1 has type I and Family 2 has type III Both families have type I. Family 1 has type III and Family 2 has type I - ✔✔-Family 1 has type III and Family 2 has type I Feedback Family 1 has carrier parents present and both males and females are affected, so it is type III, autosomal recessive. Family 2 does not have carrier parents and affected males do have unaffected daughters, so it is type I, autosomal dominant. 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? 0% 25% 50% 100% - ✔✔-50% Feedback Aa X aa => 50% Aa diseased, 50% aa normal Craniofrontonasal Dysplasia is an X-linked Dominant disorder. This condition is very rare and is caused by mutations in the Ephrin B-1 gene. Which one of the following pedigrees portrays the familial inheritance pattern of this X-linked Dominant disorder? Option 1 Option 2 Option 3 Option 4 - ✔✔-Option 4 Feedback To first differentiate between dominant and recessive, check to see if any carriers are present in any generation. Options 1 and 3 are Recessive inheritance patterns because they both contain the pattern where two parents are unaffected and have a child that is affected. Option 1 is X-linked Recessive because only males are affected in the entire pedigree. Option 3 is Autosomal Recessive because there is an affected female. Since Options 2 and 4 are both Dominant, we look to see which one has a pattern where an affected father has daughters that are all affected. Option 2 has an affected father with an affected son, so this must be Autosomal Dominant. Option 4 demonstrates an affected father passing the disease down to all of his daughters, so the answer is Option 4. Two healthy individuals give birth to a child that has Bloom Syndrome. From this information, it can be concluded that Bloom syndrome is inherited in a ________ manner. recessive dominant sex-linked - ✔✔-recessive Feedback Carrier parents are parents who do not have a particular trait, such as a disease, but that trait shows up in their offspring. The presence of carrier parents signifies that the trait is recessive. The Punnett Squares below represent three different types of dominance. Which answer choice correctly identifies the different types of Dominance? 1. A. Complete Dominance; B. Incomplete Dominance; C. Codominance 2. A. Codominance; B. Complete Dominance; C. Incomplete Dominance 3. A. Incomplete Dominance; B. Complete Dominance; C. Codominance 4. A. Codominance; B. Incomplete Dominance; C. Complete Dominance - ✔✔-3. A. Incomplete Dominance; B. Complete Dominance; C. Codominance Feedback A. Incomplete Dominance (heterozygotes are a color in-between the dominant and recessive color). B. Complete Dominance (All flowers are completely the dominant color or the recessive color). C. Codominance (heterozygotes are both the dominant and the recessive color). Match the letters with the correct names of the processes of the central dogma (Replication; Transcription; Translation). 1. A) Transcription, B) Translation, C) Replication 2. A) Translation, B) Replication, C) Transcription 3. A) Replication, B) Transcription, C) Translation 4. A) Translation B)Transcription, Replication - ✔✔-3. A) Replication, B) Transcription, C) Translation Feedback Correct! The central dogma starts with DNA, which is able to be replicated to prepare for cell division. DNA in the nucleus can also be accessed and transcribed to mRNA. The mRNA will leave the nucleus and enter the cytoplasm where it will be translated into a protein by the ribosome. Label the lettered strands of nucleic acid below. The possible words to use in labeling strands are: Coding Strand, Non-template Strand, Non-coding Strand, Template Strand, and mRNA. Note: A and B refer to one strand of DNA; C and D refer to the other strand of DNA. 1. A) Template B) Coding C) Non-template D) Non-Coding E) mRNA 2. A) Template B) Non-coding C) Non-template D) Coding E) mRNA 3. A) Non-template B) Non-Coding C) Template D) Coding E) mRNA 4. A) Non-template B) Coding C) Template D) Non-coding E) mRNA - ✔✔-4. A) Non-template B) Coding C) Template D) Non-coding E) mRNA Feedback Correct! The coding and the non-template strands are the same strand (blanks A and B). The non-coding and the template strands are the same strand (blanks C and D). If one strand of chromosome 2 has a DNA sequence that consists of this: 5' AAG CGG TAC GTA 3' What will be the composition of the complementary DNA strand? (Select all that apply) a. 5' TTC GCC ATG CAT 3' b. 3' TTC GCC ATG CAT 5' c. 5' TAC GTA CCG CTT 3' d. 3' AAG CGG TAC GTA 5' - ✔✔-b. 3' TTC GCC ATG CAT 5' c. 5' TAC GTA CCG CTT 3' Feedback Correct! All complementary base pairing must be antiparallel. The complementary strand to 5' AAG CGG TAC GTA 3' is 3' TTC GCC ATG CAT 3'. If we simply 'flip' the sequence, we get 5' TAC GTA CCG CTT 3'. Thus, these are the correct answers. During DNA replication, which of the following sequences can be used as a primer for the following DNA sequence: 3' AGT GGA TCA CTA GGC TCT 5'? (Recall that DNA replication uses RNA primers whereas PCR uses DNA primers). 5' UCA CCU AGU GAU 3' 5' TCA CCT AGT GAT 3' 3' UCA CCU AGU GAU 5' 3' TCA CCT AGT GAT 5' - ✔✔-5' UCA CCU AGU GAU 3' Feedback Correct! During DNA replication, RNA is used as a primer for the DNA polymerase. Recall that primers are complementary and antiparallel to the strand of DNA that is being copied. The complementary RNA sequence for 3' AGT GGA TCA CTA GGC TCT 5'? is 5' UCA CCU AGU GAU CCG AGA 3'. Thus 5' UCA CCU AGU GAU 3', which is contained within the sequence above (beginning at the 5' end), could serve as a complementary primer for DNA synthesis. Which of the following is the correct tRNA anticodon for the mRNA codon 5' GCA 3' ? 5' UGC 3' 5' CGU 3' 5' TGC 3' 5' CGT 3' - ✔✔-5' UGC 3' Feedback Correct. mRNA binding to tRNA is also complementary and antiparallel. The following sequence is the coding DNA strand of the collagen gene: 5' ATG GCG TTC GAA 3' What is the sequence of the corresponding mRNA? a. 3' ATG GCG TTC GAA 5' b. 5' AUG GCG UUC CUU 3' c. 5' AUG GCG UUC GAA 3' d. 5' UTG GCG TTC GUU 3' - ✔✔-c. 5' AUG GCG UUC GAA 3' Feedback Correct. The coding strand and the mRNA both go in the same direction, but do not both contain Ts. What would be the resulting mRNA sequence from a template strand with this sequence: 5'-CAG CTC GTC-3'? a. 5'-GUC GAG CAG-3' b. 5'-GAC GAG CUG-3' c. 3'-GAC GAG CUG-5' d. 3'-GUG GAG GAG-5' - ✔✔-b. 5'-GAC GAG CUG-3' Feedback Correct! Sequences that bind each other (are complementary) must also be antiparallel (running in opposite directions). The sequence that is complementary to 5'-CAG CTC GTC-3' is 3'-GUC GAG CAG-5'. Since the possible answers in the list to choose from are presented with their 5' ends on the left, we simply 'flip' the sequence to get 5'-GAC GAG CUG-3'. What is the coding strand sequence if the non-template strand sequence is 5'-AGC CTT TAA CTA-3' a. 5'-TCG GAA ATT GAT-3' b. 5'-TAG TTA AAG GCT-3' c. 3'-AGC CTT TAA CTA-5' d. 3'-ATC AAT TTC CGA-5' - ✔✔-d. 3'-ATC AAT TTC CGA-5' Feedback Correct! The names "coding strand" and "non-template strand" refer to the exact same piece of DNA. Therefore, if the non-template strand is 5'-AGC CTT TAA CTA-3', then the coding strand is exactly the same sequence (because it is the same piece of DNA: 3'-ATC AAT TTC CGA-5'. What would be the amino acid sequence that would result from this template sequence: 5'-TGC AAG CCA-3'? a. Pro Leu Trp b. Thr Phe Gly c. Gly Phe Thr d. Trp Leu Ala - ✔✔-d. Trp Leu Ala Feedback Correct! The RNA polymerase enzyme makes mRNA by antiparallel and complementary basepairing with the template strand of DNA. Therefore, the mRNA transcribed from this template would be 3'—ACG UUC GGU-5'. mRNA codons are read 5' --> 3' on the genetic code table. 'Flipping the sequence gives us 5'-UGG CUU GCA-3'. Using the genetic code table, we see that UGG encodes Trp; CUU encodes Leu; and GCA encodes Ala. What would the amino acid sequence be from this coding strand sequence: 5'-GGA AGG CCC3'? a. Gly Arg Pro b. Lys Pro Ser c. Pro Gly Arg d. Val Ala Pro - ✔✔-a. Gly Arg Pro Feedback Correct! The mRNA transcribed from the template strand is complementary and antiparallel to it. The coding strand of DNA (also called the non-template strand) is also complementary and antiparallel to the template strand. Thus, translating 5'-GGA AGG CCC-3' gives us Gly Arg Pro Proteins that promote coiling of DNA and help prevent DNA strands from "tangling" are called? Histones Enzymes Polymerase Chaperones - ✔✔-Histones Feedback Correct! The combination of DNA and histone proteins creates nucleosomes. Epigenetics. Given the following modification of the nucleosomes, discuss how the transcription and gene expression is impacted. In A the the nucleosomes are less condensed, and in B they are more tightly wound. Would A or B foster the ability to have more active transcription of genes? A (less condensed) B (tightly wound) - ✔✔-A (less condensed) Feedback Correct. When the histones of the nucleosomes have DNA wound tightly around them, less DNA is available to be bound and activated by transcription factors. Some changes to the DNA do not modify the coding sequence of the DNA but do affect its winding and unwinding from nucleosomes. These changes can increase or decrease the availability of DNA and hence, the transcription of a gene. These are called _________ changes? Genetic Epigenetic Epidermal Hypogenetic - ✔✔-Epigenetic Feedback Correct. Epigenetic changes result from modifications of the DNA that affect nucleosome spacing and, therefore, transcription. They do not alter the DNA sequence itself. Select all of the following that are involved in the process of actively transcribing a gene. 1. Epigenetic modifications to nucleosomes 2. Ribosomes 3. DNA Ligase 4. Transcription Factors 5. Promoters 6. RNA polymerase 7. DNA polymerase - ✔✔-1. Epigenetic modifications to nucleosomes 4. Transcription Factors 5. Promoters 6. RNA polymerase Feedback Correct! Epigenetic changes result in modifications of the nucleosomes that affect chromatin packing and transcription. Transcription Factors are specialized proteins that recognize specific promoter sequences and bind to them. Once bound to the promoters, the transcription factors recruit RNA polymerase to the transcription start site. RNA polymerase begins transcribing the gene sequence into a new RNA molecule. What changes can occur to the DNA (ex. methyl groups) and to the histones that will impact gene expression? Select all that apply 1. methylation 2. chromosome replication 3. acetylation 4. splicing - ✔✔-1. methylation 3. acetylation Feedback Correct! Methyl groups can be added to DNA (methylation) without altering the DNA sequence. DNA methylation changes the spacing of nucleosomes. Acetyl groups can be added to histones (acetylation) and this also changes the spacing of nucleosomes. Nucleosome spacing affects the transcription of genes. Which of the following statements about epigenetics is false? 1. Environmental stimuli control gene expression. 2. Gene expression is influenced by chemical modifications of the DNA and/or histone proteins. 3. Access to the promoter gene by transcription factors is affected. 4. The DNA sequence is permanently altered. - ✔✔-4. The DNA sequence is permanently altered. Feedback Correct! Epigenetics involves reversible changes to the DNA or to histone proteins. These changes, such as adding or removing methyl groups from DNA or acetyl groups from histone proteins, respond to environmental stimuli. Such changes can increase or decrease nucleosome spacing, which can make the promoter of a gene accessible or inaccessible. In this way, epigenetic changes influence gene expression. The LCT gene codes for Lactase, which is responsible for the breakdown of lactose. Which of the following statements could explain how Lactase activity is increased in the presence of lactose? 1. Lactose blocks RNA polymerase from binding the promoter sequence, facilitating transcription of the LCT gene. 2. Lactose prevents binding of transcription factors at the transcription start site of the LCT gene. 3. The presence of lactose causes nucleosomes to separate, exposing the LCT gene. 4. The presence of lactose causes nucleosomes to pack together tightly, exposing the LCT gene. - ✔✔-3. The presence of lactose causes nucleosomes to separate, exposing the LCT gene. Feedback Correct! In this case, the presence of lactose in the diet is an environmental stimulus. In response to this stimulus, DNA may be modified by the addition or removal of methyl groups and histones may be modified by the addition or removal of acetyl groups. These changes can alter the spacing of nucleosomes and therefore, affect gene expression. Histone proteins can be chemically modified by addition of an methyl group. If this causes nucleosomes to pack _________ , the process of ___________ is decreased at those DNA sites. 1. more loosely; transcription 2. more loosely; translation 3. more tightly; transcription 4. more tightly, translation - ✔✔-3. more tightly; transcription Feedback Correct! Changes to histone proteins or to DNA that result in tighter nucleosome packing make the DNA less available for transcription. Therefore, transcription is decreased. The LEP gene codes for an anorexigenic hormone. Interestingly, breastfeeding has been shown to alter methylation of the promoter of this gene, leading to increased LEP expression. What is likely happening in response to breastfeeding: 1. Breastfeeding increases the methylation of the promoter of the LEP gene, decreasing the spacing between nucleosomes. 2. Breastfeeding decreases the methylation of the promoter of the LEP gene, increasing the spacing between nucleosomes. 3. Breastfeeding decreases the methylation of the promoter of the LEP gene, decreasing the spacing between nucleosomes. 4. Breastfeeding increases the methylation of the promoter of the LEP gene, and the tight packing of the DNA alters binding of RNA polymerase. - ✔✔-2. Breastfeeding decreases the methylation of the promoter of the LEP gene, increasing the spacing between nucleosomes. Feedback Correct. When histones wrap up DNA, transcription will be repressed and when histones open DNA will be available for transcription. In the organization of DNA into chromosomes, DNA is wrapped around _________ to form nucleosomes. Nucleosomes are organized further to form _____________. RNA; triple helix histones; nucleus histones; chromatin promoters ; histones - ✔✔-histones; chromatin Feedback Correct! DNA is wrapped around proteins called histones to prevent tangling, and the combination of DNA and histone proteins creates nucleosomes. The histones promote coiling of the nucelosomes into a larger chromatin fiber. A mutation in the DNA that changes the sequence of a codon but does NOT change the amino acid sequence of the protein describes a _____ _____. frameshift mutation. missense mutation. nonsense mutation. silent mutation. - ✔✔-silent mutation. Feedback -If you chose 'Frameshift Mutation', this is incorrect. Errors that increase or decrease the number of nucleotides in a gene cause frameshift mutations. The insertion of an extra base or the removal of one of the bases will change which groups of three bases that the ribosome reads when it translates the message. This is said to 'shift' the 'reading frame' from the correct groups of three bases to different 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". S [Show More]

Last updated: 2 years ago

Preview 1 out of 94 pages