Exam 6 Review
Supplemental Instruction
IowaStateUniversity / Leader: / Matt C.
Course: / Biol/Gen 313
Instructor: / Dr. Myers & Dr. Vollbrecht
Date: / 04/30/2017

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

Multiple Choice

  1. Which of these is an epigenetic modification?
  2. Cytosine methylation
  3. Histone lysine acetylation
  4. Moving histones along the DNA
  5. Histone lysine methylation
  6. 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.

  1. Consider the genotypes given below. H produces standard blood types and is dominant to h, which produces the Bombay blood type. IA and IB are codominant to each other, but are each dominant to IO. What is the expected phenotypic ratio in the offspring of these two?

Hh IAIA x Hh IBIO

  1. 9:3:3:1
  2. 3:3:2
  3. 3:3:1:1
  4. 3:1
  5. None of the above.

The hh genotype is epistatic, hiding any of the IX genotypes. So there is a 3:1 ratio of H_:hh to begin with. From there, we’ll continue assuming H_ is present and now move onto the IX cross. You should be able to work out that the two genotypes that you can observe are IAIB and IAIO which each have their own phenotype. These also occur in equal proportion. Returning to the 3:1 ratio of H_:hh, we now have H_ split into two separate phenotypes: H_IAIB and H_IAIO. This splits the ratio into 1.5 : 1.5 : 1 (the 3 split in half). Since we avoid fractions in ratios, we multiply everything by 2 to get 3 : 3 : 2.

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

  1. Silent
  2. Suppressor
  3. Missense
  4. Nonsense
  5. 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).

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

Genes / Rate / Genes / Rate / Genes / Rate
A 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%
  1. A, C, and E
  2. A, B, and C
  3. B and E
  4. C and E
  5. 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.

  1. What catalyzes the formation of a peptide bond during translation?
  2. rRNA
  3. mRNA
  4. A protein in the 30S complex
  5. A protein in the 50S complex
  6. None of the above

The go-to example of a ribozyme.

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

  1. Autosomal dominant
  2. Autosomal recessive
  3. X-linked dominant
  4. X-linked recessive
  5. 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.

  1. During which phase of mitosis does the nuclear envelope break down?

  1. Prophase
  2. Metaphase
  3. Anaphase
  4. Telophase
  5. Interphase

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

  1. How does a eukaryote initiate translation?
  2. The ribosome recognizes the Shine-Dalgarno sequence.
  3. The ribosome recognizes a Shine-Dalgarno-like sequence.
  4. The ribosome recognizes several initiation consensus sequences at the same time.
  5. The ribosome scans for any start codon.
  6. None of the above

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

  1. 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 Flowers
101 / 88 / 211
  1. Since p < 0.05, reject the null hypothesis; the observed ratio differs from 1:2:1
  2. Since p > 0.05, reject the null hypothesis; the observed ratio differs from 1:2:1
  3. Since p < 0.05, accept the null hypothesis; the observed ratio does not differ from 1:2:1
  4. Since p > 0.05, accept the null hypothesis; the observed ratio does not differ from 1:2:1
  5. None of the above

Use the formula 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.

  1. 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?
  2. DNA
  3. RNA
  4. Protein
  5. DNA and RNA together
  6. 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.

  1. Which of these induces translocation in prokaryotic ribosomes?
  2. IF-2-GTP
  3. EF-G
  4. EF-Ts
  5. RF-2
  6. None of the above

Translocation is the process by which the ribosome moves along the mRNA strand. This occurs during elongation.

  1. What unwinds the DNA to make the replication bubble during replication?

  1. Helicase
  2. Primase
  3. Topoisomerase
  4. SSBs
  5. 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.

  1. 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?
  2. Transmission
  3. Molecular
  4. Population
  5. Phylogenetic
  6. 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.

  1. Which of the following is not a conclusion from Chargaff’s work?
  2. [A] + [G] = [T] + [C]
  3. [G] = [C]
  4. [A] = [T]
  5. [G] + [T] = [A] + [G]
  6. All of the above agree with Chargaff’s rules.

The first 3 are Chargaff’s rules. The fourth one just agrees with Chargaff’s rules.

  1. What is the following nitrogenous base?
  1. Cytosine
  2. Thymine
  3. Adenine
  4. Guanine
  5. This isn’t a nitrogenous base.
  1. How does telomerase solve the end-replication problem?
  2. It adds bases in an atypical 5’ to 3’ direction.
  3. It adds bases in an atypical 3’ to 5’ direction.
  4. It contains an RNA template that it extends the DNA by copying.
  5. It contains a DNA template that it extends the genomic DNA by copying.
  6. 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.

  1. How does RNAi function?
  2. RNA signals upregulate mRNA-degrading enzymes.
  3. Small RNA fragments guide RNase enzymes to target RNAs.
  4. Steroid signals increase inhibition of DNA transcription.
  5. Transcription of a viral genome inhibits translation of host cell metabolic enzyme mRNAs.
  6. 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.

  1. You are studying a lac operon that is missing the CAP-cAMP binding site. How will regulation of this operon work?
  2. The operon won’t be transcribed.
  3. The operon will be transcribed in the presence of glucose and absence of lactose.
  4. The operon will be transcribed in the presence of lactose and the absence of glucose.
  5. The operon will be transcribed in the presence of lactose, but regardless of the concentration of glucose.
  6. 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.

  1. What sort of transposon structure can cause gene deletion via intrachromosomal recombination?
  2. Any single transposon can cause this.
  3. Any two transposons can cause this.
  4. Two transposons with the same orientation are required.
  5. Two transposons with inverted orientations are required.
  6. This process can’t happen.

Two transposons going in the same direction can cause deletion of the region between them via intrachromosomal crossing over. If the transposons are in an inverted orientation, they will cause inversion of the gene between them via intrachromosomal crossing over.

  1. 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 ebony
93 / 0 / 105 / 0
  1. EE RR
  2. Ee RR
  3. Ee Rr
  4. EE Rr
  5. 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.

  1. 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

  1. Ff; 4 petals
  2. Ff; 6 petals
  3. ff; 4 petals
  4. ff; 6 petals
  5. 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.

  1. Birds have ZW-ZZ sexual differentiation. Which is the heterogametic sex in birds?
  2. Males, because they have genotype ZW
  3. Males, because they have genotype ZZ
  4. Females, because they have genotype ZW
  5. Females, because they have genotype ZZ
  6. 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.

  1. Why does X-inactivation occur?
  2. So females don’t overexpress certain genes.
  3. So males can develop male secondary sexual characteristics.
  4. So faulty X chromosomes can be silenced.
  5. So haploid male insects don’t develop female sexual characteristics.
  6. 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.

  1. Which of these modes of horizontal gene transfer is dependent upon viruses?

  1. Conjugation
  2. Transformation
  3. Transduction
  4. Transposition
  5. None of these

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

  1. 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?
  2. Negative repressible
  3. Positive repressible
  4. Negative inducible
  5. Positive inducible
  6. 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.

  1. 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 / Yellow
0 / 12 / 16
  1. ww yy x ww yy
  2. Ww yy x WW Yy
  3. Ww yy x Ww Yy
  4. ww yy x ww YY
  5. 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.

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

Y+y cc x yy C+c

  1. 1:1
  2. 1:2:1
  3. 1:1:1:1
  4. 9:3:3:1
  5. 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%.

  1. When does crossing over occur normally?
  2. During mitotic prophase
  3. During prophase I
  4. During prophase II
  5. During metaphase I
  6. This doesn’t happen in eukaryotes

Crossing over occurs during Prophase I during meiosis.

  1. In which of the following is a lariat structure formed?

  1. Poly-adenylation
  2. Splicing
  3. Telomere lengthening
  4. Attenuation
  5. None of these

The mechanism for splicing works like this: the 2’ hydroxyl at the end of the leading exon attacks the phosphate on the 3’ end of the same ribose. This breaks the end of the intron off and frees it for manipulation. The spliceosome bends the end of the intron around and attaches it at the branch point in the middle of the intron. This forms a lasso-like structure that is the lariat. The spliceosome finishes the process by breaking the end of the intron in a similar way and then attaching the end of the leading exon to the beginning of the following exon.

  1. What is sigma factor used for in prokaryotic transcription?
  2. Binding to the DNA
  3. Adding ribonucleotides to the growing mRNA
  4. Improving polymerase speed on the DNA
  5. Proof-reading
  6. 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.

  1. Indicate the 3’ carbon on this structure.
  2. A.
  3. B.
  4. C.
  5. D.
  6. None of these

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