Genetic Engineering
Mock Exam
Course Introduction
Genetic Engineering explores the principles, techniques, and applications of modifying the genetic material of living organisms. Students will learn about recombinant DNA technology, gene cloning, genome editing tools such as CRISPR/Cas9, and methods for transferring genes between species. The course examines ethical, legal, and social implications of genetic manipulation and provides case studies highlighting advancements in agriculture, medicine, and biotechnology. Laboratory exercises may include DNA extraction, PCR, gel electrophoresis, and basic genetic transformation techniques to reinforce foundational concepts in genetic engineering.
Recommended Textbook
Introduction to Genetic Analysis 11th Edition by Anthony J.F. Griffiths
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Page 2
Chapter 1: The Genetics Revolution
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Sample Questions
Q1) Which of the following features make a species suitable as a model organism?
A)small organism
B)short generation time
C)small genome
D)produce large number of offspring
E)All of the answer options are correct.
Answer: E
Q2) Wild cats (Felis silvestris)and common mice (Mus musculus)are diploid.In wild cats 2n = 38,while in common mice 2n = 40.Based on this information,we can conclude that wild cat cells have:
A)less DNA than common mouse cells.
B)smaller genomes than common mouse cells.
C)fewer DNA molecules than common mouse cells.
D)fewer genes than common mouse cells.
E)fewer sets of chromosomes than common mouse cells.
Answer: C
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3
Chapter 2: Single-Gene Inheritance
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Q1) Cystic fibrosis is an autosomal recessive condition.If the parents of a boy with cystic fibrosis have two more children,what is the probability that both of these children will be unaffected?
A)1/16
B)3/16
C)4/16
D)9/16
E)16/16
Answer: D
Q2) A plant is heterozygous at three loci.How many different gamete genotypes can it theoretically produce with respect to these three loci?
A)2
B)3
C)4
D)8
E)16
Answer: D
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Chapter 3: Independent Assortment of Genes
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Q1) A corn plant of genotype A/a;B/B;C/c;D/d;E/e;f/f;G/g is selfed.What is the probability of producing a completely homozygous individual out of all possible offspring?
A)0.000
B)0.008
C)0.013
D)0.031
E)0.062
Answer: D
Q2) Define Mendel's experimental evidence for diploid chromosome content in pea plants.
Answer: Mendel noticed that recessive traits were "hidden" in the F<sub>1</sub> progeny of his monohybrid and dihybrid crosses.When the F<sub>1</sub> plants were selfed,the recessive trait appeared again in the next generation of plants (F<sub>2</sub>).This was reproducible and followed a ratiometric pattern consistent with two genetic determinants (genes)being carried for each trait.
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5
Chapter 4: Mapping Eukaryote Chromosomes by
Recombination
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Sample Questions
Q1) In humans,the allele N causes an abnormal shape of the patella in the knee (n is the normal allele).A separate gene is concerned with finger length,and the allele B causes abnormally short fingers,whereas b gives normal length.A study focused on people who have both abnormal patellae and short fingers (they were most likely N/n <sup>.</sup><sup> </sup>B/b in genotype),having inherited the N allele from one parent and the B allele from the other parent.These N/n <sup>.</sup><sup> </sup>B/b people mated with normal spouses,producing 40 progeny classified as follows: \[\begin{array} { l l }
\text { Normal } & 3 \\
\text { Abnomal knees and fingers } & 2 \\
\text { Abromal knees only } & 17 \\
\text { Abnomal fingers only } & 18
\end{array}\]
a)Draw the chromosomes of the N/n <sup>.</sup><sup> </sup>B/b individuals,their parents,and their four types of children,showing the positions of the alleles.
b)Explain why the four types of children were in the proportions shown.
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Chapter 5: The Genetics of Bacteria and Their Viruses
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Sample Questions
Q1) A gal<sup>-</sup> mutant:
A)can make its own galactose.
B)cannot grow without galactose.
C)cannot utilize galactose as a carbon source.
D)can utilize galactose as a carbon source.
E)is resistant to galactose.
Q2) Joshua Lederberg and Edward Tatum discovered a sex-like process in bacteria using:
A)high-powered microscopy in combination with antibiotic resistant genes.
B)populations of cells with complementing auxotrophs,and the measurement of the generation of prototrophs during microbial mixing.
C)mutations in bacteria that inhibit the mating process.
D)comparison of bacterial behavior to yeast behavior.
E)chromosome sequencing and analysis.
Q3) Two labs calculated cotransduction frequencies for the genes A and B.Lab 1 calculated 0.63,and lab 2 calculated 0.47.Which lab reported the genes to be closer together?
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Chapter 6: Gene Interaction
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Sample Questions
Q1) In a flowering plant, the a/a ; B/B genotype results in yellow flowers, and the A/A ; b/b genotype gives red flowers. Wild-type flowers are orange and are observed in plants with genotype A/A ; B/B. Two possibilities have been proposed for the biosynthetic pathways of flower color:
1. yellow pigment \(\rightarrow\) red pigment \(\rightarrow\) orange pigment
2. red pigment \(\rightarrow\) yellow pigment \(\rightarrow\) orange pigment
Propose a genetic experiment that can determine which of the two possibilities is most likely correct, and explain how you would interpret your results.
Q2) Loppins can be purple or white,and this phenotype is inherited as a simple dominant/recessive trait with complete penetrance.In females,purple is dominant over white,while in males the opposite is true.If you cross a white female to a purple male,and then you "self" the F<sub>1</sub> (i.e.,you cross brothers and sisters)what phenotypic ratios would you expect in the F<sub>2</sub>? Assume that you always get a 1:1 ratio of males : females.
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8
Chapter 7: Dna: Structure and Replication
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Q1) Linus Pauling proposed at one time that the DNA molecule was,in fact,a triple helix.If so,how would this situation affect cellular division?
A)The triple-helix DNA could be replicated and passed on to daughter cells as it normally would.
B)The triple-helix DNA could be replicated normally in mitosis but not meiosis.
C)Sperm and eggs produced by triple-helix DNA would be more fertile.
D)The fact that the DNA had three nucleotide strands would make it difficult to replicate the DNA and reassemble the copies.
Q2) Imagine it is the early 1900s and the nature of genetic material is not yet known.You believe that proteins (polypeptides)are the most likely candidates for storing heritable information,rather than DNA (which you regard as some kind of macromolecule used to store hydrocarbons).Make an argument supporting this position.
Q3) What structural feature of DNA suggests a possible method of its replication?
A)The nucleotides between nucleotide strands are complementarily paired.
B)The hydrogen bonds between A-T and G-C are the same on all DNA molecules.
C)The DNA strands run anti-parallel.
D)The phosphate bonds in the backbone of all DNA molecules.
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Chapter 8: Rna: Transcription and Processing
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Q1) Suppose that you have sequenced five genes,including the putative promoter region of each gene,that are involved in the cold acclimation process in Arabidopsis.Each of the five genes is expressed only when the temperature falls below a certain temperature.You know that a specific transcription factor (CBF1)binds somewhere between -200 and -250 of these genes to initiate transcription.How could you use the sequences to determine the DNA sequence motif that is bound by CBF1?
Q2) 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:
A)simple roundworms have over 10,000 genes,so humans should have many more.
B)experimental analysis of mRNA diversity generated estimates of ~100,000 genes.
C)humans have a large genome size (3.3 \(\times\) 10<sup>9</sup> base pairs).
D)the diverse types of tissue in humans require many thousands of genes.
E)All of the answer options are correct.
Q3) A major difference between eukaryotes and prokaryotes is that eukaryotes have a nucleus,whereas prokaryotes do not.Discuss the impact of having a nucleus on the creation of mature mRNA.
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Chapter 9: Proteins and Their Synthesis
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Q1) You found a mouse gene sequence containing an open reading frame that should produce a 43 kDa protein.You express this protein in cultured mouse cells by transforming the cells with a plasmid bearing your gene of interest.When you isolate proteins from these cells and probe for your protein of interest,you find that the detected protein is 49 kDa.Why might your protein be larger than expected when analyzed by immunoblot?
Q2) Many antibiotics target prokaryotic ribosomal function,blocking translation and causing rapid death in susceptible microbial populations.You have been assigned to a structure-based drug design project for a biotechnology company.In order for an antibiotic to be effective,it must be able to block an essential ribosomal function,and the drug must have access to the portion of the enzyme or enzyme complex being targeted.What processes/enzymes might you target in your drug design?
Q3) 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.
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Page 11
Chapter 10: Gene Isolation and Manipulation
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Sample Questions
Q1) Suppose you cut two different DNAs,
\[\begin{array} { l l }
\text { one wih BamH1: } & \text { 5'-GGATCC-3' } \\ & \text { 3'-CCTAGG-5' } \\\\
\text { and one with BglII: } & \text { 5'-A GATCT-3' } \\ & \text { 3'-TCTAG A-5' } \end{array}\]
then ligate them together through their compatible sticky ends.Once joined,could you separate these two DNAs again with either restriction enzyme? Why or why not? (Each of these enzymes is cut between the first two nucleotides,at the 5? end of each strand of the palindrome.)
Q2) 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.
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Page 12
Chapter 11: Regulation of Gene Expression in Bacteria and Their Viruses
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Sample Questions
Q1) Which of the following is an example of a cis-acting element?
A)\(\beta\)-galactosidase
B)operator site
C)lacI repressor protein
D)lactose
E)permease
Q2) The lac repressor (LacI)binds to:
A)lactose and DNA.
B)RNA polymerase.
C)RNA polymerase and DNA.
D)\(\beta\)-galactosidase,permease,and transacetylase.
E)RNA and DNA.
Q3) In the presence of abundant tryptophan:
A)the attenuator stem-loop structure forms,allowing transcription to continue.
B)the attenuator stem-loop structure forms,terminating transcription.
C)the preemptor stem-loop structure forms,allowing transcription to continue.
D)the preemptor stem-loop structure forms,terminating transcription.
E)None of the answer options are correct.
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Chapter 12: Regulation of Gene Expression in Eukaryotes
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Q1) How many molecules of histones H2A,H2B,H3,and H4 (respectively)comprise a single nucleosome?
A)one each(i.e., 1, 1, 1, 1)
B)two each(i.e., 2, 2, 2, 2)
C)three each(i.e., 3, 3, 3, 3)
D)1,2,2,2
E)2,2,2,1
Q2) The ________________ is a cis-acting element that binds RNA polymerase II.
A)promoter
B)enhancer
C)promoter-proximal element
D)GC-rich box
E)All of the answer options are correct.
Q3) The enzyme responsible for adding acetyl groups to histone proteins is called:
A)histone acetylase (HA).
B)histone deacetylase (HDAC).
C)histone acetylmethyltransferase (HAT).
D)histone acetyltransferase (HAT).
E)histone methyltransferase (HMT).
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Page 14
Chapter 13: The Genetic Control of Development
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Sample Questions
Q1) The homeodomain encodes:
A)a helix-turn-helix motif.
B)an ATP-binding domain.
C)a leucine zipper.
D)a zinc-finger.
E)a protease.
Q2) The class of genes that affect patterning within each segment is the:
A)maternal-effect genes.
B)gap genes.
C)pair-rule genes.
D)segment polarity genes.
E)Hox genes.
Q3) The class of genes required that act at a double-segment periodicity is the: A)maternal-effect genes.
B)gap genes.
C)pair-rule genes.
D)segment polarity genes.
E)Hox genes.
Q4) What characteristics allowed Drosophila melanogaster (an insect)to emerge as the leading genetic model of animal development?
15
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Chapter 14: Genomes and Genomics
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Q1) 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.
Q2) 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.
E)None of the answer options are correct.
Q3) Describe the information content of a genome.
Q4) Please list three (3)approaches to reverse genetics.
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16
Chapter 15: The Dynamic Genome: Transposable Elements
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Q1) 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.
Q2) 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.
Q3) a)What is the C-value paradox?
b)What is the explanation for the C-value paradox?
Q4) Compare and contrast autonomous and nonautonomous transposable elements.
Q5) Compare and contrast IS elements,simple transposons,and composite transposons.Use diagrams to help in the comparison.
Page 17
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Chapter 16: Mutation, Repair, and Recombination
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Q1) 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)None of the answers are correct.
Q2) How do mutation and DNA damage differ?
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Page 18
Chapter 17: Large-Scale Chromosomal Changes
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Q1) A pure-breeding flowering plant with a long stem and small leaves was crossed to a pure-breeding individual of the same species that has a short stem and large leaves.The F<sub>1</sub> is comprised entirely of individuals with long stems and large leaves.When these F<sub>1</sub> individuals are testcrossed,they only produced about half as many individuals as expected.The phenotypic ratios were: 462 short stems and large leaves,450 long stems and small leaves,52 long stems and large leaves,and 50 short stems and small leaves. These 1022 plants were further crossed to a tester,and many of them displayed semi-sterility,in particular: of the 462 plants with short stems and large leaves,362 were semisterile. of the 450 plants with long stems and small leaves,90 were semisterile. of the 52 plants with long stems and large leaves,40 were semisterile. of the 50 plants with short stems and small leaves,10 were semisterile. Provide a plausible,comprehensive genetic explanation for these results.
Q2) You make meiotic chromosome preparations of two species of tetraploid plants.In one,there is normal pairing of chromosomes and tetrads (quadivalents)are observed,while in the other,there is abnormal pairing in some preparations and trivalents and univalents are seen.How would you account for the differences between these two species?
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Chapter 18: Population Genetics
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Q1) Tay-Sachs disease is inherited as an autosomal recessive.In a certain large eastern European population,the frequency of Tay-Sachs disease is 1 percent.
a)If the population is assumed to be in Hardy-Weinberg equilibrium with respect to Tay-Sachs,what is the frequency of the allele that causes Tay-Sachs?
b)What would be the frequency of heterozygotes?
c)What is the probability of two heterozygotes marrying?
d)In children of such marriages,what would be the frequency of Tay-Sachs disease?
e)What proportion of all Tay-Sachs births are produced by such marriages?
Q2) A recessive X-linked character appears in 40 percent of males and 16 percent of females in a randomly interbreeding population.Assume only two alleles are present.What are the allele frequencies? How many females are heterozygotes? How many males are heterozygotes?
Q3) Apert syndrome (acrocephalosyndactyly)results from a dominant mutant allele.Among 322,182 births to normal parents,two infants were found with this syndrome.What is the mutation rate per gamete for this genetic disease?
Q4) How is the genetic composition of a population defined?
Q5) What processes increase genetic variation in a given population?
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Page 20
Chapter 19: The Inheritance of Complex Traits
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Q1) Narrow-sense heritability (h<sup>2</sup>)is a quantification of the proportion of total variance due to:
A)additive genetic variance.
B)dominance variance.
C)environmental variance.
D)phenotypic variance.
E)total genetic variance.
Q2) Which of the following has more important value to the plant or animal breeder?
A)broad-sense heritability
B)variance
C)correlation
D)additive deviation
E)dominance deviation
Q3) The phenotypic deviation that is transmitted from parents to their offspring is known as:
A)additive genetic variance.
B)dominance variance.
C)environmental variance.
D)total genetic variance.
E)transmissible variance.
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Chapter 20: Evolution of Genes and Traits
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Q1) The melanocortin 1 receptor protein displays four amino acid substitutions between dark- and light-colored populations of pocket mice.In dark mice,these unique amino acids cause the receptor to:
A)fold inappropriately in the cell membrane.
B)incorrectly identify ligand.
C)be non-functional in regulating gene expression.
D)be internalized and degraded.
E)be constitutively active.
Q2) Some of the most striking and best understood examples of morphological divergence are found in:
A)populations of antibiotic resistant bacteria.
B)virology.
C)animal body-color patterns.
D)mouse eye-color mutants.
E)blood-type examples in mammals.
Q3) How will pleiotropy affect selection at particular loci?
Q4) Most genetic variation for many human loci lies within local populations rather than between populations or races.What does this observation tell you about human genetic evolution?
Q5) What is meant by the phrase,"The force of selection is myopic."?
Page 22
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