test 3 genetics

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IS (insertion sequence)-induced mutations are different from missense mutations in that they are:

- Not reverted by mutagens

You are studying a sample of 100 mice with the following genotypes: forty mice are D/D, twenty mice are D/d, and forty mice are d/d. In this scenario, what is the frequency of the d allele?

0.5

20. When comparing different genomes, "synteny" is defined as: A) the same genes in the same order.B) the same genes in essentially the same order.C) the same genes in reverse order. D) different genes found on similar chromosomes. E) duplication of entire genomes.

32.The functional equivalent of the TATA binding protein in prokaryotes is: A) a sigma subunit.B) a holoenzyme.C) the Rho factor. D) TFIID.E) hairpin loops.

A pure line of plant of genotype a/a;b/b; c/c;d/d; e/e (all recessive to wild type) was crossed to a wild type. One F1 individual expressed the recessive alleles d and e. This individual arose most likely from A a deletion in the quadruple homozygous recessive parent. B a deletion in the wild-type parent. C a single-point mutation in the wild-type parent position effect variegation in the F1. D a reversion in the quadruple homozygous recessive parent.

A deletion in the wild type parent

Pretend there is a segment of the human genome with the sequence:5′ CTGAATCGGGTTGGG 3′3′ GACTTAGCCCAACCC 5′In order for the top DNA strand to be replicated, which of the following primers is needed?

A primer with the sequence 5′ CCCAAC 3′

In laboratory mice, the short tail phenotype is dominant to the wild-type (long-tail) phenotype. However, crosses between any two short-tailed mice always produces mixtures of short- and long-tailed offspring. Assuming that tail length is controlled by a single locus, a likely explanation for these results that the "short tail" allele is

A recessive lethal allele

An enhancer is best described as:

A regulatory sequence in DNA that acts to promote the expression of specific eukaryotic genes

A plant trisomic for a chromosome carrying the A gene has the genotype A/a/a which of the following represents a gamete this plant could not produce?

A/A

A graduate student has just ordered a large quantity of a small molecule from a company. The student will treat bacteria with this molecule, which will bind to a specific activator protein, thus changing its shape and resulting in transcription of a target gene. This small molecule is a(n):

Allosteric effector

The "ABC" operon encodes three enzymes A, B, and C. Upstream of the operons promoter, there is an independent gene that encodes the regulatory protein R. A bacterial strain from which the gene encoding R has been completely removed is found to transcribe ABC operon even in the absence of an inducer. What can be concluded about this system?

C. The R protein likely acts as a repressor of the ABD operon transcription in the absence of the inducer

Which of the following statements regarding the development of cancer is true?

Cancers cell exhibit abnormal cell growth and will invade other host tissues

10. If the DNA template 5′-ATGCATGC-3′ were transcribed to RNA, the RNA would read:read: A) 3′ TACGTACG 5′. B) 5′ AUGCAUGC 3′. C) 5′ UACGUACG 5′. D) 3′ UACGUACG 5′. E) 5′ATGCATGC 3′.

15. The evolutionary history of a group of organisms is called a: A) homology. B) orthology. C) heterology. D) phylogeny. E) synteny.

23. The Ames test is used for determining whether a particular chemical acts as a mutagen. It does this by selecting for: A) new His- mutations in Escherichia coli.B) revertants of His- mutations in Escherichia coli.C) new His- mutations in Salmonella typhimurium.D) revertants of His- mutations in Salmonella typhimurium. E) both revertants and new His- mutations in Escherichia coli.

24. A mutation that occurred in a plant petal would be best termed: A) dominant.B) germinal.C) recessive. D) somatic. E) suppressor.

24. Why is transcription in eukaryotes more complex than in prokaryotes? A) Compared to a few thousand genes in prokaryotes, eukaryotes have larger genomes and tens of thousands of genes to be transcribed. B) Transcription and translation takes place in the same cellular compartment in prokaryotes but in different compartments in eukaryotes. C) The template for transcription in prokaryotes is simple and exists as naked DNA, whereas the eukaryotic template is organized into a highly condensed chromatin structure. D) All of the answer options are correct. E) Eukaryotes have larger genomes and more genes than prokaryotes, and their genomes are organized into a highly condensed chromatin structure.

25. Which of the following methods can be used to inactivate a gene without changing its DNA sequence? A) targeted mutagenesis B) random mutagenesis C) gene knockoutD) RNAi E) Alloftheansweroptionsarecorrect.

25. Which of the following statement/s about RNA pol II is/are FALSE? A) RNA pol II transcribes protein-encoding genes. B) General transcription factors (GTFs) help RNA pol II to position at the promoter start to initiate transcription. C) RNA pol II adds new nucleotides to the 3′ end of the transcript.D) The CTD domain of RNA pol II coordinates cotranscriptional modifications. E) RNApolIIreadsthetemplateDNAstrandina5′-to-3′directionduring transcription.

26. In E. coli, mutations arising during repair are mostly caused by: A) base excision repair.B) mismatch repair.C) recombinational repair. D) SOS repair.E) thymine dimer splitting.

28. Which of the following diseases is/are MOST likely to be caused by excessive exposure to UV light?A) Cockayne syndromeB) Huntington disease C) lung cancerD) skin cancerE) xeroderma pigmentosum

31. The Spo11 enzyme in eukaryotic cells makes:A) single-stranded cuts in the DNA to initiate replication.B) double-stranded cuts in the DNA to initiate replication.C) single-stranded cuts in the DNA to initiate recombination. D) double-stranded cuts in the DNA to initiate recombination. E) causes cancer.

Which of the following statements is true? AMutations affect only the coding regions of a gene; they are not found in noncoding regions, such as the 5? UTR .BMutations can occur in noncoding regions of a gene; however, these mutations do not affect the timing of protein expression, the amino acid sequence of a protein, or the amount of protein produced.CAny mutation, whether it occurs in the coding or noncoding regions of a gene, must produce a corresponding mutant phenotype in an organism .DMutations can occur in noncoding regions of a gene and can affect the manner in which exons are removed from an mRNA product.ENone of the answers is correct.

D.Mutations can occur in noncoding regions in noncoding regions of a gene and can affect the manner in which exons are removed from an mRNA product

11. Retrotransposons move via an intermediate that is: A) a double-stranded lollipop.B) a retrovirus.C) double-stranded RNA. D) single-stranded DNA. E) single-stranded RNA.

12. Which of the following is the most sensible order of techniques for mapping a chromosome and then locating a specific disease gene? A) contig alignment; restriction mapping; RFLP linkage analysis; chromosomal walking B) chromosomal walking; contig alignment; restriction mapping; RFLP linkage analysis C) restriction mapping; chromosomal walking; contig alignment; RFLP linkage analysis D) chromosomal walking; restriction mapping; RFLP linkage analysis; contig alignment E) restriction mapping; contig alignment; RFLP linkage analysis; chromosomal walking

14. Fragile X syndrome is caused by:A) exposure to high doses of 5-bromouracil. B) exposure to high doses of free radicals. C) an inherited microdeletion.D) spontaneous depurination.E) trinucleotide expansion.

19. RNA polymerase can transcribe in different directions on a chromosome. This is enabled because: A) genes and their associated promoters can be oriented in either direction along a chromosome. B) RNA polymerase can transcribe from either strand of DNA, synthesizing RNA in the 5′ to 3′ direction. C) the chemistry of RNA synthesis is identical regardless of which strand is transcribed. D) the 3′ end of a gene is always the template strand.E) All of the answer options are correct.

2. Before the completion of the human genome project, it was generally estimated that human DNA would contain up to 100,000 genes. We now know humans contain only around 25,000 individual genes within our genomes. This "gene count" is so surprising because:

23. Why does E. coli have several different sigma factors?A) They allow different RNA polymerases to bind to the promoters. B) They allow the different subunits of the RNA polymerase holoenzyme to bind to each other. C) There is no good reason. They all perform the same function. D) One is needed to transcribe mRNA. A second is needed to transcribe tRNA. And a third is needed to transcribe rRNA.E) They allow RNA polymerase to recognize and bind to a different subset of promoters.

30. The carboxyl tail domain of RNA polymerase II plays a key role in: A) capping of the 5′ end of a new transcript.B) recruiting capping enzymes to the RNA polymerase enzyme.C) addition of a poly(A) tail at the 3′ end of a transcript. D) splicing of introns out of RNA transcripts. E) All of the answer options are correct.

38. The role of most microRNAs within a eukaryotic cell is to:A) regulate the splicing of primary transcripts to mRNAs.B) bind with other RNAs to stabilize their secondary structure. C) associate with ribosomal proteins to facilitate translation. D) regulate RNA polymerase activity in the nucleus. E) repress the expression of genes by destroying mRNAs.

4. A small (one base pair) insertion in the middle of the coding region of a gene will cause a:A) synonymous mutation.B) silent mutation. C) nonsense mutation. D) missense mutation. E) frameshift mutation.

8. Which of the following is/are TRUE about pyrosequencing? A)Pyrosequencing uses ddNTPs as substrates. B) Pyrosequencing uses DNA polymerase and two other enzymes, sulfurylase and luciferase, in the reaction. C) PPi is released for every nucleotide incorporated into the growing strand. D) All of the answer options are correct. E)Pyrosequencing uses DNA polymerase, sulfurylase, and luciferase, and PPi is released for every nucleotide incorporated into the growing strand.

The following sequence was identified at the 3' end of a transcribed RNA in prokaryotes. What is the function of this structure shown below in transcription?

Transcription termination

26. What type of library would be most valuable in the isolation of promoter sequences found in front of a gene's transcriptional start site?

In a protein-coding gene, a mutation that changes a codon from CUU to CGU changes the encoded amino acid from a leucine to an arginine (two amino acids with different charges). This codon change would a) alter the primary structure of the protein.b) alter the secondary structure of the protein.c) affect the ability of the protein to interact with other molecules in the cell.d) alter the tertiary structure of the protein.e) All of the above.

all of these

4. A circular DNA molecule has n target sites for restriction enzyme EcoRI. How many fragments will be produced after complete digestion? A) n - 1 B) nC) n + 1 D) 2n - 1 E) 2n + 2

5. Recombinant DNA techniques typically require the action of: A) DNA polymerase and phosphatase.B) restriction enzymes and DNA ligase.C) RNA polymerase and RNA primase. D) reverse transcriptase. E) DNA helicase.

6. Sanger DNA sequencing determines the nucleotide sequence because:A) dideoxy nucleotides are incorporated into a growing DNA strand. B) dideoxy nucleotides terminate the growing DNA strand.C) PPi is incorporated into the growing DNA strand.D) PPi is released from the growing DNA strand.E) None of the answer options are correct.

a constitutive operator (Oc) results in A) no transcription.B) inducible transcription.C) transcription but no translation.D) no translation.E) constitutive transcription.

constitutive transcription

Epigenetic inheritance is defined as the inheritance of _______? A) the inheritance of nuclear DNA from one cell generation to the next.B) the inheritance of extra-chromosomal DNA from one cell generation to the next.C) the inheritance of mRNA from one cell generation to the next.D) the inheritance of chromatin states from one cell generation to the next.E) none of the above.

d) the inheritance of Chromatin states from one cell generation to the next

Which of these histone proteins would you NOT expect to find in a nucleosome "core" but rather functioning in nucleosomes packaging within chromatin?

if two populations of birds with frequencies of an allele a of 0.2 and 0.3 are forced together by unusually high winds

it is impossible to tell

In squash, the dominant epistasis is observed. It supresses the expression of pigment producing genes. Yellow squash pigment alleles is found to be dominant over green squash allele. When white squash crosses with green squash plants, all F1 progeny are white. When selfing F1 white squash, white, yellow and green squash plants are produced in F2. What is the possible ratio of white, yellow, and green squash plants in F2?

- 12:3:1

Imagine you are a student in the laboratory of Luria and Delbrück. You have plated samples from bacterial cultures 1 and 2 and added the T1 phage to these dishes. After culture, you notice that in the culture 1 plate, two T1-resistant bacterial colonies are present. In the culture 2 plate, 15 T1-resistant bacterial colonies are present. You conclude that:

- A resistance mutation occurred earlier in culture 2 than in culture 1

A drug company hoping to synthesize large amounts of insulin is faced with two options: either isolate and purify insulin from cows or produce human insulin in E.coli via cloning. The company chose the latter option. Why?

- All of the above

A researcher using a computer program to search for potential genes in the rabbit genome. She has identified several different open reading frames (ORFs) during this process. Each ORF likey has which of these following characteristics ?a)a 5 start codonb) a 3 stop codonc) it would be "gene-sized"d) it would contain sense codonse) all of the above

- All of the above

A scientist is culturing bacteria in media containing lactose, but not glucose. What statement (s) is (are) likely true of these cells?

- All of these statements are true

Sanger DNA sequencing determines the nucleotide sequence because:

- Dideoxy nucleotides terminate the growing strand of DNA

A researcher is studying a population of flies composed of 100,000 individuals. This population demonstrates positive assortative mating and does not consist of any subpopulations. Furthermore, all genotypes within this fly population are equally viable. This fly population cannot be Hardy-Weinberg equilibrium because A) All genotypes in this population are equally viable B) It is not infinitely large in size C) This population is not comprised of any subpopulations D) This population does not demonstrate random mating E) Both B and D

- It is not infinity large and this population does not demonstrate random mating

Glutamine is encoded by CAA and CAG codons. If a normal CAA codon is mutated to CAC, the codon will produce histidine instead of glutamine. This is a______ mutation A/ NONSENSE AND MISSENSE B. NONSYNOMONOUS AND TRANSITION C. TRANSITION AND NONSENCE D. TRASNVERSION AND MISSENCE

- Missense and transversion

Nondisjunction in a paTERNAL meiocyte at meiosis IIA man is found to be karyotypically 47, XXY. The presence of an extra Y chromosome most likely results from: Nondisjunction in a maTERNAL meiocyte at meiosis I B Nondisjunction in a maTERNAL meiocyte at meiosis II C. Nondisjunction in a paTERNAL meiocyte at meiosis I D. Nondisjunction in a paTERNAL meiocyte at meiosis II

- Nondisjunction in a paTERNAL meiocyte at meiosis II

Which of the following statements is/are true about the P elements of Drosophila? P elements are examples of eukaryotic DNA transposons.P elements are mobilized when females with these elements are crossed to males that lack them.P elements are silenced in M cytotypeP elements are examples of eukaryotic DNA transposons that are mobilized when females with these elements are crossed to males that lack them.P elements are examples of eukaryotic DNA transposons that are silenced in M cytotype.

- P elements are examples of eukaryotic transposons

Deducing the protein encoding genes for genomic sequences in eukaryotic organisms involves the following steps except

- Predictions based on the formation of hairpin loops

If the frequency of the recessive allele that causes phenylketonuria (PKU) when homozygous is about 0.01 European and American populations, what is the estimated heterozygote frequency in these populations? (Assume that people mate randomly with respect to PKU, and that the mutant allele does not have any effects on fitness)? A) About 0.99B) About 0.0198C) It is impossible to make an estimate.D) About 0.0001E) About 0.01

0.0198

in a human population, the genotype frequencies at one locus are 0.5 AA, 0.4 Aa and 0.1 ea A) 0.20 B) 0.32 C) 0.50 D) 0.70 E) 0.90

0.7

An A/a; B/b dihybrid is test crossed and about 3⁄4 of the progeny phenotypically resembles the dihybrid parent, while 1⁄4 resembles the tester parent. If the dihybrid paret was selfed, what would be the expected phenotypic ratio in the progeny? .9:3:4 B.12:3:1 C.9:3:3:1 D.9:7 F.15:1 E.13:3

15:!

The red fox has 17 pairs of large, long chromosomes. The arctic fox has 26 pairs of smaller shorter chromosomes. What do you expect to be the chromosome number in somatic tissues of the hybrid?

1. A mutation that changes the codon UAA to the codon UAG in a mammalian cell line is called a(n):A) synonymous mutation.B) suppressor mutation. C) nonsense mutation. D) missense mutation. E) antisense mutation.

11. Which of the following is/are FALSE concerning the comparisons between genetic, physical, and cytogenetic maps?A) The distance between two linked markers is the same in genetic and physical maps because crossing over occurs with equal frequency along the entire length of the chromosome.B) Restriction maps, contig maps, and STS maps are examples of physical maps.C) In physical maps, the distances between markers are given in megabases (Mb) where 1 Mb is approximately equal to 1 cM.D) The banding patterns of chromosomes created by different staining techniques are used in constructing cytogenetic maps.E) In genetic maps, the distances between various markers (for example, genes or RFLPs) are given in map units or centiMorgans.

13. In E. coli, a region flanked by two repeats of a sequence such as GTGGTGAA is prone to:A) deletions.B) duplications. C) frameshift mutations. D) missense mutations. E) reversion.

14. Transposable elements that transpose via an RNA intermediate is known as: A) class 1 elements. B) class 2 elements. C) class 3 elements. D) alpha elements. E) D elements.

15. In vitro (in a test tube) translation of a synthetic RNA with the repeating sequence (AGA)20 would produce which of the following? A) polypeptides containing arginine, polypeptides containing glutamic acid, and polypeptides containing lysine B) polypeptides containing arginine C) polypeptides containing a mixture of amino acids (mostly arginine, glutamic acid, and lysine) D) no translation product because the translation start site was not included on the RNA E) polypeptides starting with methionine, followed by arginine additions

16. The most closely related genes either within a single organism or between two different organisms are called: A) homologs. B) orthologs. C) paralogs. D) heterologs. E) alleles.

17. In D. melanogaster, when P cytotype (lab stock) females are crossed to M cytotype (wild) males, the resulting F1 progeny are: A) normal.B) defective.C) normal:defective in a 1:1 ratio. D) normal:defective in a 1:2:1 ratio. E) normal:defective in a 3:1 ratio.

17. In a chromosome, which of the following is TRUE?A) RNAs of different genes can be transcribed off either DNA strand, but always 5' to 3'. B) RNAs of different genes can be transcribed off either DNA strand, but always 3' to 5'. C) The RNAs of all genes are synthesized 5' to 3' off the same DNA strand.D) The RNAs of all genes are synthesized 3' to 5' off the same DNA strand. E) Different genes can be transcribed off either strand, some in the 5' to 3' direction and some in the 3' to 5' direction.

18. Which of the following statements is/are TRUE about the P elements of Drosophila?A) P elements are examples of eukaryotic DNA transposons.B) P elements are mobilized when females with these elements are crossed to males that lack them.C) P elements are silenced in M cytotype.D) P elements are examples of eukaryotic DNA transposons that are mobilized when females with these elements are crossed to males that lack them.E) P elements are examples of eukaryotic DNA transposons that are silenced in M cytotype.

20. The -10 and -35 boxes found in most bacterial promoter elements were discovered by: A) comparing the DNA sequences in front of highly expressed genes and determining nucleic acids that were common (consensus). B) the finding that RNA polymerase begins translation when a 10-nucleotide followed by a 35-nucleotide repeat of guanine ("G") are encountered. C) comparing bacterial promoters to known eukaryotic promoters. D) examining data from the human genome project.E) Robert Koch as he characterized the microbe that causes tuberculosis.

21. The "transcriptome" is defined as:A) the sequence and expression patterns of all transcripts.B) the sequence and expression of all proteins.C) the complete set of all physical interactions (i.e., protein/DNA or protein/protein). D) the complete set of all metabolites.E) None of the answer options are correct.

29. E. coli cells that have null mutations in the gene encoding adenine methylase will have:A) a higher spontaneous mutation rate than wild-type E. coli.B) the same spontaneous mutation rate as wild-type E. coli. C) a lower spontaneous mutation rate than wild-type E. coli.D) more double-strand breaks in their DNA than wild-type E. coli. E) fewer double-strand breaks in their DNA than wild-type E. coli.

29. For protein-encoding genes, six general transcription factors (GTFs) function to: A) identify a gene's promoter, facilitating RNA polymerase II binding.B) terminate transcription at the end of an open reading frame.C) assemble the RNA polymerase subunits. D) arrest DNA replication.E) recruit the ribosome to newly synthesized RNAs.

3. A corn plant is homozygous for a mutant allele that results in no pigment in the seed (i.e., white). The mutant is caused by Ds insertion that often exits late in seed development, when there is an active Ac element in the genome. If there is NO active Ac element, the seeds of this plant will be: A) no pigment (i.e., white).B) pigmented all over.C) white with small spots of pigment. D) white with large spots of pigment. E) weakly pigmented.

33. Which of the following mutations is/are MOST likely to result in cell overgrowth and (eventually) may lead to cancer?A) a gain-of-function mutation in one allele of a proto-oncogeneB) a gain-of-function mutation in one allele of a tumor suppressor gene C) a loss-of-function mutation in one allele of a proto-oncogeneD) a loss-of-function mutation in one allele of a tumor suppressor gene E) a null mutation in one allele of a proto-oncogene

39. siRNAs are distinct from microRNAs (miRNAs) in that siRNAs: A) silence their own expression. B) are smaller than microRNAs. C) are derived from rRNA sequences. D) silence gene expression. E) are made of double-stranded DNA.

4. Template strand DNA and encoded RNA are:A) complimentary of one another with antiparallel orientation.B) complimentary of one another but share the same 5′-3′ orientation. C) identical sequences with the exception of U substituted for T. D) different with regard to the inclusion of introns.

5. An autonomous element: A) requires no other elements for its mobility.B) requires one additional element(s) for its mobility. C) is only found in corn (maize).

5. In a haploid fungus, a transversion arises in the coding region of a gene necessary for the production of the amino acid leucine. The transversion substitutes the last base in the codon UAC (encoding cysteine), resulting in the stop codon UAG. This substitution will most likely cause: A) the production of a truncated (shortened) protein.B) the production of a truncated (shortened) mRNA.C) the production of a shortened mRNA and of a shortened protein. D) no detectable phenotype.E) inability to grow on medium that lacks cysteine.

7. A plant is homozygous for a point mutation in gene B. This plant produces wild-type B protein in wild-type amounts, but a more detailed analysis reveals that the B mRNA produced by this plant is two nucleotides shorter than wild type. The mutation is most likely a two base-pair deletion: A) downstream of the STOP codon, in the last exon of gene B. B) in an intron of gene B away from the splice sites.C) in the open reading frame of gene B.D) that removed a splice site of gene B. E) within the promoter of gene B.

The initial discovery of the spliceosome resulted from experiments focused on: A) autoimmunity in Lupus patients. B) the isolation of large RNA/protein complexes from cells. C) careful analysis of electron microscopy images from the cell nucleus. D) RNAs from plant cells. E) the cell biology of Huntington's disease.

11. The fluctuation test of Luria and Delbruck showed that:A) a selecting agent can affect mutation rate in E. coli.B) mutations can arise spontaneously before exposure to the selecting agent.C) mutations in E. coli occur at a relatively high frequency.D) the mutation rate in E. coli fluctuates greatly from one generation to the next. E) the T1 phage can act as a mutagen as well as a selecting agent in E. coli.

11. The spliceosome includes both protein and a functional type of RNA known as: A) transfer RNA (tRNA).B) small nuclear RNAs (snRNAs).C) micro RNAs (miRNAs). D) small interfering RNAs (siRNAs). E) ribosomal RNAs (rRNAs).

11. We know that DNA and RNA (each with four nucleotide components) both use a three nucleotide genetic code and 64 codons (43 = 64). Knowing that the minimum number of codons for the genetic code is 21 (20 amino acids, 1 stop codon), what codon size would be required if only three nucleotides were present in the genome? A) 2 nucleotides per codon B) 3 nucleotides per codon C) 4 nucleotides per codon D) 5 nucleotides per codon E) 6 nucleotides per codon

12. In E. coli, a region of a gene with repeats of the sequence CTGG will be prone to: A) deletions.B) frameshift mutations.C) missense mutations. D) reversion.E) triplet expansion.

12. The first eukaryotic transposable elements to be characterized at the molecular level were identified within the genes of which organism? A) corn B) yeast C) E. coli D) mouse E) human

14. Degeneracy in the genetic code is best illustrated by: A) stop codons. B) threonine codons. C) tryptophan codons. D) methionine codons. E) operator sequences.

15. RNA synthesis is always 5' to 3' because:A) the unwinding of the double-stranded DNA can only move one direction.B) nucleotides can only be added to an available 3'-OH group on the transcript terminus. C) nitrogenous bases cannot pair up in the 3' to 5' direction.D) the structure of ATP restricts 3' to 5' polymerization into RNA.E) RNA synthesis can move in the 3' to 5' direction.

15. Transposable elements that are comprised of DNA transposons are known as: A) class 1 elements. B) class 2 elements. C) class 3 elements. D) alpha elements. E) D elements.

16. In D. melanogaster, when M cytotype (lab stock) females are crossed to P cytotype (wild) males, the resulting F1 progeny are: A) normal.B) defective.C) normal:defective in a 1:1 ratio. D) normal:defective in a 1:2:1 ratio. E) normal:defective in a 3:1 ratio.

16. The amber mutations initially discovered by Sydney Brenner and colleagues was eventually shown to include a(n): A) unusual amino acid that generated an amber color in the affected bacteria. B) premature stop codon within the gene's open reading frame. C) insertion mutation that creates a translational frameshift. D) previously unknown signal for speeding the translation process. E) a deletion mutation that creates a translational frameshift.

16. The role of tRNA is to:A) serve as an intermediate in the decoding of genes.B) act as transporters bringing amino acids to the site of protein synthesis. C) serve as general translational components of the ribosome.D) facilitate splicing of pre-messenger RNAs.E) facilitate protein trafficking in protein secretion.

17. Genes that have been inherited from a common ancestor are called: A) homologs. B) orthologs. C) paralogs. D) heterologs. E) alleles.

19. During mutagenic treatment with nitrous acid, an adenine deaminates to form hypoxanthine, which bonds like guanine. The mutational event would be: A) AT to CG. B) AT to GC. C) AT to TA. D) GC to AT. E) GC to TA.

22. The "proteome" is defined as:A) the sequence and expression patterns of all transcripts.B) the sequence and expression of all proteins.C) the complete set of all physical interactions (i.e. protein/DNA or protein/protein). D) the complete set of all metabolites.

22. The sigma factor protein's role in transcription in E. coli includes which of the following?A) forms part of the core enzyme required for transcription initiationB) helps the holoenzyme to bind to the promoter C) contributes to the proof reading activity of RNA polymerase D) plays a role in transcription terminationE) All of the answer options are correct.

22. Which of the following statements is/are FALSE about Tc1 elements found in C. elegans? A) Tc1 elements are retrotransposons. B) Tc1 elements can transpose only in somatic cells. C) Tc1 elements were used to study the RNAi based silencing mechanisms. D) Single Tc1 element can trigger the silencing of all copies of Tc1 elements in the C. elegans genome. E) Noneoftheansweroptionsarecorrect.

23. Which of the following statements is/are TRUE about the miniature inverted repeat transposable elements (MITEs) ?A) MITEs are examples of retrotransposons.B) MITEs are examples of DNA transposons. C) MITEs can attain very high copy numbers in humans. D) MITEs contain transposase gene.E) None of the answer options are correct.Answer: B

25. If an incorrect base is incorporated during DNA synthesis and is not corrected by DNA polymerase, it can be corrected by postreplication repair. Postreplication repair does NOT involve which of the following?A) detection of the mismatch B) photoreactivation repair C) a process similar to excision repairD) recognition of the methylation status of the DNA strands E) recombinational repair

27. Which of the following diseases is/are LEAST likely to be caused by excessive exposure to UV light?A) Cockayne syndromeB) Huntington disease C) lung cancerD) skin cancerE) xeroderma pigmentosum

28. A key characteristic of bacterial RNAs that is not observed with eukaryotic RNAs is that:A) transcription generates RNAs that contain both introns and exons.B) transcription can occur in the same cellular region as translation. C) a prerequisite for translation initiation is RNA processing.D) the genetic code used by bacteria is different from other organisms. E) bacterial RNAs are generated as double stranded.

3. A mutation does not affect the length of a gene but results in an abnormally short protein. The mutation is most likely of a type called: A) silent.B) nonsense. C) missense. D) frameshift. E) deletion.

30. E. coli cells that have null mutations in the gene encoding mutH will be defective in: A) base excision repair.B) mismatch repair.C) recombinational repair. D) SOS repair.E) thymine dimer splitting.

34. Proto-oncogenes are genes that:A) always mutate into cancer-promoting oncogenes.B) encode very important proteins involved in processes such as the control of cell cycle. C) encode very important proteins that promote apoptosis.D) in their recessive form can act to promote cell cycle progression.E) normally promote oncogenesis.

4. The alpha-helix structure is stabilized by: A) covalent bonds between adjacent functional groups. B) hydrogen bonds along the peptide backbone. C) association with helicase stabilizing enzymes. D) noncovalent interactions with water. E) cationic elements found in water solutions.

4. Which of the following features do bacterial and corn transposons not have in common? A) Both may cause unstable mutations.B) Both may carry drug resistance genes in natural populations.C) Both may have inverted repeats. D) Both may move to new loci. E) Both may cause rearrangements.

5. Which of the following constitutes the primary structure of a protein? A) the folding of a polypeptide chainB) the linear sequence of amino acids in a polypeptide chainC) the polypeptide chains stacked on top of each other D) a pleated sheetE) severalpolypeptidesubunits

6. A nonautonomous element:A) requires no other elements for its mobility.B) requires additional element(s) for its mobility. C) is only found in corn (maize

6. The "RNA World" theory suggests that:A) RNA is the intermediate between DNA and proteins.B) RNA was the genetic material in the first living cells.C) many RNAs are capable of self-splicing.D) eventually, RNA will become the genetic material of most living organisms. E) RNA viruses are more dangerous than DNA viruses.

7. Bacterial transposon structure can be thought of as:A) IS sequences flanked by inverted drug-resistance genes.B) drug-resistance gene(s) flanked by IS elements.C) drug-resistance gene(s) flanked by a pair of mu () phage.D) a mu () phage flanked by two IS elements.E) resistance gene(s) flanked by inverted resistance transfer factors (RTFs).

7. Two gametes, each carrying a mutant recessive allele for a different gene/enzyme in the adenine biosynthetic pathway, come together to form a diploid embryo. The individual derived from this embryo will display a: A) lethal phenotype in media lacking adenine supplementation.B) wild-type phenotype, capable of synthesizing adenine.C) a weak adenine requirement for survival (intermediate phenotype). D) a profound defect in adenine synthesis as two enzymes are missing. E) embryonic lethality.

9. Pulse-chase experiments provided evidence that:A) RNA is made of ribose sugars rather than deoxyribose.B) RNA is an information-transfer intermediary between DNA and protein. C) many viruses have RNA genomes.D) RNA is capable of catalyzing biological reactions.E) RNA is synthesized in short pulses of transcriptional activity.

Following splicing, the conserved adenosine nucleotide within the intron displays an unusual array of bonds, including: A) a phosphate attached to the 4-carbon of the adenine nitrogenous base. B) phosphodiester bonds attached at three places on the ribose (2′, 3′, 5′). C) additional methyl (-CH3) groups added to the adenine. D) a DNA strand attached to the spliced RNA. E) attachment to an arginine amino acid.

The conserved adenine nucleotide in introns serves as the: A) site for intron recognition for the spliceosome. B) branch point for the formation of the intronic "lariat." C) key point for ribosome assembly and initiation of translation. D) initial site of intronic RNA digestion and removal.

1. Which of the following scientists discovered the Ac-Ds transposable elements in maize? A) Marcus Rhoades B) Rollins EmersonC) Barbara McClintockD) George BeadleE) All of these scientists made the discovery.

10. Four clones (A, B, C, and D) of human genomic DNA are tested for sequence-tagged sites 1 through 5. A shows 2 and 3; B shows 2 and 5; C shows 1 and 5; D shows 3 and 4.What is the order of the clones in their contig?A) ABCD B) BDAC C) CBAD D) ACBD E) DCAB

10. In the 1940s, George Beadle and Edward Tatum developed the one-gene-one- polypeptide hypothesis by mutating the genes of Neurospora crassa (bread mold). They identified mutants that were unable to synthesize amino acids, such as arginine. When multiple mutant strains were identified for a single trait (adenine synthesis deficient), they showed that: A) all the mutants encoded the same single enzyme. B) genes were transcribed before being translated. C) each step in the biochemical pathway was mediated by an enzyme encoded by a single gene. D) the biosynthesis of arginine could occur, even with mutated enzymes. E) separate genes/enzymes performed identical steps in the pathway.

12. These functional RNAs function to regulate protein production of specific genes. A) transfer RNA (tRNA)B) small nuclear RNAs (snRNAs)C) micro RNAs (miRNAs) D) small interfering RNAs (siRNAs) E) ribosomal RNAs (rRNAs) Answer: C

13. Eukaryotic retrotransposons such as Ty1 and Copia are flanked by: A) direct repeats (DRs). B) inverted repeats (IRs).C) long terminal repeats (LTRs).D) long inverted repeats (LIRs).E) None of the answer options are correct.

14. What is meant by "codon bias"? A) Codons are specific for the amino acids they specify. B) Certain viruses like to infect particular DNA sequences in our genome. C) Some organisms will use specific codons for amino acids they specify. D) Certain viruses like to infect particular DNA sequences in our genome, and some organisms will use specific codons for amino acids they specify. E) Noneoftheansweroptionsarecorrect.

16. In a classical genetic experiment, Newcombe spread E. coli cells on complete medium. After several generations of growth, he replica-plated the colonies onto two plates with complete medium plus a selective agent. Colonies on the first replica were left untouched, while on the second replica, they were respread, allowing bacteria from each colony to be "moved" to a new location on the plate. Significantly more mutants after spreading (resistant to the selective agent) were observed than if they had not been respread. The experiment was used to: A) demonstrate how to screen for environmental mutagens.B) demonstrate that in some cases mutations are caused by the selective agent itself. C) provide evidence that mutations occur in the absence of the selective agent.D) provide evidence that mutation occurs in prokaryotes as well as in eukaryotes. E) show a direct correlation between the amount of the selective agent used and the number of resistant mutants.

18. Genes that are related by gene-duplication events within the same genome are called: A) homologs. B) orthologs. C) paralogs. D) heterologs. E) alleles.

18. Which of the following acts before the others? A) tRNA alignment with mRNAB) aminoacyl-tRNA synthetaseC) RNA polymerase D) ribosome movement to the next codon E) amino acid chain elongation

19. LINEs differ from retrotransposons in that LINEs do not: A) encode transposase. B) encode reverse transcriptase.C) contain LTRs.D) transpose in a replicative manner. E) contain the transposase gene.

19. The principle of parsimony refers to:A) the most complicated explanation involving the smallest number of evolutionary changes.B) the most complicated explanation involving the largest number of evolutionary changes.C) the simplest explanation involving the smallest number of evolutionary changes. D) the simplest explanation involving the largest number of evolutionary changes. E) explanations that support your hypothesis.

2. A corn plant is homozygous for a mutant allele that results in no pigment in the seed (i.e., white). The mutant is caused by Ds insertion that often exits late in seed development, when there is an active Ac element in the genome. The seeds of this plant will be: A) no pigment (i.e., white).B) pigmented all over.C) white with small spots of pigment. D) white with large spots of pigment. E) weakly pigmented.

20. The spontaneous reversion rate for a chemically induced mutation is 1 × 10-8. For EMS , the rate is 0.9 × 10-8, and for acridine, it is 2 × 10-5. What change was involved in the original mutation? A) an AT to CG transversion B) an AT to GC transition C) a frameshift mutationD) a GC to AT transition E) a GC to TA transversion

20. What percentage of the human genome is derived from transposable elements? A) less than 5% B) 25% C) 50% D) 75% E) nearly 100%

21. The -10 and the -35 boxes found in bacterial gene promoters function to: A) recruit DNA polymerase activity for DNA replication before cell division. B) activate transcription at the -35 region of a gene.C) orient RNA polymerase at a gene's transcription start site. D) identify the boundaries between introns and exons.E) identify the boundaries for DNA unwinding during transcription.

21. The grasses (such as barley, rice, sorghum, and corn) vary widely in the size of their genomes, yet they are descended from a common ancestor. What accounts for the variation in DNA content (i.e., the total amount of DNA)? A) large-scale duplication of genesB) large-scale loss of genesC) accumulation of LTR retrotransposons D) loss of LTR retrotransposonsE) None of the answer options are correct.

22. A Neurospora nonsense mutation known to be UAG is treated with two compounds, hydroxylamine and 5-bromouracil. Hydroxylamine (HA) causes G·C → A·T transitions, while 5-bromouracil causes T·A → C·G transitions. Which of these agents do you expect to induce revertants? A) Both agents are expected to induce revertants.B) Both agents are expected to induce revertants, but hydroxylamine will be more effective.C) 5-bromouracil only is expected to induce revertants.D) Hydroxylamine only is expected to induce revertants.E) Neither hydroxylamine nor 5-bromouracil agents are expected to induce revertants.

23. The "interactome" is defined as:A) the sequence and expression patterns of all transcripts.B) the sequence and expression of all proteins.C) the complete set of all physical interactions (i.e., protein/DNA or protein/protein). D) the complete set of all metabolites.E) None of the answer options are correct.

27. In eukaryotic species, three separate RNA polymerases transcribe different categories of genes. RNA polymerase I is known to primarily transcribe:A) messenger RNAs.B) protein-encoding genes. C) most ribosomal RNAs.D) transfer RNAs and small nuclear RNAs. E) genes responsible for DNA replication.

3. The spliceosome functions to:A) insert introns into mRNA sequences before translation.B) create noncoding sequences in genes to enhance gene stability. C) remove noncoding introns from transcribed RNAs.D) control translation, ensuring that only exons are translated.E) inhibit transcription of noncoding DNA regions.

31. What is the function of the TATA binding protein?A) aids in the removal of introns from eukaryotic pre-mRNAB) allows prokaryotic RNA polymerase to bind to the promoter of genes C) allows eukaryotic RNA polymerase II to bind to the promoter of genes D) helps termination factors bind and terminate transcription.E) All of the answer options are correct.

34. In what cellular compartment are introns removed from pre-mRNA to make mature mRNA? A) cytoplasmB) endoplasmic reticulum C) nucleusD) mitochondriaE) Golgi apparatus

5. In what cellular compartment are introns removed from pre-mRNA to make mature mRNA? A) cytoplasmB) endoplasmic reticulum C) nucleusD) mitochondriaE) Golgi apparatus

6. In a haploid fungus, a small in-frame deletion arises in the coding region of a gene necessary for the production of the amino acid leucine. The deletion removes the three base pairs corresponding to a UAC codon (encoding cysteine). This small deletion will most likely cause: A) the production of a truncated (shortened) protein.B) the production of a truncated (shortened) mRNA.C) the production of a shorter mRNA and of a shorter protein. D) no detectable phenotype.E) inability to grow on medium that lacks cysteine.

7. The experimental value of a "pulse-chase" cellular labeling experiment using radioactive uracil is that: A) all RNAs within a cell are labeled and easily detected.B) the radioactivity allows for easy purification of cellular nucleic acids.C) all RNAs created during the pulse are labeled and detectable, revealing their stability and localization.D) the spliceosome function is easily seen with radiolabeled RNAs in the cellular cytosol.E) cells adapt to the pulse of radioactivity, revealing nucleic acid repair mechanisms.

8. A biochemical pathway making flower pigments shows the following sequential color conversions (each arrow represents an enzyme-catalyzed step). Colorless yellow blue red ABCA plant is homozygous for mutations in enzyme "B," inactivating the active site for this enzyme. The resulting flowers will be:

8. A point mutation in a gene's promoter will most likely cause the production of: A) a shortened mRNA and a truncated protein.B) a shortened mRNA and a wild-type protein.C) reduced amounts of mRNA and protein. D) reduced amounts of mRNA and wild-type amounts of protein. E) wild-type amounts of mRNA and reduced amounts of protein.

8. Hybridization of single-stranded wild-type DNA with DNA from mutations caused by IS elements characteristically shows (through electron microscopy):A) chi structures.B) unpaired tails. C) a single-stranded loop representing IS DNA.D) a single-stranded loop representing wild-type DNA. E) theta structures.

The Hardy Weinberg Principle tells us that ) if, in a given population, p+q=1 then the population is in equilibrium B) in any large population, the frequency of genotypically A/A individuals is equal to the square of the frequency of the A alleleC) in a randomly mating population, in the absence of mutations, selection, migrations, and genetic drift, genotypic frequencies can be predicted from allelic frequenciesD) a and cE) a, b, and c

C. In a randomly mating populations - in the absence of mutations, selection, and genetic drift- genotypic frequencies can be predicted from allelic frequencies

You are interested in studying a gene called CFTR because mutations in this gene cause cystic fibrosis. You have made a line of mice that lack the mouse CFTR gene. These mice are unable to clear bacteria from their lungs, so they get lung disease. You put a normal human CFTR gene into some of these mice and discover that the mice with the human gene are able to clear bacteria from their lungs and no longer get lung disease. From this experiment, you can conclude that: A) The DNA sequences of the mouse CFTR gene and human CFTR gene are identical. B) The amino acid sequences of the mouse CFTR protein and the human CFTR protein are identical. C) The mouse CFTR gene and human CFTR gene encode proteins that can serve a similar function. D) Both answers B and C are true.C

C. The mouse cftr gene and human CFTR gene encode proteins that can serve similar functions

Lysine and arginine are two amino acids with similar R groups and similar biochemical behavior. A point mutation results in the substitution of lysine for arginine and has no detectable effect of that protein. This is an example of:

CONSERVATIVE SUBSTITUTION

1. Formation of the peptide bond could be described as: A) a spontaneous reaction between two peptides, forming a covalent bond. B) a ribosome-catalyzed process occurring in the nucleus. C) a tRNA catalyzed process. D) a ribosome-catalyzed dehydration reaction (H20 released). E) transcription, a key process in gene expression.

10. A mutant plant with white flowers exists that lacks red anthocyanin pigment, normally made by enzyme P. Indeed the petal tissue lacks all detectable activity for enzyme P. Despite the lack of enzyme activity, a study of homozygous mutant cells using antibodies against the wild-type enzyme demonstrated that the cells homozygous for the mutation still had the enzyme (i.e., the antibody showed the presence of the enzyme). Which statement could explain these results? A) The mutant had another gene the researcher was not aware of which produced an enzyme that could function like the mutated enzyme. B) The mutant cells likely had large scale chromosomal mutations that resulted in the expression of some gene similar to the gene for enzyme P. C) The mutant allele must have had a nonsense mutation that resulted in complete formation of a non-functioning enzyme. D) The mutant allele must only have missense mutations that simply knocked out enzyme function yet the enzyme would still bind to the antibody. E) Noneoftheanswersarecorrect.

10. Retroviruses replicate using: A) DNA polymerase.B) RNA polymerase. D) is found only in plants. E) None of the answer options are correct. D) is found only in plants. E) None of the answers are correct. C) restriction endonuclease. D) reverse transcriptase.E) topoisomerase.

12. Five nutritional mutants in Neurospora were independently isolated. They all require compound F to grow. Intermediate compounds in the biosynthesis of compound F are known and tested for their ability to support the growth of each mutant. The results are given in the table below, where "+" indicates growth and "0" indicates no growth. CompoundsMutants ABCDEF --------------------------- 1 000+0+ 2 0+0+0+ 3 00000+ 4 0+++0+ 5 ++++0+ Assuming a linear pathway, what is the order of the compounds A through F? A) EABCDF B) ECBDAF C) ECDBAF D) EACBDF E) EBACDF

13. The use of proflavin-induced mutations (like FCO) in the rII genome was found to experimentally support a three-nucleotide explanation for the genetic code because: A) an insertion mutation could be suppressed by a deletion mutation. B) the rII genes are easier to mutate than other genes. C) proflavin was an inexpensive mutagenic compound. D) a gene with three insertions or three deletions could sometimes provide a functional protein. E) the rII gene encoded a protein with exactly the correct number of amino acids, as would be explained by a three nucleotide genetic code.

15. Imagine that exposure to the antibiotic streptomycin promotes mutations causing streptomycin resistance in E. coli. If you repeated Luria and Delbruck's test using streptomycin instead of the T1 phage, what results would you expect?A) None of the 20 individual cultures would show any streptomycin-resistant colonies. B) The 20 individual cultures would show a high variation in the number of streptomycin-resistant colonies.C) The 20 individual cultures would show a high variation in the size of streptomycin- resistant colonies.D) The 20 individual cultures would show comparable numbers of streptomycin-resistant colonies.E) The 20 individual cultures would show streptomycin-resistant colonies of unusually large size.

18. After mutagen treatment, a molecule of 2-aminopurine (an adenine analogue) incorporates into DNA. During replication, the 2-AP protonates. The mutational event caused by this will be: A) AT to CG. B) GC to AT. C) AT to TA. D) AT to GC. E) GC to CG.

2. A mutation changes a codon from AAA (encoding lysine) to AGA (encoding arginine) in yeast, but no mutant phenotype is detected when the mutant strain is plated and grown on minimal or complete medium. This type of mutation is called: A) synonymous. B) suppressor. C) nonsense.D) missense. E) frameshift.

2. The three-dimensional structure of an active hemoglobin (heterotetramer) structure would be referred to as its: A) primary structure.B) secondary structure. C) tertiary structure.D) quaternary structure. E) quinternary structure.

21. A missense mutation in Neurospora will revert by treatment with nitrous acid, but not by hydroxylamine. Hydroxylamine (HA) causes only G·C → A·T transitions, while nitrous acid causes both G·C → A·T and A·T → G·C transitions. The original mutation (not the reversion) must have been: A) AT to CG.B) AT to GC.C) AT to TA. D) GC to AT. E) GC to TA.

26. In eukaryotic species, three separate RNA polymerases transcribe different categories of genes. RNA polymerase III is known to primarily transcribe:A) messenger RNAs.B) protein-encoding genes. C) most ribosomal RNAs.D) transfer RNAs and small nuclear RNAs. E) genes responsible for DNA replication.

3. Fibrous proteins contain a structure that is: A) rounded and compact. B) globular. C) only comprised of secondary structure. D) linear/extended. E) random and variable.

8. Which of the following is evidence that RNA was a message-carrying intermediary between DNA and protein? A) A hydroxyl group is present on ribose.B) RNA is single stranded and thus cannot be copied by semiconservative replication in a manner similar to DNA.C) RNA structure includes a molecular code, proving that it carries a genetic message. D) RNA is produced in the nucleus (with DNA) and then migrates to the cytosol, the location of protein synthesis.E) Pulse-chase experiments revealed RNA to be exclusively localized to the nucleus.

9. A transversion somewhere within gene "B" in a laboratory plant results in the production of a mutant mRNA that is much longer than the wild type. This transversion is most likely located: A) downstream of the STOP codon, in the last exon of gene "B." B) in an intron of gene "B" away from the splice sites.C) in the open reading frame of gene "B."D) within a splice site of gene "B." E) within the promoter of gene "B."

a mutant plant with white flowers exists that lacks red anthocyanin pigment, normally made by enzyme P. Indeed the petal tissue lacks all detectable activity for enzyme P. Despite the lack of enzyme activity, a study of homozygous mutant cells using antibodies against the wild-type enzyme demonstrated that the cells homozygous for the mutation still had the enzyme. Which statement could explain these results? A.The mutant had another gene the researcher was not aware of which produced an enzyme that could function like the mutated enzyme. B. The mutant cells likely had large scale chromosomal mutations that resulted in the expression of some gene similar to the gene for enzyme P. C. The mutant allele must have had a nonsense mutation that resulted in complete formation of a non-functioning enzyme D. The mutant allele must only have missense mutations that simply knocked out enzyme function yet the enzyme would still bind to the antibody E.None of the answers is correct

D. the mutant allele must only have missense mutations that simple knocked out enzyme function yet the enzyme would still bind to the antibody

1. Which of the following is/are TRUE for RNA compared to DNA?A) RNA has ribose sugar in its nucleotides, rather than the deoxyribose found in DNA.B) RNA is usually single stranded and can make more complex three-dimensional molecular shapes than double-stranded DNA.C) RNA contain the bases A, G, C, and U, whereas DNA contain the bases A, G, C, and T.D) RNA can catalyse biological reactions, but DNA cannot.E) All of the answer options are correct.

13. Deducing the protein encoding genes from genomic sequences involves: A) ORF detection.B) direct evidence from cDNA sequences. C) predictions of binding site.D) predictions based on codon bias.E) All of the answer options are correct.

13. Which of the following is/are TRUE about functional RNA? A) Functional RNAs do not encode proteins and are active as RNA. B) Some functional RNAs encode proteins. C) Examples of functional RNA include mRNA, tRNA, and rRNA. D) Examples of functional RNA include tRNA, rRNA, snRNA, and miRNA. E) Functional RNAs do not encode proteins, and examples include tRNA, rRNA, snRNA, and miRNA.

14. Which of the following mRNA codons would form a codon-anticodon base pairing interaction with the 3′-UAG-5′ tRNA anticodon?A) 3'- ATC-5'B) 5'- GAU-3' C) 5'-ATC-3' D) 3'-AUC-5' E) 5'-AUC-3' Answer: E

17. The rare enol form of thymine pairs with guanine. If a thymine enolization occurs during replication, what would be the mutational event? A) AT to TA B) CG to AT C) CG to GC D) GC to TA E) TA to CG

17. Which could be an anticodon for the amino acid isoleucine? A) UAA B) UAG C) UAU D) UAA and UAG E) UAA, UAG, and UAU

24. Which of the following statements is/are TRUE about reverse genetics?A) Reverse genetics analysis starts with an altered phenotype in an organism.B) Reverse genetics analysis starts with a known DNA sequence, mRNA, or protein.C) Reverse genetics can be performed by random mutagenesis, targeted mutagenesis,or by phenocopying.D) All of the answer options are correct.E) Reverse genetics analysis starts with a known DNA sequence, mRNA, or protein, and it can be performed by random mutagenesis, targeted mutagenesis, or by phenocopying.

32. People who develop the hereditary form of retinoblastoma inherit: A) a pair of dominant alleles causing retinoblastoma.B) a pair of normal (non-retinoblastoma) alleles.C) a pair of recessive alleles causing retinoblastoma. D) a single dominant allele causing retinoblastoma. E) a single recessive allele causing retinoblastoma.

33. Which of the following is/are role(s) of the 5′ cap?A) The cap helps the RNA polymerase find the promoter and initiate transcription. B) The cap plays a role in the removal of introns.C) The cap acts as a binding site for the ribosome.D) The cap protects the RNA from degradation.E) The cap acts as a binding site for the ribosome and protects the RNA from degradation.

35. Mice that are homozygous for a complete deletion of the wild-type gene "T" develop multiple tumors in a variety of tissues very early in life. Gene "T" is most likely:A) an activated oncogene.B) a dominant tumor suppressor gene. C) an oncogene.D) a proto-oncogene.E) a tumor suppressor gene.

6. How many nucleotides would be expected for a gene coding for a protein with 300 amino acids? A) 300 B) 100 C) 600 D) 1200 E) 900

9. IS-induced mutations are different from missense mutations in that they are: A) nonrevertible. B) less severe.C) more severe.D) nonpolar.E) not reverted by mutagens.

9. The best evidence to prove that a candidate gene is a disease gene is finding: A) a start and stop codon.B) a CpG island upstream.C) that the gene is expressed. D) the homologous gene in many similar animals. E)a mutation in the homologous sequence from a disease sufferer

9. While one gene usually specifies one enzyme, which of the following is/are NOT true? A) One gene can specify a single enzyme if that enzyme contains a single type of polypeptide chain. B) One gene can specify parts of two enzymes if the enzymes have one type of polypeptide chain in common. C) One enzyme may be determined by two genes if the enzyme has two different types of polypeptide chains. D) One mutation may cause multiple nutritional requirements if it interferes with the synthesis of an intermediate common to several pathways.E) Two mutations in the same gene can never have different effects, even when studied in great biochemical detail.

A mutant yeast strain can convert galactose to UDP galactose, but it is unable to carry out the next step of galactose metabolism because it cannot make the Gal10 enzyme. All of these are possible explanations for the phenotype EXCEPT __________. A) A mutation in the protein-coding region of the GAL10 gene. B) A deletion of the UAS of the GAL10 gene. C) A deletion affecting the GAL4 protein-binding site upstream of the GAL10 gene D) A mutation affecting the GAL4 gene that makes its activation domain nonfunctional E) None of the above

E. A mutation in the protein coding region of the Gal 80 gene OR D

A student is growing yeast cells in galactose rich media. Which of these statements is likely true of these cells? ) Gal3 will be bound to the activation domain of Gal4.b) Gal80 will be bound to the activation domain of Gal4.c) Gal3 will bind to Gal80, and no protein will be bound to the activation domain of Gal4.d) The GAL1gene will not be transcribed in these cells.e) Gal4 will be "inactive" in these cells.

Gal 3 will bind to Gal80, and no protein will be bound at the activation domain of Gal4

The immediate source for an F' plasmid is a(n):

HFR bacterial cell

Why are transposable elements inserted into introns not typically problematic to an organism?

Introns are spliced out of the pre-mRNA

Which of these is not an example of gender specific epigenetic silencing?

Positive effect variegation

Chargaff's Rule does not hold for which of the following genome types? Yeast Bacteria Single-stranded DNA virus Invertebrates Archaeobacteria

Single stranded DNA virus

The "ABC" operon encodes three enzymes: A,B, and C. Upstream of the operon's promoter, there is an independent gene that encodes the regulatory protein R. A bacterial strain from which the gene encoding R has been completely removed is found to transcribe the ABC operon even in the absence of an inducer. What can be concluded about this system?

The R protein likely acts as a repressor of ABC operon transcription in the absence of an inducer.

Approximately 45 percent of the human genome is derived from transposable elements, such as LINES and SINES. The following statements explain how it is possible for us to survive with the high percentage of transposable elements except:

The active transposable elements that are capable of increasing in copy number remain active by host regulatory mechanisms

the cDNA for a eukaryotic gene B is 1000 nucleotide pairs long. a cDNA clone is used to isolate a genomic clone of gene B, and the gene is sequenced. From start to stop codon, the gene is found to be 2000 nucleotide pairs long. The most probable reason for the discrepancy is that

The gene has 1000 nucleotide pairs of introns.

What structural feature of DNA suggests a possible method of its replication? A)The nucleotides between nucleotide strands are complementarily paired. B)The DNA strands run anti-parallel. C)The hydrogen bonds between A-T and G-C are the same on all DNA molecules. D)The phosphate bonds in the backbone of all DNA molecules.

The nucleotides between nucleotide strands are complementary paired

Transposable elements can cause corn kernels to be spotted if they jump in or out of the color genes during the development of the kernel. However, the size of the spots can vary greatly. In words and diagrams, describe the difference between the two kernels shown below at the molecular and developmental levels.

The transposition event occurs earlier in kernel A

A researcher would initiate a genetic screen using a transposon "mutagen" because a) unlike other mutagens, transposons will not harm the recipient cell.b) the transposon will insert ubiquitously in nongenic DNA; the researcher can concludethat the remaining DNA is genic.c) the transposon may insert into a gene, resulting in a mutant phenotype and providing a molecular "tag" of the gene.d) transposons are less toxic to work with than chemical mutagens.

The transposon may insert into a gene; resulting in a mutant phenotype and providing a molecular tag of the gene

In the following pedigree the shaded symbols represent a rare X-linked disease of the blood. The proband (person of interest, indicated with an arrow) has 45 chromosomes and is sterile. Propose a single mechanism to account for both the sterility and the blood disease in this female.

The woman is likely demonstrating a case of Turner syndrome

A scientist studying the attenuation of the tryptophan operon would find which result when there is an abundance of tryptophan?

Transcription will terminate when segments 3 and 4 of the leader mRNA base pair with each other

1. Which of the following sequences includes a clear eight base pair palindrome? A) 5′-CCGATCGATCCC-3′B) 5′-TGGGGTTTT G-3′C) 5′-GGGGAAAA-3′ D) 5′-GATCCTAG-3′E) 5′-AACCAACCAA-3′

12. Perhaps the most common use of plasmid vectors is to:A) enable the amplification of cloned DNA within a bacterial host cell. B) isolate viral DNA from infected cells.C) insert restriction enzyme cut sites into cDNAs.D) generate recombinant proteins.E) manipulate the cell biology of bacterial cells.

16. A wild-type Aspergillus strain is transformed with a plasmid carrying a hygromycin resistance allele, and cells are plated on hygromycin. One resistant colony showed an aberrant type of aerial hyphae. When crossed to wild type, the progeny were 1/2 normal hyphae, Hyg sensitive, and 1/2 aberrant hyphae, Hyg-resistant. The probable explanation is that: A) the plasmid was inserted in a gene for normal hyphal development. B) the plasmid interfered with hyphal development. C) a mutation arose in a gene for hyphal development on the plasmid. D) a mutation arose in a gene for hyphal development on a recipient chromosome. E) the recipient was a heterokaryon carrying some mutant nuclei.

21. The DNA sense strand for a particular amino acid is 5′-ATG-3′. What RNA sequence would be transcribed for this codon, what tRNA anticodon would recognize it, and what amino acid would be added in response to this codon? A) 5′-CAU-3′ (RNA sense); 3′-GUA-5′ (tRNA anticodon); histidine B) 3′-AUG-5′ (RNA sense); 3′-UAC-5′ (tRNA anticodon); valine C) 5′-CAU-3′ (RNA sense); 3′-AUG-5' (tRNA anticodon); histidine D) 5′-UUU-3′ (RNA sense); 3′-AAA-5′ (tRNA anticodon); phenylalanine E) 5′-AUG-3′ (RNA sense); 3′-UAC-5′ (tRNA anticodon); methionine

23. Bacterial artificial chromosomes are comprised of: A) components of the bacterial F plasmid and can carry large inserts (200,000 bp). B) fosmid DNA sequences, enabling the transfer of DNA between bacteria. C) viral lambda DNA sequences, enabling the transfer of DNA between bacteria. D) a bacterial chromosome, with some artificial components. E) recombinant proteins isolated from specific phage viruses.

24. The "wobble" base is a tRNA site where: A) inosine nucleotide can sometimes be found. B) mutations often create severe mutations in the encoded protein. C) the amino acid is closely located to facilitate its addition to the new protein. D) aminoacyl-tRNAs read the tRNA specificity. E) tRNAsdisplaystructuralinstability.

25. Evidence that amino acids were "illiterate," or passively added to the newly synthesized protein at the direction of the tRNA to which they are attached, was derived from experiments with: A) nickel hydride, altering the amino acids attached to charged tRNAs to determine if they were still utilized by the ribosome. B) proflavin-based mutations on subsets of anticodons to determine if the mutations affect the charging of tRNAs. C) abnormal forms of the aminoacyl-tRNA transferase enzymes. D) chemically synthesized, novel amino acid structures. E) eukaryotic ribosomes combined with prokaryotic tRNAs.

28. The anticodon on the tRNA molecule:A) binds to the mRNA in a complementary fashion. B) is oriented and written in the 5′3′ direction. C) is a catalytic part of protein synthesis.D) is the same for all tRNA molecules.E) containsamino-acyl-tRNAsynthetase.

31. The ribosome in eukaryotes can be described as being composed of: A) rRNAs and over 50 different proteins. B) a 50S subunit and a 30S subunit, forming a 70S ribosome. C) proteins in quaternary structure, binding tRNAs within the nucleus. D) a large enzyme complex that is found only on the endoplasmic reticulum membrane. E) rRNAs, protein subunits, ATP, and aminoacyl-tRNA synthetase.

33. The A site, P site, and E site each control ______________ (in order) during translation. A) binding incoming tRNAs (A), retention of the peptide chain during elongation (P), exit of deacylated tRNAs (E) B) activation of tRNAs (A), protease-based release of the protein product (P), exit of empty rRNAs (E) C) activation of ribosome activity (A), processing of amino acids (P), elongation of the peptide chain (E) D) amino acid bond transfer (A), protein processing (P), and elongation of the peptide chain (E) E) binding incoming tRNAs (A), retention of the peptide chain during elongation (P), exit of deacylated tRNAs (E) as well as activation of ribosome activity (

38. The proteome for humans is about 70,000 proteins whereas only 21,000 protein- coding genes have been discovered. Which statement could explain this? A) Splicing of eukaryotic mRNA can result in the production of more than one protein from a single gene. B) Many of our proteins are actually produced by symbiotic organisms in our bodies. C) Splicing of eukaryotic mRNA can result in the production of more than one protein from a single gene and many of our proteins are actually produced by symbiotic organisms in our bodies. D) We do not know enough about the human genome to explain this fact.

a partial diploid of genetoppes I+ p...

a inductive production b galactodolase

When scanning a cDNA sequence for the open reading frame of a gene, a researcher will look for all of the following except

a promoter sequence

which of the following chromosomal aberrations can result in masking recessive alleles resulting in their expression? A) deletionsB) duplications C) inversionsD) translocations

a. deletions

A drug company hoping to synthesize large amounts of insulin is faced with two options: either isolate and purify insulin from cows or produce human insulin in E. coli via DNA cloning. The company chooses the latter option. Why?

all of the above

Deducing the protein encoding genes for genomic sequences in eukaryotic organisms involves:

all of the above

Which of these is likely a functional domain of a regulatory protein that modulates the transcription of specific genes in a eukaryotic cell?a) a DNA-binding domainb) a domain that allows the regulatory proteinto interact with CBP or a mediator complexc) a domain that influences chromatin condensationd) a domain that acts as a sensor of physiological conditionse) All of the above.

all of the above

A researcher is studying transposable elements in violets and identifies elements with properties similar to that of Ac in maize. Importantly, this new violet transposable element can mobobilize independently (and this process does not require the action of any additional elements). This is an example of a (n) _______

autonomous element

18. Gel electrophoresis can be used to separate DNA or RNA molecules based on their: A) charge. B) molecular weight. C) solubility in buffer. D) All of the answer options are correct. E) charge and molecular weight.

20. In Neurospora, a linear biochemical pathway synthesizes an amino acid Z. E1, E2, and E3 are enzymes that catalyze the three reactions. E1 E2 E3W X Y Z Null mutants for the enzyme E2 gene will grow on minimal medium supplemented with compounds: A) W or X.B) Y or Z.C) W or X or Y or Z. D) Z only.E) W only.

20. In the Sanger sequencing method, the use of dideoxy adenosine triphosphate stops nucleotide polymerization opposite: A) A's in the template strand. B) T's in the template strand. C) G's in the template strand. D) C's in the template strand. E) any base selected randomly in the template strand.

21. This linear vector is comprised of bacteriophage genome components and can carry cloned DNA up to 15 kb in size. A) pUC19B) lambdaC) fosmidD) bacterial artificial chromosome E) T-vector

22. The aminoacyl-tRNA synthetase enzymes are responsible for: A) synthesizing tRNA molecules in the nucleus from rRNA genes. B) adding amino acids to appropriate tRNAs (charging the tRNA). C) regulating the process of amino acid synthesis. D) catalyzing the enzymatic step of peptide bond formation during translation. E) matching each tRNA with the appropriate mRNA codon.

22. The value of the beta-galactosidase gene in cloning could best be described as the gene: A) that provides valuable restriction enzyme cut sites.B) that "reports" if a recombinant product is formed.C) that confers resistance to antibiotics. D) that facilitates easier transfer of the DNA into bacterial cells. E) thatencodesaspecialDNAligaseenzyme.

25. In the study of genetics, a "library" denotes: A) the information carried in a digital database, such as GenBank. B) a collection of DNA fragments isolated from a particular group of cells (or tissue) representing the genetic content of those cells. C) a complex collection of restriction enzymes gathered for the purpose of restriction enzyme mapping. D) all the proteins present in a particular cell type, collected by protein extraction. E) a collection of books that explore the topic of genetics.

26. How does a nonsense suppressor mutation prevent amber mutants from terminating their polypeptides prematurely?A) The mutation turns the amber codon back into a wild-type codon.B) The mutation alters a tRNA so that it reads the amber codon and inserts an amino acid. C) The mutation alters the release factors that would halt synthesis. D) The mutation results in a wobble that allows synthesis to continue. E) The mutation replaces the amber codon with an ochre codon.

28. When seeking to identify the gene that is responsible for a particular cellular characteristic, scientists will generate cells carrying a recessive mutation that perturbs the characteristic of interest and then try to identify or "clone" the wild-type copy of that gene through functional complementation. This type of complementation could be described as: A) intensification of the recessive mutant trait via transferred plasmid DNA carrying the gene of interest. B) reversal (rescue) of the recessive mutant trait via transferred plasmid DNA carrying the gene of interest. C) identification of two different genes that both regulate the same cellular trait. D) a biochemical phenomenon where an enzymatic process is eliminated by plasmid DNA expressing a gene of interest. E) mutation of a gene of interest to complement an opposing biochemical pathway.

32. Homologous recombination in yeast facilitates the: A) segregation of genes during meiosis. B) targeted replacement of a gene in a living yeast cell. C) amplification of homologous yeast genes. D) independent assortment of separate gene alleles. E) expression of similar gene sequences via one promoter.

34. Embryonic stem (ES) cells have the unique ability to:A) form a mouse pup without any additional cells.B) combine with other stem cells to generate a chimeric mouse pup. C) grow in the presence of gancyclovir.D) eliminate specific genes in response to chemical cues.E) dividemorerapidlythanothercelltypes.

35. In bacteria, the Shine-Dalgarno sequence is found on the mRNA and is recognized by the ________________________ to reveal __________________________. A) the 16S rRNA; the translation STOP codon B) the 30S subunit; the translation START codon C) initiator tRNA; the translation START codon D) ribosome A site; the translation START codon E) amber tRNA; the translation STOP codon

7. The polymerase chain reaction requires ssDNA primers that:A) anneal to the same strand of template DNA, though at distant sites.B) anneal to opposite strands of template DNA at distant sites, with their 3′ ends directed toward each other.C) anneal to opposite strands of template DNA at distant sites, with their 5′ ends directed toward each other.D) anneal to each other to prime DNA polymerization.E) hybridize only to DNA within the open reading frame of a selected gene.

Gal80 inhibits GAL gene expression by: A) Specifically binding to GAL4 and preventing DNA-binding B) Specifically binding to GAL4 and preventing activation C) Specifically binding to GAL3 and preventing DNA-binding D) Specifically binding to GAL3 and preventing activation E) Inducing an allosteric change in GAL3

has a pedigree and says the propositus

b. the woman is likely demonstrating turner

A researcher is studying a population of flies comprised of 100,000 individuals. This population demonstrates positive assortative mating and does not consist of any subpopulations. Furthermore, all genotypes within this fly population are equally viable. This fly population cannot be Hardy-Weinberg equilibrium because ... A) All genotypes in this population are equally viable B) It is not infinitely large in size C) This population is not comprised of any subpopulations D) This population does not demonstrate random mating E) Both B and D

both b an d

Which of these epigenetic changes could shut down transcription of a particular genes?

both c and d

. Which of the following statements about PCR is/are FALSE?A) PCR reaction contains DNA template, primers, deoxyribonucleotide triphosphates, and heat- resistant DNA polymerase.B) Each PCR cycle includes three steps: denaturation, annealing, and extension.C) After each PCR cycle, amount of DNA increases in a linear manner.D) One limitation in PCR is that it requires prior knowledge of the sequence or at least part of the sequence to be amplified.E) PCR is a very sensitive technique and can amplify target sequences present in extremely low copy numbers.

1. The genome of Hemophilus influenzae, the first free-living organism to have its genome sequenced, is how large? A) 0.5-megabase B) 1.0-megabase C) 1.8-megabase D) 12-megabase E) 18-megabase

13. A plasmid vector has a gene for erythromycin resistance (EryR) and a gene for ampicillin resistance (AmpR). The Amp gene is cut with restriction enzyme, and donor DNA treated with the same enzyme is added. What genotype of cells needs to be selected to show evidence of transformation? A) AmpREryRB) AmpRErySC) AmpSEryRD) AmpSErySE) None of the answer options are correct.

15. A radioactive probe is generated using the actin gene from yeast. This probe is used in a Southern blot analysis of EcoRI digested genomic DNA from the ciliatedprotozoan Tetrahymena thermophila. The autoradiogram shows a single-labeled band of 4 kb in size. This means that the Tetrahymena actin gene is: A) 4 kb in size.B) the same size as the yeast gene.C) present in one copy in the genome.D) present in four copies in the genome.E) identical in sequence with the yeast gene.

17. The cDNA for a eukaryotic gene B is 900 nucleotide pairs long. A cDNA clone is used to isolate a genomic clone of gene B, and the gene is sequenced. From start to stop codon, the gene is found to be 1800 nucleotide pairs long. The most probable reason for the discrepancy is that: A) the mRNA broke during cDNA synthesis. B) the gene is present as a tandem duplication. C) the gene has 900 nucleotide pairs of introns. D) the genomic clone is not really gene B, just a related gene. E) there was a sequencing error.

2. The human genome is comprised of how many base pairs (bp)? A) 1 billion B) 2 billion C) 3 billion D) 4 billion E) 5 billion

24. A cDNA library is produced from ______________ and is valued in the analysis of __________________. A) chromosomal DNA digested with restriction enzymes; chromosomal organization B) chromosomal DNA digested with restriction enzymes; the promoter and regulatory elements of a particular geneC) a DNA copy of mRNA isolated from a group of cells; the protein-encoding open reading frame of a gene (no introns) D) a DNA copy of mRNA isolated from a group of cells; the promoter and regulatory elements of a particular gene E) protein sequences isolated from particular tissues; enzyme function encoded by a particular gene

29. In Southern and Northern blotting, the probe being used to analyze DNA or RNA identifies the target sequence via: A) attaching to the gene target via antibody-like protein-to-protein interactions. B) a DNA ligase-mediated attachment to the target DNA or RNA fragments. C) annealing between complementary sequences on the probe and the gene target. D) an enzymatic step that includes crosslinking between similar nucleic acid sequences. E) amplification during the polymerase chain reaction.

29. The ribosome is the primary site of: A) oxidative phosphorylation. B) protein packaging. C) protein synthesis. D) cellular respiration. E) amino acid storage.

3. A linear DNA molecule has n target sites for restriction enzyme EcoRI. How many fragments will be produced after complete digestion? A) n - 1 B) n C) n+1 D) 2n - 1 E) 2n + 2

3. The strategy for obtaining a genomic sequence can be divided into four steps: 1) overlap contigs for complete sequence, 2) sequence each fragment, 3) cut many genome copies in to random fragments, and 4) overlap sequence reads. What is the order of the four steps? A) 1, 2, 3, and 4 B) 4, 3, 2, and 1 C) 3, 2, 4, and 1 D) 3, 2, 1, and 4 E) 2, 4, 3, and 1

30. Transgenic plants can be generated using T-DNA plasmid carrying a gene of interest. To get the DNA into the plant cells, the researchers: A) use a syringe needle to inject the DNA into pollen before fertilization. B) use a syringe needle to inject the DNA into groups of cells in the plant's root tissue. C) co-cultivate bacteria with T-DNA and plant cells, resulting in DNA transfer. D) add a mild detergent to cultures of plant cells to open holes in the cell wall. E) remove the plant cell's normal DNA and replace it with the T-DNA.

31. Detection of a transgenic plant can be accomplished in a number of different ways, but often the initially transformed cells (in tissue culture) are enriched for the presence of a transgene by: A) screening groups of cells using the polymerase chain reaction. B) examining if the phenotype of the cells has been altered. C) placing the potentially transformed cells in the presence of a drug that inhibits the division of non-transgenic cells. D) microscopically examining the cells and isolating those that have "extra" DNA. E) None of the above. Transgenesis cannot be selected for in tissue culture.

33. Thymidine kinase is valuable in the targeting of mouse genes because:A) it enhances the efficiency of gene replacement.B) it suppresses expression of the target gene.C) it allows researchers to enrich cells for those that have undergone homologous recombination. D) transgenic cells are resistant to gancyclovir. E) transgenic cells are sensitive to gancyclovir.

34. Macrolides are antibiotics that kill bacteria by: A) inhibiting tRNA charging. B) releasing the protein before translation is complete. C) blocking the exit tunnel for "empty tRNAs." D) competing for tRNA binding in the "A site." E) causing tRNAs to be destroyed.

35. A transgenic mouse is different from a knockout mouse in that:A) transgenic mice have DNA added to their genome.B) transgenic mice do not have DNA added to their genome.C) the transgene is inserted at a random (ectopic) site in the genome. D) a knockout mouse is more healthy than a transgenic mouse. E) theknockoutmouseissterile.

37. Signal peptidase activity is responsible for: A) directing defective proteins to the ubiquitin-proteasome pathway. B) adding a signal phosphate group to some proteins, regulating their activity. C) removing a hydrophobic secretion signal from the N-terminus of a protein. D) adding a signal glycosylation modification to a new protein. E) destroying peptide bonds in defective proteins.

9. Which of the following is NOT true of a cDNA clone?A) cDNAs are generally copies of mRNA from expressed genes.B) Reverse transcriptase activity is required for cDNA cloning.C) cDNA clones lack exons, as introns are spliced together before cloning.D) cDNAs are typically smaller than the chromosomal sequence for a particular gene. E) cDNAs can be fused to a promoter to enable expression of the encoded gene.

In humans, males have only one X chromosome and females have two, yet housekeeping X chromosome genes are expressed in roughly equal amounts in males and females. How can this be explained?

c. one X chromosome in females is inactivated

When comparing the three key models of DNA replication, the model that included the synthesis of a brand new double-stranded DNA molecule from an original molecule was named: dispersive replication conservative replication semiconservative replication liberal replication none of the above

conservative

10. Identification of mRNA and initiation of priming for cDNA synthesis is accomplished by: A) purifying only cytosolic RNAs before initiating the process.B) coupling cDNA synthesis to exon splicing.C) detecting the 5′ cap sequence to initiation cDNA synthesis. D) use of oligo-dT to prime cDNA synthesis from the polyA tail. E) column purification of mRNA sequences.

14. Assume that a cosmid will carry inserts of about 50 kb and that cosmids are used to clone a 3 Mb (megabase) genome. Assuming you are particularly lucky and have no duplication in your library, what is the smallest number of cosmid clones you would need for a "complete" genomic library? A) 3000 clones B) 600 clones C) 300 clones D) 60 clones E) 30 clones

18. DNA probes are often synthesized based on knowledge of the protein produced by a gene. What might be a potential problem of synthesizing DNA probes in this manner? A) The deduced DNA probe would be missing introns associated with prokaryotic B) Because codons in DNA are redundant, knowing which codon to assign an amino acid may be difficult. C) The deduced DNA probe would be missing introns associated with eukaryotic genes. D) Because codons in DNA are redundant, knowing which codon to assign an amino acid may be difficult, as well as the deduced DNA probe would be missing introns associated with eukaryotic genes. E) Noneoftheansweroptionsarecorrect.

2. In recombinant DNA technology, DNA is most often cut using: A) DNA ligase. B) DNA polymerase. C) DNA gyrase. D) restriction endonucleases. E) terminal transferase.

23. The "wobble" base is less important than the other two nucleotides in a codon and is found: A) at the 5′ end of the RNA codon's sense strand. B) at the 3′ end of the mRNA codon's antisense strand. C) at the 3′ end of the tRNA anticodon. D) at the 5′ end of the tRNA anticodon. E) within a tRNA hairpin loop structure.

32. Which of the following are types of posttranslational processing that will not occur in a prokaryote?A) addition of a phosphate groupB) removal of a phosphate group C) addition of a co-factorD) cleavage of a mitochondrial targeting sequence E) None of the answer options are correct.

36. During the initiation of prokaryotic translation, a defect in IF3 function would generate: A) inefficient association of the 30S and 50S subunits. B) proteolysis of the 50S subunit. C) an inability of tRNA and mRNA binding to the 30S subunit. D) premature association of the 30S and 50S subunits. E) normal progression of translation (no defect).

5. "Traditional" whole genome shotgun (WGS) sequencing employs which of the following techniques? A) the construction of genomic libraries B) cloning of DNA in bacterial systems C) Sanger dideoxy DNA sequencingD) All of the answer options are correct.

7. Pyrosequencing determines the nucleotide sequence because:A) dideoxy nucleotides are incorporated into a growing DNA strand. B) dideoxy nucleotides terminate the growing DNA strand.C) PPi is incorporated into the growing DNA strand.D) PPi is released from the growing DNA strand.E) None of the answer options are correct.

8. The intermediate temperature cycle (72°C) in the polymerase chain reaction enables: A) hybridization between the template DNA and the PCR primers.B) denaturation of the template DNA to enable primer access to matching sequence. C) activation of the primers to being the amplification procedure. D) activity of the thermostable DNA polymerase enzyme.E) a reduction of contamination in the amplification reaction.

a prokaryotic operon is compossed of a series of adjacent genes under the control of The same start codon The same stop codon The same promoter and start codon The same promoter and operator

d. the same operator and promotor

11. Which of the following about plasmid vectors is/are TRUE?A) Plasmid vectors must have convenient restriction sites/polylinker sites.B) Plasmid vectors must have an origin of replication.C) Plasmid vectors must carry a selectable marker (drug resistance) for selection. D) Plasmid vectors can accept DNA inserts as large as 10 kb to 15 kb.E) Alloftheansweroptionsarecorrect.

19. Francis Crick proposed "adapter" molecules to serve as an intermediary between mRNA and the synthesis of protein. Key to any model of how RNA is "read" in the process of protein synthesis would be a consideration of how: A) amino acids are selected for addition to the new protein. B) the RNA code was read by the translation enzyme. C) peptide bonds are formed. D) one codon could select a specific amino acid. E) All of the answer options are correct.

19. Which of the following about the Sanger sequencing method is/are TRUE? A) Sanger sequencing uses dideoxynucleotide triphosphates to terminate DNA polymerization. B) Dideoxynucleotides lack 2′ and 3′ OH groups in the ribose sugar. C) All four deoxynucleotide triphosphates are used in a Sanger sequencing reaction. D) Fluorescent dyes can be used to automate Sanger sequencing. E) Alloftheansweroptionsarecorrect.

27. A tRNA with the anticodon 3′-ACC-5′ would carry the amino acid: A) phenylalanine.B) tyrosine.C) serine. D) threonine. E) tryptophan.

27. Foreign DNA can be introduced into a cell using which of the following method/s? A) transformationB) projectile gunC) injection D) virus infection E) Alloftheansweroptionsarecorrect.

30. All of the following are associated with ribosomes except: A) decoding center.B) peptidyltransferase center.C) exit site. D) groove for RNA binding. E) groove for DNA binding.

36. Which of the following organisms can be made transgenic by injection of DNA into the gonad, generating extrachromosomal arrays of the transgene in some egg cells? A) mouse (Mus musculus)B) yeast (Saccharomyces cerevisiae)C) bacteria (Escherichia coli)D) humans (Homo sapiens)E) roundworm(Caenorhabditiselegans)

4. The term coverage, when applied to genomics, means sequencing: A) every base pair of the entire genome.B) every base pair of a single chromosome.C) the same gene in multiple individuals. D) the same gene in multiple species. E) the same base pair multiple times.

A student is studying two types of chromatin. The first type is "open" or loosely packed and would be referred to as euchromatin. The second type is "closed" and tightly packed and would be considered

heterochormatin

a population is reproductively isolated. which of the following is primarily responsible for the introduction of new alleles a.) Independent assortment during meiosisb.) Inbreedingc.) Genetic driftd.) Germ line mutationse.) Somatic cell mutations

mutations in the germ line and mutations in germ line

When comparing the three key models of DNA replication, the model that included the separation of the two strands of the original DNA (template and using those strands as templates to synthesize two new DNA strands is called: directional replication. dispersive replication. conservative replication. semiconservative replication

test 3 genetics

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Completing the application, underwriting, and delivering the policy (primerica)