The Biochemical Basis of Heredity Practice Test (page 3)
Review the following concepts if needed:
- Nucleic Acids for Genetics
- Protein Structure for Genetics
- Genetic Information Flow for Genetics
- Defining the Gene for Genetics
- Protein Synthesis for Genetics
- DNA Replication for Genetics
- Mutations and DNA Repair for Genetics
The Biochemical Basis of Heredity Practice Test
For each of the following definitions, give the appropriate term and spell it correctly. Terms are single words unless indicated otherwise.
- The method of DNA replication in which each strand of the double-helical molecule serves as a template against which a complementary new strand is synthesized.
- A genetic locus that serves as a recognition site for RNA polymerase attachment.
- A group of three nucleotides in mRNA that together specify an amino acid.
- A short RNA sequence onto which DNA polymerase III adds deoxyribonucleotides during bacterial DNA replication.
- Development of a wild-type (normal) trait in an organism or cell containing two different mutations combined in a hybrid diploid or a heterokaryon.
- A spiral secondary structure in parts of many peptide chains, constituting the secondary level of organization. (One or two words.)
- The process whereby RNA is synthesized from a DNA template.
- The single-stranded pieces of DNA produced by discontinuous replication of double-stranded DNA. (Two words.)
- Regions within an eukaryotic primary transcript that are removed during processing of mRNA.
- Removal or reversal of damaged DNA by a light-dependent enzyme.
Choose the one best answer.
- A genetic unit that codes for the amino acid sequence of a complete polypeptide chain is most closely related to
- an anticodon
- a promoter
- a gene
- a codon
- a homotrimer
- Without referring to a table of mRNA codons, solve the following problem. Given a hypothetical segment of antisense strand DNA 3'-GGC AAC CTT GGC 5', the corresponding polypeptide segment could be
- Given the antisense strand DNA codon 3' TAC 5', the anticodon that pairs with the corresponding mRNA codon could be
- 3' CAT 5'
- 5' AUG 3'
- 3' UAC 5'
- 5' GUA 3'
- none of the above
- Which of the following is not a characteristic of cellular RNA?
- contains uracil
- is single-stranded
- is much shorter than DNA
- serves as template for its own synthesis
- contains ribose
- A mutation in the codon UCG to UAG (Table 3-1) is be described as
- a missense mutation
- a neutral mutation
- a silent mutation
- a nonsense mutation
- a frame shift mutation
- An amino acid that cannot participate in alpha-helical formation is
- more than one of the above
- A coding system in which each word may be coded by a variety of symbols or groups of letters (e.g., the genetic code) is said to be
- An amino-acyl synthetase is responsible for
- formation of a peptide bond
- attaching an amino group to an organic acid
- causing a peptide chain to form secondary and higher structural organizations
- movement of tRNA molecules from A to P sites on a ribosome
- joining an amino acid to a tRNA
- The step in the flow of genetic information wherein DNA is copied into mRNA is called
- reverse transcription
- The growing polypeptide chain is first attached to a tRNA molecule in which site of the ribosome?
- A site
- B site
- P site
- anticodon loop
- active site
DNA and Protein Synthesis Questions
- Given a single strand of DNA …3'-TACCGAGTACTGACT-5' … , construct
- the complementary DNA chain,
- them RNA chain that would be made from the given strand, and
- the polypeptide encoded (use Table 3-1).
- Which strand is the sense strand?
- In 1928, before DNA was recognized as the hereditary material, F. Griffith performed a series of experiments in which he infected mice with two different strains of pneumonia-causing bacteria (S. pneumonia). He had in his laboratory two strains of this bacterium; the S strain was virulent (disease-causing) and had a smooth colony appearance and the R strain was not virulent and had a rough colony appearance.
- Experiment 1: Mice infected with the S strain became sick and died.
- Experiment 2: Mice infected with the R strain did not become sick.
- Experiment 3: Mice infected with heat-killed S bacteria did not become sick.
- Experiment 4: Mice infected with a mixture of heat-killed S bacteria and live R bacteria became sick and died.
- How can the results of Experiment 4 be explained?
- Do these results provide evidence that DNA is the hereditary material?
- If the ratio (A+G)/( T + C) in one strand of DNA is 0.7, what is the same ratio in the complementary strand?
- The bacterial virus, phage T4, is a simple infectious agent that is composed of a DNA core and a protein coat. It infects a bacterial cell by attaching to its surface and then injecting material into the cell. This material then directs the synthesis of new viral particles. Using radioactive phosphorous (32P) and radioactive sulfur (35S), design an experiment using this virus to show that DNA is the hereditary material. (Hint: see Example 12.1.)
- How many different mRNAs could specify the amino acid sequence met-phe-ser-pro?
- If the coding region of a gene is estimated to consist of 450 nucleotide base pairs (bp), how many amino acids would the corresponding polypeptide chain contain?
- Given the hypothetical enzyme below with regions A, B, C, and D (* = disulfide bond; hatched area = active site), explain the effect of each of the following mutations in terms of the biological activity of the mutant enzyme:
- nonsense in DNA coding for region A,
- silent in region D,
- deletion of one complete codon in region C,
- missense in region B,
- nucleotide addition in region C.
- A large dose of ultraviolet irradiation can kill a wild-type cell even if the DNA repair system is unsaturated. How?
- A single base addition and a single base deletion approximately 15 bases apart in the mRNA specifying the protein lysozyme from the bacterial virus T4 caused a change in the protein from its wild-type composition … lys-ser-proser-leu-asn-ala-ala-lys… to the mutant form… lys-val-his-his-leu-met-ala-alalys…
- From the mRNA codons listed in Table 3.1, decipher the segment of mRNA for both the original protein and the double mutant.
- Which base was added? Which was deleted?
- If the DNA of an E. coli has 4:2 × 106 nucleotide pairs in its DNA, and if an average gene contains 1500 nucleotide pairs, how many genes does it potentially possess?
- The DNA of bacterial phage lambda has 1:2 × 105 nucleotides. How many proteins of molecular weight 40,000 could be coded by this DNA? Assume a molecular weight of 100 for the average amino acid.
- Acridine dyes can apparently cause a mutation in the bacteriophage T4 by the addition or deletion of a base in the DNA chain. A number of such mutants have been found in the rII region of T4 to be single base addition type (+) or deletion type (–) mutants. A normal or wild-type rII region produces a normal lytic period (small plaque size) in the host bacterium Escherichia coli strain B. Phage T4 mutants in the rII region rapidly lyse strain B, producing a larger plaque. Several multiple mutant strains of T4 have been developed. Determine the lytic phenotypes (large plaque or small plaque) produced by the following rII, single base mutations in E. coli B assuming a triplet codon (suppose the mutant sites are close together):
- ( + )
- ( + )( – )
- ( – )( – )
- ( + )( – )( + )
- ( + )( + )( + )
- ( – )( – )( + )
- ( – )( – )( – )
- ( + )( + )( + )( + )
- ( + )( – )( + )( + )( + )
- The "dotted" gene in maize (Dt) is a "mutator" gene influencing the rate at which the gene for colorless aleurone (a) mutates to its dominant allele (A) for colored aleurone. An average of 7.2 colored dots (mutations) per kernel was observed when the seed parent was dt/dt, a/a and the pollen parent was Dt/Dt, a/a. An average of 22.2 dots per kernel was observed in the reciprocal cross. How can these results be explained?
- Assuming no intensity effect is operative, which individual would carry fewer mutations: an individual who receives 25 roentgens in 5 h or an individual who receives only 0.5 roentgen per year for his or her normal lifetime (60 years)? In terms of percentage, how many more mutations would be expected in the individual with the higher total dosage?
- If the mutation rate of a certain gene is directly proportional to the radiation dosage, and the mutation rate of Drosophila is observed to increase from 3% at 1000 roentgens to 6% at 2000 roentgens, what percentage of mutations would be expected at 3500 roentgens?
- A number of nutritional mutant strains were isolated from wild-type Neurospora that responded to the addition of certain supplements in the culture medium by growth (+) or no growth (0).Given the following responses for single-gene mutants, diagram a metabolic pathway that could exist in the wild-type strain consistent with the data, indicating where the chain is blocked in each mutant strain.
- Point mutations correlated with amino acids in the active site in widely separated regions of a gene can render its enzymatic or antibody product inactive. What inference can be made concerning the three-dimensional structure of the active sites in such proteins?
- A nonsense point mutation in one gene can sometimes be at least partially suppressed in its phenotypic manifestation by a point mutation in a different gene. Offer an explanation for this phenomenon of second-site suppression.
- In addition to the kind of mechanism accounting for second-site suppression of nonsense mutations (see previous problem), give two other possible mechanisms for this type of suppression of missense mutations.
- Interallelic in vitro complementation has been observed in alkaline phosphatase enzymes and other proteins. How can a diploid heterozygote bearing two point mutations within homologous genes result in progeny with normal or nearly normal phenotypes (complementation)?
- Why are most mutations in protein-coding genes recessive to their wild-type alleles?
- Bacterial cells that are sensitive to the antibiotic streptomycin (strs) can mutate to a resistant state (strr). Such "gain of function" mutations, however, occur much less frequently than "loss of function" mutations such as mutation from the ability to make the amino acid histidine (his+) to the inability to do so (his–), or mutation from the ability to metabolize the sugar lactose (lacþ) to the inability to do so (lac–). Formulate a hypothesis that explains these observations.
DNA Repair Questions
- Exposure to UV light results in the formation of thymine dimers between adjacent thymines in a single DNA strand. Describe how repair of this damage might lead to a mutation.
- A new mutant strain of E. coli has been identified that is observed to produce progeny containing spontaneous mutations in many different genes at a higher rate than wild-type E. coli. The mutant is described as having a "mutator" phenotype. Hypothesize about the wild-type function of the mutated gene and why it has this mutant phenotype.
- alpha helix
- Okazaki fragments
DNA and Protein Synthesis
- (a) 5'-ATGGCTCATGACTGA 3' (b) 5'-AUGGCUCAUGACUGA 3' (c) met-ala-his-asp-(stop) (d) The complementary strand shown in the answer for part (a) is the sense or untranscribed strand.
- (a) The R bacteria have become "transformed" into virulent bacteria by obtaining material from the dead S strain cells. (b) No, the material from the dead S strain could be DNA, RNA, protein, or any other cellular material.
- In one population of a bacteriophage, the 32P radioisotope can be incorporated into DNA; and in a second phage population, the 35S radioisotope can be incorporated into proteins (some amino acids have sulfur in their side chains). These two populations of phage can be used to infect E. coli cells in separate experiments. If DNA is the hereditary material, then it should be injected into the cell. Thus, the 32P radioisotope label should be found within the bacterial cells, once the empty phage particles are removed from the outside of the cell. If protein is the hereditary material, the 35S radioisotope label will be found within the bacterial cells.
- 1 × 2 × 6 × 4 = 48
- Approximately 150 amino acids
- (a) The protein would be slightly shorter than normal. Since region A does not seem to interact with other portions of the polypeptide chain, the mutant enzyme should still function normally (barring unpredicted interaction of the side chain of the mutant amino acid with other parts of the molecule). If a nonsense mutation had occurred in region D, however, a very small chain would have been produced that would be devoid of a catalytic site, because proteins are synthesized beginning at the NH2 end. (b) Same sense mutants produce no change in their polypeptide products from normal. (c) The polypeptide would be one amino acid shorter than normal. Since region C does not seem to be critical to the tertiary shape of the molecule, the mutant enzyme would probably function normally. (d) An incorrect amino acid would be present in region B. As long as its side chain did not alter the tertiary shape of the molecule, the mutant enzyme would be expected to function normally, (e) A frame shift mutant in region C is bound to create many missense codons (or perhaps a nonsense codon) from that point on through the carboxyl terminus, including the enzymatic site. Such a protein would be catalytically inactive.
- Nonfunctional DNA fragments might be produced if DNA replication occurs before all of the critical repairs have been made.
- 2800 genes
- 50 proteins, assuming only one strand of the DNA is transcribed into mRNA
(b) G was added, A was deleted (bold)
- (a), (c), (d), (e), (g), (i) = large plaque, (b), (f), (h), (j) = small plaque
- Seed parent contributes two sets of chromosomes to triploid endosperm; one Dt gene gives 7.2 mutations per kernel, two Dt genes increase mutations to 22.2 per kernel
- 20% more mutations in the individual receiving 0.5 roentgen per year.
- The polypeptide chain folds into a configuration such that noncontiguous regions form portions of the catalytic or antibody-combining sites.
- The suppressing mutation could be in that portion of a gene specifying the anticodon region of a tRNA molecule. For example, a tyrosine suppressor gene changes the anticodon of tRNAtyr from 3'-AUG-5' to 3'-AUC–5', thereby allowing it to recognize UAG mRNA nonsense codons. If the genes for tRNAtyr exists in multiple copies and only one of the tRNAtyr genes was mutated to a suppressor form, there would still be other normal (non-suppressor) tRNAtyr genes to make some normal proteins. The efficiency of suppression must be low to be compatible with survival of the organism.
- (1) A change in one of the ribosomal proteins in the 30S subunit could cause misreading of the codon-anticodon alignment, resulting in substitution of an "acceptable" (although perhaps not the normal) amino acid in a manner analogous to the misreading induced by the antibiotic streptomycin. In a cell-free system with synthetic poly-U mRNA, streptomycin causes isoleucine tRNA to be substituted for that of phenylalanine tRNA. (2) A mutation in a gene coding for an amino-acid-activating enzyme (amino-acyl synthetase) causes a different amino acid to occasionally be attached to a given species of tRNA. For example, if AUU (isoleucine mRNA codon) is mutated to UUU (phenylalanine mRNA codon), its effect may be suppressed by the occasional misattachment of isoleucine to tRNAphe by a mutant amino-acyl synthetase that is less than 100% specific in its normal action of attaching phenylalanine to tRNAphe.
- Such proteins are normally homopolymers (quaternary complexes consisting of two or more identical polypeptide chains). If two mutant polypeptide chains contain compensating amino acid substitutions, they may aggregate into a heterodimer that exhibits at least partial enzymatic activity.
- Wild-type alleles usually code for complete, functional enzymes or other proteins. One active wild-type allele can often cause enough enzyme to be produced so that normal or nearly normal phenotypes result (dominance). Mutations of normally functioning genes are more likely to destroy the biological activities of proteins. Only in the complete absence of the wild-type gene product would the mutant phenotype be expressed (recessiveness).
- Loss of function can potentially occur by point mutations at a number of sites within a cistron coding for a given fermentation enzyme or in any gene coding for one of the multiple enzymes in a common biosynthetic pathway such as those in histidine synthesis. The loss of such an indispensable function is lethal. Streptomycin distorts ribosomes, causing misreading of the genetic code. Only a limited number of changes in the ribosomal proteins or rRNA could render the ribosome immune from interference by streptomycin and still preserve the way these components normally interact with mRNA, tRNA, initiation factors, etc., during protein synthesis.
- Repair of thymine dimers occurs using excision repair. This type of repair involves using the remaining template (after excision) to synthesize a new, replacement strand. This synthesis, especially when there is a lot of DNA damage, can be error-prone, thus leading to mutations.
- The wild-type function of this gene may be in a DNA repair function. When mutated, the cell cannot repair its DNA efficiently, leading to a higher rate of spontaneous mutations across the genome.
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