Bio 97 Quiz Questions
The difference between ATP and the nucleoside triphosphates used during DNA synthesis is that A) the nucleoside triphosphates have the sugar deoxyribose; ATP has the sugar ribose. B) ATP contains three high-energy bonds; the nucleoside triphosphates have two. C) ATP is found only in human cells; the nucleoside triphosphates are found in all animal and plant cells. D) triphosphate monomers are active in the nucleoside triphosphates, but not in ATP. E) the nucleoside triphosphates have two phosphate groups; ATP has three phosphate groups.
A) the nucleoside triphosphates have the sugar deoxyribose; ATP has the sugar ribose. ATP is a ribose nucleotide triphosphate. DNA or Deoxyribose Nucleic Acid is synthesized by polymerization of deoxyribose nucleotide triphosphates.
DNA within the cell is wrapped around nucleosomes to form chromatin. When chromatin is in the form of a 30 nm fibril, how many copies of H1 and H2A histone proteins would be found in a stretch of chromatin containing 20 nucleosomes? A) 40 H1, 20 H2A B) 20 H1, 40 H2A C) 40 H1, 40 H2A D) 10 H1, 30 H2A E) 20 H1, 20 H2A
B) 20 H1, 40 H2A From lecture 5 you know that nucleosomes are octomers made up of 2 copies of H2A/H2B and H3/H4. In the 30 nm chromatin fibril, each nucleosome is associated with a single H1 molecule that helps the chromatin fold into the solenoid structure. Therefore 20 nucleosomes will have 20 H1 molecules and 40 H2A.
A part of an mRNA molecule with the following sequence is being read by a ribosome: 5' CCG-ACG 3' (mRNA). The anticodon loop of the first tRNA that will complement this mRNA is: A) 3' UGC 5' B) 3' GGC 5' C) 5' UGC 3' D) 5' ACG 3' E) 5' GGC 3'
B) 3' GGC 5' The tRNA anti-codon will be anti-parallel and complementary to the codon - 3'-GGC-5' base pairs with 5'-CCG-3'
The pattern of transcription for a pair of identical twins (Tweedledee and Tweedledum) is very similar at the age of 10. However by the time they are 50, transcription of the leptin gene is much higher in Tweedledee than in Tweedledum. This could be due to what change in Tweedledee A) Spontaneous mutation in their father's sperm cell B) Histone acetylation at the site of Tweedledee leptin gene C) DNA methylation at the site of Tweedledee's leptin gene D) DNA phosphorylation at the site of Tweedledum's leptin gene E)A crossing over event during meiosis in their mother's egg cell
B) Histone acetylation at the site of Tweedledee leptin gene The twins have identical genetic make ups, so changes in transcription are due to epigenetic modifications. Histone acetylation and phosphorylation increase transcription while methylation reduces gene expression
The strands that make up DNA are antiparallel. This means that A) the twisting nature of DNA creates nonparallel strands. B) the 5' to 3' direction of one strand runs counter to the 5' to 3' direction of the other strand. C) base pairings create unequal spacing between the two DNA strands. D) one strand is positively charged and the other is negatively charged. E) one strand contains only purines and the other contains only pyrimidines.
B) the 5' to 3' direction of one strand runs counter to the 5' to 3' direction of the other strand. The two DNA strands are arranged so that there 5' and 3' ends are in opposite directions. This anti-parallel orientation is essential so that the A:T and G:C base pairs are oriented correctly to so that they can form hydrogen bonds with each other.
The polymerase chain reaction (PCR) is used to amplify fragments of DNA. During a PCR reaction, what breaks hydrogen bonds between nucleotides? A) DNA gyrase B) DNA helicase C) 95 degree C incubation D) Taq DNA polymerase E) 55 degree C incubation
C) 95 degree C incubation In a PCR reaction hydrogen bonds between the bases are broken by heating the solution to 95 degrees, allowing the two DNA strands to separate. DNA helicase carries out this function in a living cell, not in PCR. DNA gyrase relieves the tension of coiled DNA molecules. 55 degrees is not hot enough to break all the hydrogen bonds and melt (separate the strands) of a long DNA molecule.
Cytosine makes up 32% of the nucleotides in a sample of DNA. Approximately what percentage of the nucleotides in this sample will be Thymine? A) 16% B) 42% C) 32% D) 18% E) It cannot be determined from the information provided.
D) 18% If cytosine = 32%, Guanine will = 32% for a total of 64% A + T = 100 - 64 = 36% S Since A = T (because they base pair) T = 36/2 = 18%
A butterfly species that is diploid, has either blue or orange spots on its wings as the consequence of a gene on chromosome 12, and either long or short antennae as the result of a second gene on chromosome 19, as shown below. A female with a paternal set of one orange and one long gene chromosome and a maternal set comprised of one blue and one short gene chromosome is expected to produce which of the following types of eggs after meiosis? A) All eggs will have maternal types of gene combinations. B) All eggs will have paternal types of gene combinations. C) Half the eggs will have maternal and half will have paternal combinations. D) Each egg has a one-fourth chance of having a blue long genotype. E) Each egg has a one-fourth chance of having a blue orange genotype.
D) Each egg has a one-fourth chance of having a blue long genotype Blue and orange are two alleles of a gene on chromosome 12, while long and short are alleles of a gene on chromosome 19. The female is heterozygous for both genes. Each egg therefore has a ½ chance of having blue and a ½ chance of having the long allele = ¼ chance of being blue long. Blue and orange are alleles, so no egg will have both alleles (d) and all eggs will have a mix of maternal and paternal gene combinations.
What is the correct order of phases in a single cycle of the polymerase chain reaction? A) Denature, elongate, anneal B) Anneal, denature, elongate C) Elongate, anneal, denature D) Elongate, denature, anneal E) Denature, anneal, elongate
E) Denature, anneal, elongate In a typical PCR reaction cycle the DNA is first heated to separate the two strands. The DNA is then cooled to the annealing temperature to allow the PCR primers to bind to their complementary sequences on the template strands. The temperature is now raised to 72 deg C to allow Taq polymerase to elongate the primer and synthesize a new strand of DNA. This cycle is then repeated 20-40 times.
Cytosine makes up 32% of the nucleotides in a sample of mRNA. Approximately what percentage of the nucleotides in this sample will be Uracil? A) 18% B) 42% C) 32% D) 16% E) It cannot be determined from the information provided.
E) It cannot be determined from the information provided. Since RNA is single stranded, knowing the percentage of one base does not allow you to predict the percentage of any other base
Suppose you provide an actively dividing culture of E. coli with radioactive thymine. What would you expect to see if the cell replicates its DNA and divides once in the presence of the radioactive base? A) One of the daughter cells but not the other would have radioactive DNA B) Neither daughter cell would have radioactive DNA. C) All of the bases of the DNA would be radioactive D) DNA in both daughter cells will be radioactive E) Radioactive thymine will pair with non-radioactive guanine in both daughter cells
E) Radioactive thymine will pair with non-radioactive guanine in both daughter cells Initially the E.coli bacteria were grown in normal media so the cells would be unlabelled. When the bacterium copies its DNA by semi-conservative replication, radioactive thymine will be incorporated into the newly synthesized strands. Following division, both daughter cells will have DNA with one radioactive and one non-radioactive strand and be radioactive.
A new nucleotide monomer is discovered that has a phosphate group, ribose sugar, and a nitrogen base with a single carbon ring. Which type of nucleic acid is this likely to form and what known nucleotides (specify the base) is this novel nucleotide likely to pair with?
Since it is a base with a single carbon ring, it is a pyrimidine that will pair with purine. The novel nucleotide has a ribose sugar. Therefore it will form RNA. It has a base with a single carbon ring. Therefore it is a PYRIMIDINE. It will base pair with PURINE nucleotides such as ADENINE and GUANINE that have 2 carbon ring structures
A man with achondroplasia and a normal woman have six children and all are male. What are the odds that their next child will be a girl?
The correct answer is 50%. The sex of the child has no connection with achondroplasia. Also, the probability of each event is independent of it's history and therefore despite having 6 boys in a row, the chances of a 7th is still 50%
During meiosis I, when does homologous chromosome pairing and recombination occur? a. prophase I b. pro-metaphase I c. metaphase I d. anaphase I e. telophase I
a. prophase I During prophase of meiosis 1, replicated homologous chromosomes pair so that crossing over can occur. Chromosomes begin to align too the spindle in pro-metaphase, are present at the center of the spindle in metaphase and separate and migrate to the spindle poles during anaphase. Telophase marks the end of mitosis.
During which stage or phase of the cell cycle does the cell actively transcribe and translate all the protein products necessary for normal cellular structure and function? a. M phase b. G 1 c. G 2 d. S e. G0
b. G1 In Lecture 8 you learned that cells actively transcribe and translate the proteins necessary for cellular function in Gap 1 or G1 phase. During G2 the cell prepares for mitosis. DNA synthesis/replication occurs during S phase and there is minimal transcription during M phase. Cells in G O are quiescent and are considered to have exited the cell cycle
Meiosis, in contrast to Mitosis: a. initiates in eggs after fertilization b. requires pairing of replicated homologous chromosomes c. occurs in somatic cells d. has a single interphase e. results in 4 diploid daughter cells
b. requires pairing of replicated homologous chromosomes A key feature of meiosis is that the replicated homologous chromosomes pair up with each other during Prophase of the first meiotic division. This is necessary for synapsis and recombination to occur, as well as for the segregation of homologous chromosomes during Metaphase of Meiosis 1
By convention, when the P value for the difference between the observed experimental outcomes and the expected outcome is less than 5 percent (< 0.05), the experimental results are considered to be a. within normal expected range. b. statistically significantly and different from the expected outcome. c. not significant. d. less than one standard deviation from the mean. e. equal to the mean.
b. statistically significantly and different from the expected outcome.
Which of the following is not true of a codon? a. It consists of three nucleotides. b. It never codes for more than one amino acid. c. It extends from one end of a tRNA molecule. d. It is the basic unit of the genetic code. e. It may code for the same amino acid as another codon does.
c. It extends from one end of a tRNA molecule. The codon is present in the mRNA not the tRNA - the tRNA has an anti-codon that is complementary to the codon and base pairs with it
Each of the following sentences is a modification of the sentence THECATATETHERAT. Which of them is analogous to a single substitution mutation? a. THECATATETHEFATRAT b. THERATATETHERAT c. THETATETHERAT d. THECATATTHERAT e. CATATETHERAT
c. THETATETHERAT in a single substitution mutation, one of the three letters in a codon is changed. THECATATETHEFATRAT - Insertion of 3 bases (one codon) with retention of the reading frame CATATETHERAT - deletion of 3 bases (one codon) THECATATTHERAT - deletion of 1 base and frameshift THETATETHERAT - deletion of 2 bases and frameshift
Typically, methylation of nucleosome N-terminal tails leads to a. removal of the protein components of the chromatin from the DNA. b. relaxed packaging of the chromatin and increased transcription. c. tighter packaging of the chromatin and reduced transcription. d. increased amounts of euchromatin relative to heterochromatin. e. activation of topoisomerase
c. tighter packaging of the chromatin and reduced transcription. Post-translational methylation of the C-terminal tails of nucleosomal histone proteins results in tighter packing of the nucleosomes and reduction in transcription. It does not remove proteins or activate topoisomerase
Most people with the dominant mutant polydactyly allele have extra digits, but at least 25% have the normal number of digits. What is the genetic explanation for this observation? a. dominant negative allele b. temperature-sensitive allele c. lethal allele d. partial dominance e. incomplete penetrance
e. incomplete penetrance All individuals carry the dominant allele that causes polydactyly but only a subset of then actually show the phenotype. This is an example of incomplete penetrance and is likely due to either interactions with alleles of other genes in the genome or gene-environmental interactions.
At which phase are centrioles beginning to move apart in animal cells? a. telophase b. anaphase c. prometaphase d. metaphase e. prophase
e. prophase During prophase the mitotic spindle begins to form and the centrioles begin to move apart from each other.
