Genetics Final Exam

It is acidic.

It contains deoxyribose.

It is found in cell nuclei.

It contains phosphate.

It contains proteins.

RNA

DNA

protein

carbohydrate

none of the above

phosphodiesters

proteins

nucleotides

nucleosides

polymers

cell membrane

endoplasmic reticulum

vacuoles

chromosomes

None of the above

a deoxyribose sugar

a nitrogenous base

a phosphate

A and B only

All of the above

5' → 3'

3' → 5'

5' → 2'

2' → 5'

none of the above

protein

DNA

RNA

rRNA

none of the above

guanine to adenine

adenine to thymine

cytosine to adenine

uracil to cytosine

none of the above

2.0 angstroms.

3.4 angstroms.

20 angstroms.

34 angstroms.

none of the above

2.0 angstroms.

3.4 angstroms.

20 angstroms.

34 angstroms.

none of the above

Ionic

Covalent

Hydrogen

Electro-ionic

none of the above

A

B

D

Y

Z

thymine

uracil

adenine

cytosine

guanosine

conservative

semiconervative

dispersive

transferal

none of the above

Hydrogen

Ionic

Phosphodiester

Electrostatic

None of the above

a single continuous linear double helix.

a double helix replicated semiconservatively.

a double helix replicated conservatively.

single-stranded DNA.

None of the above

a single continuous linear double helix.

a double helix replicated semiconservatively.

a double helix replicated conservatively.

a triple helix replicated semiconservatively.

None of the above

Initiation

Elongation

Termination

Translation

Translocation

Initiation

Elongation

Termination

Translation

Translocation

Intiation

Elongation

Termination

Translation

Translocation

Primases

Helicases

Topoisomerases

Telomeres

Ligases

one

two

three

four

none of the above

Primase

Helicase

Topoisomerase

Telomerase

Ligase

no

one

two

several

many

methylases

capping proteins

ligases

telomerases

single-stranded binding proteins

redundancy

repair enzymes

precision of replication machinery

DNA polymerase proofreading mechanism

Restriction endonucleases

Translation

Transcription

Recombination

Transformation

None of the above

Allelic exchange

Gene conversion

Crossing over

Recombination

DNA replication

homologous nonsister chromatids

homologous sister chromatids

heterologous nonsister chromatids

heterologous sister chromatids

None of the above

instruction from normal cell-cycle program

X-rays

Chemical change

Physical change

Radiation

nitrogen

carbon

sulfur

phosphorous

iodide

15%

20%

30%

35%

none of the above

5' AATTCGCTTA 3'

3' AATTCGCTTA 5'

5' TAACGCTTAA 3'

5' TAAGCGAATT 3'

3' TAAGCGAATT 5'

abc, aBc, AbC, aBC

abc, abc, ABC, ABC

aBc, aBc, AbC, AbC

abC, abc, ABc, ABC

Abc, Abc, aBC, aBC

anaphase

interphase

prophase

metaphase

none of the above

homoduplex

heteroduplex

homotriplex

heterotriplex

none of the above

chromosome length will gradually increase

chromosome length will gradually decrease

chromosome length will stay constant

chromosomes will fail to dissociate after replication

none of the above

Methylases

Recombinases

Topoisomerases

Ligases

Primases

radioactivity

gel filtration

ion exchange

centrifugation

none of the above

radiolabeled phosphate

calcium chloride

radiolabeled nitrogen

sodium acetate

cesium chloride

5%

10%

25%

40%

45%

15 map units

35 map units

Either 15 map units or 35 map units, depending on the order of the genes

The map distance between A and C can not be predicted from this data

12 m.u.

13 m.u.

25 m.u.

50 m.u.

75 m.u.

10%

25%

50%

75%

100%

ABCD

ADBC

ABDC

BADC

CADB

spindle fiber

synaptic junction

synaptomnemal complex

chiasma

none of the above

D and E are on different pairs of chromosomes.

D and E are linked and exactly 50 map units apart.

D and E are linked and at least 50 map units apart.

either A or B

either A or C

The crossover would have to occur between the A locus and the centromere and involve two homologous (non-sister) chromatids.

The crossover would have to occur between the A locus and the end of the chromosome and involve two homologous (non-sister) chromatids.

The crossover would have to occur on the same chromosome arm as the A locus and involve two homologous (non-sister) chromatids.

The crossover would have to occur on the same chromosome as the A locus and involve two homologous (non-sister) chromatids.

The crossover would have to occur between the A locus and the centromere and involve two sister chromatids (not homologous) chromatids.

The crossover would have to occur between the A locus and the centromere locus and involve two homologous (non-sister) chromatids.

The crossover would have to occur between the A locus and the B locus locus and involve two homologous (non-sister) chromatids.

The crossover would have to occur between the B locus and the end of the chromosome locus and involve two homologous (non-sister) chromatids.

A double crossover would have to occur, with one crossover between the A locus and the centromere and a second crossover between the A and B loci locus and both crossovers would have to involve two homologous (non-sister) chromatids.

No crossover in a somatic cell could generate a twin spot.

Two homologous chromatids break and rejoin at random sites along the chromosome.

The genetic information on one chromatid is replaced by copying genetic information from a homologous chromatid without there being any physical exchange between the chromosomes.

Two homologous chromatids break and rejoin at precisely the same site along the chromosome so that there is no loss or gain of material on either product.

It is not known what occurs during crossing over.

ACDBE

AECDB

EACDB

EADCB

CDABE

Chiasmata are randomly distributed along chromosomes.

All chromosome pairs have the same number of chiasmata.

A single chromosome pair always has the same number of chiasmata in every meiotic cell.

Chiasmata are spaced along a chromosome arm more regularly than would be expected by chance.

Chiasmata are spaced more irregularly along a chromosome arm than would be expected by chance.

If the allele frequencies in a population are given by p and q, the genotype frequencies are given by p2, 2 pq and q2.

The allele frequencies in a population will not change over time.

If the allele frequencies in a population are given by p and q, the genotype frequencies are given by p2 and q2. d. The first and third answers are correct.

The first and second choices are correct.

In populations with high selection pressure, the allele frequency is low.

the allele is recessive and deleterious.

The first and second choices are correct.

in populations with a high frequency of the allele, selection pressure is relatively low.

decrease

increase

maintain

Allele frequencies will tend to move towards fixation or loss.

Allele frequencies will tend to move toward a stable equilibrium.

Allele frequencies may initially hover at an unstable equilibrium, but will eventually change.

the frequency of the allele will stay high for a long time, then decline slowly.

the allele declines in frequency until it is eliminated.

the frequency of the allele will be unchanged;

the population will remain in Hardy-Weinberg equilibrium. the frequency of the allele will decline rapidly, and then stabilize at very low frequency.

both alleles are maintained at a frequency different from that predicted by Hardy-Weinberg principles based on the strength of selection against the recessive allele.

the selective advantage enjoyed by the deleterious allele in the heterozygote exactly

genetic diversity in the population can be maintained indefinitely, in spite of selection acting against the allele. balances the selective disadvantage suffered by homozygous recessive individuals.

The first two choices are both correct.

The first three choices are correct.

genetic variation among populations will decline.

the most common outcome will be that the most common allele will become fixed (will increase to a frequency of 1) in the population.

both alleles are maintained in the population at frequencies different from those predicted by Hardy-Weinberg principles.

the genetic variation within a population is maintained over time

The mutation rate for new loss-of-function mutations in the CFTR gene is too low for the prevalence of the disease to be explained by mutation/selection balance.

Pier and his colleagues have found, in 11 European countries, an association between the severity of typhoid fever outbreaks and the frequency of the delta-F508 allele (the most common loss-of-function mutation) a generation later.

Pier et al. engineered cells homozygous for functional CFTR alleles, homozygous for a common loss-of-function allele, and heterozygous for the two. The loss-of-function homozygotes were virtually impervious to invasion by typhoid fever-causing bacteria; heterozygotes were more vulnerable, but accumulated 86% fewer bacteria than did the dominant homozygotes.

All of the choices above are correct.

Salmonella typhi bacteria manipulate their host cells, causing them to express more CFTR protein on their membranes.