Multiple Choice · Web viewMultiple Choice Why does a cloning vector need an origin? To provide a promoter site for transcribing the cloned gene. To facilitate conjugation. To allow

FINAL EXAM REVIEWSupplemental InstructionIowa State University

Leader: KalycaCourse: BIOL 313

Instructor: Myers/ VollbrechtDate: 04/26/2017

Multiple Choice

1. Why does a cloning vector need an origin?a. To provide a promoter site for transcribing the cloned gene.b. To facilitate conjugation.c. To allow the vector to be replicated within the bacterium.d. To promote excision by a virus to allow transduction to occur.e. None of the above.

When we transform, only a handful of cells will take up the plasmid vector. Because of this, we need an origin of replication where the bacterial replication machinery can access the plasmid and make more to distribute some into daughter cells.

2. What can SNPs help determine?a. The genetic basis of a disease.b. The relatedness of two individuals.c. An evolutionary phylogenetic tree.d. Chromosome mapping.e. Multiple of the above are true.

All of these are uses of SNPs, in fact.

3. Given the following data, determine the variance in clutch size.Nest 1 Nest 2 Nest 3 Nest 4 Nest 5 Nest 6

Clutch Size 7 2 2 4 5 4

a. 1.5b. 1.8 c. 3.0d. 3.6e. None of these.

Solve for the mean first: (7+2+2+4+5+4)/6 = 4.

Then use the variance formula:

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4. Which of the following could be the sequence on one strand of dsDNA at a restriction site?

a.b. ACTAGTc. CGTTGCd. AATAAA

e. GGGGGGf. Multiple of these.

Restriction sites are palindromic sequences. A palindromic sequence is one that is the same 5’ to 3’ on the parallel and anti-parallel strand.

5’-ACTAGT-3’3’-TGATCA-5’

If we flip the bottom around so we can read it 5’ to 3’, we can see that it’s the same.

5. In a population of 200, there are 32 individuals with genotype TT for gene locus T. Which of the following genotype counts could exist at Hardy-Weinberg equilibrium in this example?

a. 32 ttb. 96 Ttc. 64 Tt

d. 72 tte. Multiple of these.

Since we can treat this as being in HW equilibrium, we can calculate the proportion of T – we’ll refer to this as p – just from the number of homozygous TT individuals; we don’t need the number of heterozygotes. To calculate the value of p, recognize that the

proportion of homozygotes is equal to p2. So, if p2= 32200

=0.16, then p=0.4. Since

1=p+q, we can solve for q, the proportion of allele t. It follows that q=0.6. Now to solve for the number of heterozygotes and tt homozygotes, recognize that the proportion of heterozygotes is 2pq and tt homozygotes is q2. So: 2 (0.4 ) (0.6 )=0.48 and (0.6 ) (0.6 )=0.36. To determine the expected amount in the population, multiply the total population by these expected proportions: (0.48)(200) = 96 Tt and (0.36)(200) = 72 tt.

6. Approximate the narrow-sense heritability of trait W given the following.VI VG VD

Variance 62 224 106

a. Less than 0.26b. 0.28c. 0.47

d. Greater than 0.75e. Not enough information is

given.

This is a terribly cruel question, but it involves all components of variance. First, to solve for narrow-sense heritability we need VA. Since VG=VA+VD+VI, VA=VG-VD-VI = 224 –

106 – 62 = 56. Now, narrow-sense heritability is defined as h2=V A

V P, but we don’t have

VP. VP is defined as VP = VG+VE, but we also don’t have VE. Well let’s plug in what we

do have: h2=V A

V P=

V A

V G+V E= 56

224+V E. Now, VE is 0 or greater. Narrow-sense heritability

will be at a maximum when VE is 0 and will be smaller at higher values of VE. So if VE is

0 then h2= 56224+0

=0.25. So narrow-sense heritability will be, at most, 0.25, but will

probably be less.

7. Beginning with two inbred flies with extreme wing dysmorphia, a cross was made producing F1 progeny with intermediate average wing length. Specifically, the original

parent flies had wing lengths of 0.6 mm and 2.0 mm, respectively, and the F1 flies had wing lengths of 1.3 mm. In the ensuing F1 cross producing the F2 generation, 300 individuals grew to adulthood and exhibited wing lengths varying from 0.7 to 1.9 mm. How many genes would you expect to be involved in controlling this trait?

a. Less than 3b. 3 or morec. 4 or less

d. More than 4e. This can’t be determined.

This is a tough way to do this question, but it’s also on the practice test and I decided it was worth having the method to do this explained somewhere. So to start out, recognize that the phenotypes of the two parents are the homozygous extremes. One has all the addition alleles and the other is missing all of them. This is what is implied by the “inbred” description. So we know what our homozygotes look like and we know from previous material that the cross of two homozygotes like these will result in full heterozygotes, which is supported by the question.

However, when we look at our F2 offspring, we don’t see any homozygotes. We will infer from this that the population is too small to expect to see homozygotes; we infer that the expected proportion of the population of homozygotes is so small that we would expect to see less than 1 homozygote individual in a population of 300 as given in the

question. That proportion, 1 out of 300 is equal to 1

300=0.0033. Now we have a formula

we can use to determine the expected number of genes. That formula is ( 14 )

n

=r where n

is the number of genes and r is the ratio of one of the homozygotes in the population – what we just calculated. If we plug that number in and use logarithms, we find that n = 4.11. So 4 is too few genes to expect to not see a homozygote in this population. So more than 4 is the answer.

You can practice this question on the 2016 practice exam.

8. Which of these requires a model predicted from recombination rates to use?a. Illumina sequencing.b. Pyrosequencing.c. Shotgun sequencing.d. Map-based sequencing.e. Multiple of the above.

This is the defining characteristic of map-based sequencing. You do end up getting a true DNA sequence in base pairs, but you match the base pair sequence with a gene map from recombination rates to get the full map.

9. When performing a test with a microarray, radioactively-labelled NTPs are added to a culture broth of growing cells. The cells are subsequently lysed and the lysate is washed over the microarray and radioactivity is measured. What kind of data are you collecting?

a. Transcriptomic.b. Genomic.c. Metabolomic.

d. Proteomic.e. Multiple of these.

NTP is how we abbreviate ribonucleotides. So NTPs are used in making RNAs and mRNAs specifically for this question. Since the microarray will be determining the amount of mRNA present, it is a measure of the trascriptome.

10. You perform a PCR reaction and sequence the results afterword. Unfortunately, the sequence exhibits several point mutations in comparison to the template DNA. What is the likely explanation for this?

a. Your template didn’t melt fully during the denaturation phase.b. Your primers didn’t anneal properly resulting in nonspecific binding by the

polymerase.c. Your polymerase has low fidelity (accuracy).d. You accidentally introduced a DNA contaminant to the PCR mix that was also

cloned.e. Multiple of the above are equally likely.

Point mutations indicate a problem with our elongation phase, where we use our polymerase. While it may be difficult to recognize that, it should be recognized that issues in denaturation or annealing would lead to limited to no polymerization. As a take-away, make sure you know what each stage in PCR does; why we do each step.

11. You are studying the heritability of dorsal fin height in salmon. You select tall-finned individuals from a population with mean fin height of 7 cm. The selected group has a mean fin height of 10 cm and after breeding, the offspring have a mean fin height of 8 cm. What can you tell given this information?

a. Narrow-sense heritability is 0.3b. Broad-sense heritability is 0.3c. Narrow-sense heritability is 0.7d. Broad-sense heritability is 0.7e. Multiple of the above

Here, we’re using the formula R = h2S where S is the selective pressure, R is the response to that pressure and h2 is the narrow-sense heritability. We get S from the difference between the selected group mean trait value and the mean trait value of the general population. We get S from the difference between the mean offspring trait value and the mean trait value of the general population. So for this question:

h2=RS= 8−7

10−7=1

3=0.33

12. Probe DNAs are small DNA fragments that are either radioactively- or fluorescently-labeled. What can we do with probe DNA?

a. Identify cells that have taken up a plasmid vector.b. Measure transcription levels.c. Determine the presence of a desired genomic DNA sequence.d. Test for specific homologies between species.e. Multiple of the above could be done with probe DNA.

In fact, all of these are possible uses of probe DNA. Though, the most common use and the use most likely to show up on a test is answer c.

13. What is useful about dideoxynucleotides?a. They stop DNA polymerization.

b. They release pyrophosphate.c. They can form atypical base-pairs.d. They can self-cyclize, promoting mRNA degradation.e. None of the above are helpful uses of dideoxynucleotides.

These are ddNTPs. We use these in Sanger sequencing to randomly stop polymerization. That polymerization limiting creates fragments of different sizes and we can use these different sizes to determine sequence.

14. You have two strains of orchid. One is an inbred line that has variance in height of 5.1 cm. The other is a wild-bred line that has variance in height of 7.3 cm. What is the most influential determiner of orchid height given this information?

a. Additive genetic variance since the wild-bred line has higher variance than the inbred line.

b. Genetic variance in general. We can’t isolate the increased wild-bred variance to just additive effects.

c. Phenotypic variance since that is what is actually measured above.d. Environmental variance since the difference between the inbred and wild-

bred variances are less than the total variance of the inbred.e. This can’t be determined given the above.

Since the inbred line is inbred, it has no genetic variance. So all of the variance in that line is due to environmental variance. The wild-bred line gives us a more typical phenotypic variance – one that includes genetic variance. However, this line experiences the same environmental variance as the inbred line, so genetic variance only accounts for about 2.2 cm of the total 7.3 cm variance. This means that environmental variance at 5.1 cm accounts for the majority of the variance in plant height.

15. In a line of lab mice, the trait for curled or normal ears is governed by a single trait with simple dominant/recessive interactions between the two alleles. However, 5 other gene products have an influence on the ear-curl gene, affecting the degree of ear-curling on a range from total epistatic shielding of the ear-curl allele down to no influence on ear curl. These affects are observed in a wide range of environmental conditions. Which of the following likely has the highest value when discussing ear-curl variance?

a. VA

b. VD

c. VI

d. VE

e. This can’t be determined.

The situation described will have next to no additive variance since the trait isn’t governed by an additive allele. It will have some dominance genetic variance, but much more of the influence on this trait is apparently due to gene-gene interactions (the 5 genes affecting the ear-curl), so VI is quite high. Additionally, it’s clarified that the environment doesn’t appear to have much of an effect over these interactions, so VE shouldn’t be too high.

16. When cloning a gene of interest using a plasmid vector, what step follows restriction digestion?

a. PCRb. Ligationc. Transformationd. Transduction

e. Transposition

We’ll start by cloning the gene with PCR. Then we digest it with restriction enzymes to make sticky ends that will match the plasmid. Ligation follows to bind the gene fragment into the plasmid vector. Finally, we transform the plasmid vector into the competent cells.

17. You perform a microarray test similar to what we discussed in class. Red-fluorescing NTPs were given to cancerous cells for growth while green-fluorescing NTPs were given to normal cells for growth. When the mRNAs from these cells are purified and washed over the microarray, what is indicated by a microarray point that appears blank?

a. Both cells transcribed the gene at that point.b. Neither cell transcribed the gene at that point.c. The fluorescent tag got in the way of the base-pairing.d. It indicates the absence of the gene in both cell types.e. The microarray will never have a blank point on it.

A red spot indicates a gene that is mostly transcribed in a cancerous cell. A green spot indicates a gene that is mostly transcribed in a noncancerous cell. A yellow spot is intermediate and indicates transcription in both types of cells. A blank spot indicates no transcription in either cell type.

18. You are working with a breed of frogs that emit loud, prolonged calls during mating season. To see if the trait is heritable or not, you select a group of long-calling frogs from the general population and breed them. The original average call duration was 150 seconds. The mean call duration of the selected group was 200 seconds. If the mean offspring call duration is 190 seconds, what is the narrow-sense heritability and how heritable does this value make the trait?

a. 0.75; highly heritable.b. 0.8; highly heritable.c. 0.75; poorly heritable.

d. 0.95; highly heritable.e. None of the above.

Like before: h2=RS=190−150

200−150=40

50=0.8. Heritabilities over 0.5 are mostly governed by

additive heritability. With heritability below 0.5, environmental or other effects are more important than additive effects.

19. When attempting to overexpress a gene in E. coli, you add in a high-copy vector that leads to many more mRNA transcripts being produced, however; the change in phenotype doesn’t match the change in transcription levels. What explains this best?

a. The cell recognizes that too much of that transcript is being produced and degrades the excess.

b. Your measurement of transcript levels must be incorrect.c. You targeted the wrong gene.d. Transcription levels don’t perfectly correlate to expression levels.e. Your cells probably died.

That is the important result to take away from this. We try to use the transcriptome to approximate the proteome, but it isn’t perfect. The amount of transcript doesn’t match the observed expression levels fairly often because there are so many levels of regulation between the two.

20. Which of the following is an assumption made about populations in Hardy-Weinberg equilibrium?

a. The population is very large.b. Mating occurs randomly.c. There is no selection acting on the population.d. Only two of the above are true.e. All three are true.

Memorize these. Not much else to do

These next two questions will be discussed more on Friday, but we’ll give a brief introduction to them now.

21. An extended drought over the course of many years causes the nutty seeds released by an island’s trees to become harder and more difficult to crack. As a result, the island’s single bird species experiences selective pressure preferring stronger beaks that can crack the seed coat. Due to this, strong-beaked birds flourish on the island, though they are still able to breed with their weaker-beaked cohabitants. What has occurred here?

a. Evolutionb. Speciationc. Genetic driftd. Mutatione. None of the above

Prolonged changes in allele proportions in a population is the definition of evolutionary change we will use.

22. A small band of birds colonize an island whose previous bird population was killed by a storm. While the original birds had a mix of brown and black coats, only brown-coated birds remained after a few generations. It appears that there wasn’t any selective pressure one way or the other, so what happened here?

a. The founder effectb. Mutationc. Genetic driftd. Gene flowe. None of the above

The founder effect is when a small group colonizes a new area and the new colonial population’s gene frequencies will reflect that original colonizing group more so than the population they came from. However, genetic drift is the issue present here. Genetic drift occurs most significantly in small populations. Genetic drift is where random chance causes a change in allelic frequency, regardless of selective pressures. Genetic drift often pushes small populations towards homozygosity and the loss of alleles.

Introduction: The material on this exam is from exams 1 through 4.

Multiple Choice

23. Which of these is an epigenetic modification?a. Cytosine methylationb. Histone lysine acetylationc. Moving histones along the DNA

d. Histone lysine methylatione. All of the above

All of these are examples of epigenetic modifications. An epigenetic modification is any change that affects expression levels that doesn’t arise from a change to the actual bases of the DNA.

24. Which of these mutations is most likely to produce an inactive protein?

a. Silentb. Suppressorc. Missense

d. Nonsensee. Multiple of the above are

equally likely.

Nonsense mutations usually cause the most significant changes to the protein because a stop codon is prematurely introduced. Silent mutations produce no change and suppressor mutations undo previous deleterious mutations. Missense mutations can decrease protein activity, but are less likely to do so then with nonsense mutations (missense mutations just change one amino acid to another).

25. Given below is a table of recombination rates. Which of the following is a possible linkage group?

Genes Rate Genes Rate Genes RateA and B 49% B and C 48% C and E 50%A and C 13% B and D 50% D and E 49%A and D 7% B and E 12%A and E 50% C and D 6%a. A, C, and Eb. A, B, and Cc. B and E

d. C and Ee. None of the above

Unlinked genes will have recombination rates close to 50% meaning they appear to assort independently. Linked genes will have much lower recombination rates. The two linkage groups in this question include A, C, and D as one and B and E as the other.

26. What is the mode of inheritance demonstrated in this pedigree?

a. Autosomal dominantb. Autosomal recessivec. X-linked dominant

d. X-linked recessivee. Y-linked

It clearly isn’t dominant since the parents aren’t affected, but can pass on the trait. Since females are affected, it isn’t Y linked. So now we know it’s either X-linked recessive or autosomal recessive. If the gene is X-linked recessive, then we know that the original father can’t be a carrier, since he would otherwise be affected, so the mother would have to be a carrier. However, none of their daughters could be affected because the father can’t pass on any affected alleles. Only the sons would be affected in the F1 if it was X-linked. Can it be autosomal recessive? Yes. But it requires both of the parents to be heterozygous carriers which is possible.

27. From the above pedigree, what is the probability that the child of IV1 and IV3 would have the trait of interest?

a. 2/3b. 1/2c. 1/12d. 1/16e. 1

IV 1 has a ½ chance being a heterozygote while IV3 has a 2/3 chance of being a heterozygote. For the most part, these two have very similar pedigree backgrounds (grandmothers with the trait). The difference, however, comes from the fact that IV3 has brother with the trait and so this clearly tells us both parents are heterozygotes. For IV1 we don’t know if the parent from outside the family is heterozygous or not give the family information. Since IV3 has a brother then that means only 2 of the 3 genotypes left from the two heterozygote cross is heterozygous. That is to find the probability that those two are heterozygotes but now we have to consider what’s the probability of two heterozygotes having an affected child and in this case we just have to do a normal Punnett square between heterozygotes ---- the probability of an affected child turns out to be ¼. To find the overall probability of this event we just multiply everything together: Probability of child having trait = (1/2)(2/3)(1/4) = 1/12

28. During which phase of mitosis does the nuclear envelope break down?a. Prophaseb. Metaphasec. Anaphase

d. Telophasee. Interphase

Prometaphase is the specific phase where this occurs, but some biologists don’t recognize a separate prometaphase and just combine prometaphase into prophase.

29. How does a eukaryote initiate translation?a. The ribosome recognizes the Shine-Dalgarno sequence.b. The ribosome recognizes a Shine-Dalgarno-like sequence.c. The ribosome recognizes several initiation consensus sequences at the same time.d. The ribosome scans for any start codon.e. None of the above

Eukaryotes scan for a start codon. Prokaryotes recognize the Shine-Dalgarno sequence.

30. You cross two purple-flowered plants and obtain the following offspring counts. You believe that petal color is governed by one gene locus with two, codominant alleles. As such, you expect a 1:2:1 ratio in the offspring. Perform a chi-squared test to see if the data supports this conclusion.

Blue Flowers Red Flowers Purple Flowers101 88 211

a. Since p < 0.05, reject the null hypothesis; the observed ratio differs from 1:2:1b. Since p > 0.05, reject the null hypothesis; the observed ratio differs from 1:2:1c. Since p < 0.05, accept the null hypothesis; the observed ratio does not differ from

1:2:1d. Since p > 0.05, accept the null hypothesis; the observed ratio does not differ

from 1:2:1e. None of the above

Use the formula ∑ (Observed−Expected)2

Expected to get your chi-squared value. You can

calculate expected using the ratio predicted (1:2:1) and the given population. Also note that these don’t have to be in the right order. Even though the ratio is written as 1:2:1 and we list the data classes as blue, red, purple, this doesn’t mean that we are expecting a 1 blue : 1 red : 2 purple ratio. We’re looking for any ratio that matches 1:2:1 so we’re going to try to match 1 blue : 2 purple : 1 red. Using that to establish our expected, our chi-squared value is 2.06. Now we need to know our degrees of freedom which is just the number of classes in our data set minus 1. Our classes are red, blue, and purple, so we have 3 classes. So degrees of freedom is equal to 2. Now use the table above and find where the chi-squared value falls between (the lower part of the table). Trace this up to

see where our p-value falls between (between 0.50 and 0.25 here). The question answers establish a p of 0.05 as our critical limit. If our p-value is less than that p-value then we reject our null hypothesis and decide that our proposed ratio is wrong. If our p-value is greater than the critical p-value (as is the case here) then we fail to reject the null hypothesis. Though, we’ll just consider this as accepting our null hypothesis and claiming that our ratio is accurate.

31. In the set of experiments performed by Avery and MacLeod, they tried to identify the “transforming principle” discovered by Griffith. To do so, they treated a slurry of dead virulent cells with either protease, DNase, or RNase and then checked to see if they could transform nonvirulent cells to virulent ones. What would you propose to be the transforming principle if only protease prevented transformation?

a. DNAb. RNAc. Proteind. DNA and RNA togethere. None of the above

Which enzyme prevents transformation indicates the compound that is the transforming principle – our genetic material. In the actual experiment, DNase completed this since DNA is our genetic material. However, in this case we’re saying that protease prevented transformation which would indicate that protein is the genetic material since protease degrades proteins.

32. What unwinds the DNA to make the replication bubble during replication?a. Helicaseb. Primasec. Topoisomerase

d. SSBse. Polymerase

Helicase unwinds DNA. Primase adds an RNA primer, topoisomerase cleaves, spins, and religates the DNA to relieve tension, and SSBs bind single-stranded DNA to keep it separate. Polymerase makes the DNA strand.

33. A group of researchers hypothesizes that a heritable disease is passed on via maternal effect and attempts to use the ancestral tree of a patient to verify this. What branch of genetics does this study belong to?

a. Transmissionb. Molecularc. Populationd. Phylogenetice. None of the above

Transmission genetics studies heredity; how things are passed on. Molecular genetics is concerned with mechanisms and chemicals. Population genetics follows evolutionary changes.

34. What is the following nitrogenous base?

a. Cytosineb. Thyminec. Adenine

d. Guaninee. This isn’t a nitrogenous base.

35. How does telomerase solve the end-replication problem?a. It adds bases in an atypical 5’ to 3’ direction.b. It adds bases in an atypical 3’ to 5’ direction.c. It contains an RNA template that it extends the DNA by copying.d. It contains a DNA template that it extends the genomic DNA by copying.e. None of the above.

Telomeres are made up of G-rich repeat sequences of about 6-8 bp. Telomerase has an RNA that will match with the free single-stranded end of the DNA. Telomerase is actually a reverse transcriptase and will use its RNA as the transcript to make more DNA on the longer strand. It does this in repeated cycles until the longer ssDNA strand is long enough that normal lagging strand synthesis can occur.

36. How does RNAi function?a. RNA signals upregulate mRNA-degrading enzymes.b. Small RNA fragments guide RNase enzymes to target RNAs.c. Steroid signals increase inhibition of DNA transcription.d. Transcription of a viral genome inhibits translation of host cell metabolic enzyme

mRNAs.e. None of the above.

There are many ways by which RNAi can act, but all of them involve small RNA molecules being used to inhibit transcription or prevent translation of mRNA molecules by acting on DNA or RNA.

37. You are studying a lac operon that is missing the CAP-cAMP binding site. How will regulation of this operon work?

a. The operon won’t be transcribed.b. The operon will be transcribed in the presence of glucose and absence of lactose.c. The operon will be transcribed in the presence of lactose and the absence of

glucose.d. The operon will be transcribed in the presence of lactose, but regardless of the

concentration of glucose.e. The operon will always be transcribed.

The lac operon needs an activator. This is why glucose levels must also be low for the operon to begin synthesis normally.

38. You are working with two unlinked genes in Drosophila melanogaster. The first controls body type and has two alleles. Dominant E produces normal body color while recessive e produces ebony body color. The second controls eye color and also has two alleles. Red eyes arise from the dominant allele R, while sepia eyes arise from the recessive allele r. You testcross a female fly with red eyes and normal body color and observe the following offspring counts. What is the female’s genotype?

Red and normal Red and ebony Sepia and normal Sepia and ebony93 0 105 0

a. EE RRb. Ee RRc. Ee Rr

d. EE Rre. None of the above

Offspring body color is always normal, so the tested parent must by homozygous dominant. However, the eye trait is split between dominant and recessive, so the tested parent must be heterozygous.

39. In a breed of flowering plant, the number of flower petals is controlled by one gene locus. Allele F produces 6 petals while allele f produces 4 petals. If the phenotype of this trait is governed by paternal imprinting, what will be the genotype and phenotype of the offspring of the following cross, female listed first?

ff x FF

a. Ff; 4 petalsb. Ff; 6 petalsc. ff; 4 petals

d. ff; 6 petalse. None of these

Imprinting means that the imprinted parent’s allele doesn’t function. Since it paternal imprinting in this example, the father’s allele (F) doesn’t affect offspring phenotype even though it shows up in the genotype. This means that the mother’s allele (f) governs the offspring phenotype while also being a part of the genotype. Note: the Ff offspring of this cross can pass on both alleles as activated versions if the offspring is female. Male offspring will pass on two inactivated alleles because the locus is still paternally imprinted.

40. Birds have ZW-ZZ sexual differentiation. Which is the heterogametic sex in birds?a. Males, because they have genotype ZWb. Males, because they have genotype ZZc. Females, because they have genotype ZWd. Females, because they have genotype ZZe. None of the above

The heterogametic sex has two different sex chromosomes. In XY-XX sexual differentiation, this is the male. In ZW-ZZ sexual differentiation, this is the female.

41. Why does X-inactivation occur?a. So females don’t overexpress certain genes.b. So males can develop male secondary sexual characteristics.c. So faulty X chromosomes can be silenced.d. So haploid male insects don’t develop female sexual characteristics.e. None of the above.

Dosage compensation is the problem being solved here. Remember that X-inactivation also leads to the mosaic expression effect in females.

42. Which of these modes of horizontal gene transfer is dependent upon viruses?a. Conjugationb. Transformationc. Transduction

d. Transpositione. None of these

Conjugation is contact-mediated, transformation requires naked DNA in the soup around the bacteria, and transduction is virus-mediated.

43. An operon in E. coli is governed in the following manner. A small molecule enters the cell and induces a change in a protein bound to the operator. The protein detaches from the operator and transcription stops. What is the manner of regulation of this operon?

a. Negative repressibleb. Positive repressiblec. Negative inducibled. Positive induciblee. None of the above

Positive/negative is determined by the protein (regulator). If the protein binds and causes transcription to go, it’s an activator and the system is positive. If the protein binds and causes transcription to stop, it’s a repressor and the system is negative. Inducible/repressible is determined by the signal molecule. If the signal molecule enters and transcription starts, the molecule is an inducer and the system is inducible. If the signal molecule enters and transcription stops, the molecule is a co-repressor and the system is repressible.

44. Consider the following. You perform a cross of a plant that produces striped squash and one that produces yellow squash. After observing the offspring as described below, what do you propose to be the genotypes of the parents?

White Striped Yellow0 12 16

a. ww yy x ww yyb. Ww yy x WW Yyc. Ww yy x Ww Yy

d. ww yy x ww YYe. None of these

We know the genotype of the striped parent: ww yy. That’s the only genotype that makes striped melons. We know part of the yellow parent phenotype too: ww Y_, since we only pass through to yellow with both of those conditions met. So the rest of this problem is just a testcross for the Y locus. Since we’re split between striped and yellow offspring, the yellow parent must have a Yy genotype. So the cross is ww yy x ww Yy. To have a white plant, we need a W_ phenotype and

if we have a W_ phenotype, we no longer care what the Y gene does because W_ is epistatic and hides Y.

45. Consider the following. If you perform the cross listed, what is your expected phenotypic ratio?

Y+y cc x yy C+c

a. 1:1b. 1:2:1c. 1:1:1:1d. 9:3:3:1

e. None of the above

Since parent 1 can only gives a c allele and parent 2 can only give a y allele, all offspring start off with _c _y. From there, 50% of offspring will get Y+ from parent 1 while the other 50% get y. From parent 2, 50% of offspring get C+ while the other 50% get c. The likelihood, then, of Y+C+, of Y+c, of yC+, and of yc are 25%.

46. When does crossing over occur normally?

a. During mitotic prophaseb. During prophase Ic. During prophase IId. During metaphase Ie. This doesn’t happen in

eukaryotes

Crossing over occurs during Prophase I during meiosis.

47. What is sigma factor used for in prokaryotic transcription?a. Binding to the DNAb. Adding ribonucleotides to the growing mRNAc. Improving polymerase speed on the DNAd. Proof-readinge. Sigma factor is used in eukaryotes, not prokaryotes

Sigma factor allows the RNA polymerase holoenzyme to attach to DNA. Sigma factor actually slows down the polymerase initially, though.

48. Indicate the 3’ carbon on this structure.

a. A.b. B.c. C.d. D.e. None of these

A is a 5’ carbon. The second A on the right side that should be a D is a 2’ carbon.

49. Which of the following is a bond represented in this diagram?

a. Ionicb. Peptidec. Phosphodiesterd. Van der Waals

A.

C.

B.

A.

e. None of these are represented in this diagram.

Phosphodiester bonds hold DNA and RNA together.

50. Why does lagging strand synthesis produce Okazaki fragments?a. Because the replication bubble opens 5’ to 3’ along the leading strand.b. Because waiting to synthesize the lagging strand would cost more energy.c. Because DNA polymerase only has 5’ to 3’ polymerizing activity.d. Because the eukaryotic genome isn’t circular.e. All of the above.

These all taken together fairly thoroughly explain why lagging strand synthesis occurs in fragments, especially in eukaryotes. Prokaryotes also usually synthesize the lagging strand in fragments, but this is less universal.