Genetics 301 Final Examination
Spring 2003
May 9, 2003

Version A
120 points possible

30 Multiple Choice Questions (Choose the best answer)

1. What would be the frequency of AABbCc individuals from a mating of two AaBbCc individuals?

    1. 1/64
    2. 1/32
    3. 1/16
    4. 1/8
    5. 3/16


2. The stage of meiosis in which sister chromatids separate from each other is:

    1. prophase I
    2. metaphase I
    3. prophase II
    4. metaphase II
    5. anaphase II


3. Which of the following components is present in a mature eukaryotic messenger RNA (ready to be transcribed) but not in a prokaryotic RNA?

    1. transfer RNA
    2. 3’ Poly A tail
    3. mitochondrial DNA
    4. stop codon
    5. codons coding for the protein to be produced


4. A lysogenic strain of E. coli contains:

    1. a vector of yeast or bacterial origin which is used to make many copies of a particular DNA sequence
    2. a bacterial chromosome with a human gene inserted
    3. a bacterial chromosome with the F factor inserted
    4. a human chromosome with a transposable element inserted
    5. a bacterial chromosome with a phage inserted


5.Replication of DNA in eukaryotes:

    1. takes place in a "conservative" manner
    2. takes place in a "dispersive" manner
    3. does not involve the production of "Okazaki fragments"
    4. involves more than one origin of replication per chromosome
    5. takes place only in the 3’ to 5’ direction


6.. Enhancer sequences

    1. are present in the genome of many animal and plant species
    2. are found in prokaryotes but not in eukaryotes
    3. were identified as the integration sites for bacterial viruses
    4. represent integration sites for transposable elements
    5. represent the termination signals for transcription


7.Apoptosis refers to:

    1. the cells giving rise to and derived from a specific cell are known
    2. a process under which some cells die as a normal step in development
    3. the regulatory genes for the organism have been genetically mapped
    4. cell components in the membrane involved in signaling have been isolated
    5. cell components in the nucleus involved in signaling have been isolated


8. Zinc finger proteins and helix-turn-helix proteins are:

    1. types of DNA-binding proteins
    2. involved in the control of translation
    3. components of ribosomes
    4. part of the hemoglobin in blood cells
    5. bound to transfer RNA during replication


9.Gene regulation in prokaryotes:

    1. is often influenced by hormones
    2. is usually regulated by tRNA binding to promoter sites
    3. often involves control by differential splicing of mRNA
    4. often involves control of the transciption of multiple genes on a single messenger RNA
    5. often involves control by varying the number of gene copies


10. Zebrafish are advantageous for developmental genetic research because:

    1. it only has a single cell and can be used to study the cell cycle
    2. they are a vertebrate species which is easily to raise and in which development can be easily visualized outside the female
    3. the entire cell lineage is known and can be followed
    4. it is easily reared and studied for a mammal
    5. their genome size is even larger than that of humans


11.A homeotic mutation is one which:

    1. is present in only one form in an individual
    2. substitutes one body part for another in development
    3. results in development of a tumor
    4. is wild type at one temperature and abnormal at another
    5. leads to increased body size in an organism


12. Assuming that the level of glucose is low, a mutation in the operator of the lac operon in E. coli, preventing binding of the repressor to the operator, should result in:

    1. constitutive expression of the lac operon genes only on the same chromosome (in cis)
    2. constitutive expression of the lac operon genes on both the same chromosome (in cis) and lac operons on other chromosomes (in trans).
    3. lack of expression or reduced expression of the lac operon genes under all circumstances
    4. expression of the genes only when lactose is present
    5. expression of the genes only when lactose is absent


13.An "open reading frame" refers to::

    1. a segment of DNA lacking promoter sequences ("TATA" box)
    2. a segment of DNA lacking repetitive elements
    3. a segment of DNA containing no intron splice signals
    4. a segment of DNA in one of three possible frames in which no stop codons are evident


14. The study of allozymes

    1. uses hybridization to detect specific DNA restriction fragments in genomic DNA
    2. is used to determine whether a gene is transcribed in specific cells
    3. measures the transfer frequency of genes during conjugation
    4. is used to detect genetic variation at the protein level.
    5. is used to amplify genes for producing useful products


15.Artificial chromosome vectors for cloning

    1. can generally accommodate larger inserts than phage vectors can
    2. can only be propagated in mammalian cells
    3. replicate in E. coli and have maximum insert sizes of about 40 kb
    4. are most useful for studies of cDNA rather than genomic DNA
    5. burst bacteria and form plaques on a "lawn" of bacteria


16.In calculating the likelihood of a match for DNA markers in criminal and paternity cases:

    1. highly uniform markers with low variability among individuals are most often used
    2. variations in the amino-acid-coding regions of genes are most often compared
    3. the individual’s racial/ ethnic group is not considered in the calculation
    4. the likelihood of a match within the person’s ethnic/racial group for each marker is considered
    5. a single marker is usually sufficient to draw a high-probability conclusion

17.The polymerase chain reaction or PCR is a technique that

    1. was used to demonstrate DNA as the genetic material
    2. measures whether DNA is composed of all old, half old, or all new strands
    3. uses short DNA primers and a thermostable DNA polymerase to replicate many copies of a specific DNA sequence in the laboratory
    4. measures the ribosome transfer rate during translation
    5. detects the concentration of polymerases involved in replication


18. Digesting DNA with a restriction enzyme that recognizes a 4 bp sequence will cut the DNA on average into pieces of:

    1. 16 bp long
    2. 64 bp
    3. 256 bp
    4. 1,024 bp
    5. 4,096 bp


19. "Blue-white selection" of plasmids allows:

    1. selection for plasmids which carry a repetitive DNA element
    2. easy identification of plasmids which carry an insert
    3. replication of transfer RNA within the bacterial cell
    4. insertion of centromeres into ribosomes lacking them
    5. pieces of DNA from different sources to hybridize to each other and to be joined together


20. QTL analysis is performed by:

    1. associating traits with the inheritance of a large number of genetic markers in a cross.
    2. digesting DNA with several restriction enzymes and identifying regions which are not cut by any enzyme
    3. determining which genes are expressed at a developmental stage
    4. determining the base composition (%GC) of a DNA sample
    5. determining the most rapidly-evolving parts of genes


21.Assuming Hardy-Weinberg equilibrium, the genotype frequency of heterozygotes, if the frequency of the two alleles at the gene being studied are 0.8 and 0.2, will be:

    1. .80
    2. 0.64
    3. 0.48
    4. 0.32
    5. 0.16


22. A circular piece of double-stranded DNA is initially cut with a restriction enzyme, yielding a single fragment 12 kb long. When this single linear fragment is then cut with a second restriction enzyme in two locations, the center and 1 kb from one end, how many DNA fragments would be detectable using gel electrophoresis?

    1. 3 fragments: 6, 5 and 1 kb
    2. 3 indistinguishable fragments each 4 kb long
    3. 2 fragments, 5 and 1 kb long
    4. 2 fragments: 5.5 and 1 kb long
    5. 4 fragments: 1, 2 3 and 6 kb.


23. Mitochondrial DNA studies in humans and Neanderthal samples led to the conclusion that :

    1. it is more variable in females than in males
    2. Neanderthals are a distinct lineage such that all humans are closer to each other than any are to Neanderthals.
    3. African populations have less mitochondrial sequence variation than do other human groups
    4. DNA samples older than 10,000 years will never be successfully sequenced.
    5. nuclear genes usually evolve faster than mitochondrial genes do


24. A broad-sense heritability of .85 (85%) would mean that :

    1. only a few genes were involved with determining the trait value
    2. many genes are involved with determining the trait value
    3. most of the variation in the trait is environmentally influenced
    4. the total variance in the trait is large
    5. most of the variation in the trait is genetically influenced


25. If the narrow-sense heritability of a trait is 0.4, the population mean is 40, and the mean of the selected parents is 50, the expected mean of the progeny will be:

    1. 40
    2. 42
    3. 44
    4. 46
    5. 48


26. Which of the following is one of the assumptions of the Hardy-Weinberg principle?

    1. random mating
    2. superior fitness of heterozygotes
    3. significant migration of individuals
    4. high mutation rate at a specific gene
    5. genetic drift


27. If the selection coefficient against two homozygotes (AA and A’A’) are 0.3 and 0.7, respectively, and they are the only two alleles present in the population, the equilibrium frequencies of the A and A’ alleles will be:

    1. 0.7 and 0.3
    2. 0.6 and 0.4
    3. 0.5 and 0.5
    4. 0.4 and 0.6
    5. 0.3 and 0.7


28. Heterozygote advantage refers to::

    1. genetic crosses being most useful when performed in the heterozygote
    2. the fact that it is easier to detect heterozygotes than homozygotes with molecular tests
    3. higher linkage being found in heterozygotes
    4. higher inversion rates being found in heterozygotes
    5. heterozygotes being of superior genetic fitness in a population


29.If the frequency of males affected with an X-linked recessive condition in a human population is .01 (one in a hundred), what will be the expected frequency of affected females?

    1. .0001
    2. .001
    3. .02
    4. .01
    5. .05


30. The following genotypes are found in a population:

AA 55

Aa 30

aa 15

What are the allele frequencies of A and a?

    1. A = 0.85 and a = 0.15
    2. A = 0.75 and a = 0.25
    3. A = 0.70 and a = 0.30
    4. A = 0.60 and a = 0.40
    5. A = 0.55 and a = 0.45