Molecular Biology – 6 Hours

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At least 2 questions from this section will be on the final exam

SAMPLE QUESTIONS FOR THE FINAL EXAM Question 1. Ferritin is a protein involved in the storage of iron inside cells. To prevent toxic accumulation of too much iron inside cells, the intracellular level of ferritin is tightly regulated. To study the regulation of ferritin synthesis, mammalian cells are grown with or without iron in the culture medium. Note that iron in the culture medium is rapidly transported inside cells. a) Upon addition of iron to the culture medium, the intracellular concentration of ferritin mRNA is unchanged but the concentration of ferritin protein increases. How do you think ferritin expression is regulated? Briefly explain. The regulatory sequence given below is found in the ferritin mRNA between the cap structure and the start codon. 5’-GGGUUUCCGUUCAACAGUGCUUGGACGGAAACCC-3’ Mutations within in this sequence are used to study the regulation of ferritin expression. The following observation are made: • ferritin expression is high, independent of the iron concentration, when (i) the entire region is deleted, or

(ii) the region located upstream of the underlined sequence is deleted or (iii) the underlined sequence is replaced with a random sequence.

• ferritin expression remains iron-dependent when this region is replaced by the following sequence: 5’-GGGCUCAGGUUCAACAGUGCUUGGACCUGAGCCC-3’. Note that the sequence differences are indicated in bold.

b) Explain why these observations suggest that both sequence and structure of the 5’ end of ferritin mRNA are important for the regulation of ferritin expression.

c) Ferritin translation becomes iron-independent when the regulatory sequence is moved from the 5’ side (upstream of the open reading frame) to the 3’ side (downstream of the open reading frame) of ferritin mRNA. Which step of ferritin translation do you think is affected by the intracellular level of iron?

d) IRP is a protein involved in the regulation of ferritin expression. Anti-IRP antibodies attached to sepharose beads are added to a cell extract, then the extract is centrifuged to separate the pellet fraction (containing the sepharose beads ) from the supernatant fraction. If the cells are cultured in the absence of iron, ferritin mRNA is found together with IRP in the pellet. In contrast when cells are cultured in the presence of iron ferritin mRNA remains in the supernatant fraction while IRP alone is found in the pellet. Briefly explain the likely role of IRP in the regulation of ferritin expression. Question 2. You are studying the development of a newly discovered insect. Like drosophila, it undergoes a stage in early larval development where the eve gene is expressed in a pattern of 7 stripes. You are particularly interested in stripes 2 and 5. The following figures show the organization of the cis-acting elements that control the expression of the eve gene in seven stripes. On Figure 1 (top part), the seven boxes represent segments of DNA or UAS (upstream activating sequences) responsible for the activation of the eve gene in each of the seven stripes. The UAS responsible for gene expression in stripes 2 and 5 contain binding sites for five distinct proteins acting as transcriptional regulators (pA, pB, pC, pD, and pE) (Figure 1 top part). The quantitative

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distribution of these five eve-gene regulators throughout the embryo (from the anterior to the posterior pole) is also shown on the lower part of this figure.

Figure 1: Organization of the regulatory region of the eve gene and distribution of its transcriptional regulators in the embryo. a) Transcription regulators pA, pB, pC and pD seem to be important for establishing eve expression in stripe 2. They all have binding sites (A,B,C &D) in the cis-acting segment controlling expression in stripe 2. Based on the distribution of these 4 transcription regulators in the embryo, determine which of the 4 regulators are most likely to be activators and which one are most likely to be repressors of eve gene expression in stripe 2. Briefly explain. b) The UAS responsible for gene expression in stripe 5 (UAS5) has one binding site for pA (A) and two binding sites for pE (E1 and E2). To study the binding of pE to UAS5, radiolabeled DNA fragments containing UAS5 or mutant forms of UAS5 are mixed with protein pE and are submitted to gel electrophoresis under conditions that do not disrupt protein-DNA interactions. An X-ray film is then placed in contact with the gel to visualize the radioactive DNA. The following figures show the results of these experiments.

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Figure 2: DNA-protein binding assay. Protein was added at low (+) or high concentration (+++) to radiolabeled DNA segments containing both wild type E-binding sites (E1E2), or wild type E1 and mutant E2 (E1E2*), or mutant E1 and wild type E2 (E1*E2). The mutations in E1 and E2 are known to abolish protein-DNA interactions. In the first lane, 0 indicates that no protein was added to the radiolabeled DNA. E1 and E2 have each a single protein E binding site. Propose a model explaining the differences observed between the addition of a low and a high amount of protein pE and the difference between the protein-binding properties of E1 and E2. c) Briefly explain how proteins pA and pE lead to eve expression in stripe 5. Be sure to account for how the posterior (right) side of stripe 5 is formed. (hint: part b should help you answering this question). Question 3: Scientists have introduced in bacteria a wild type gene coding for GFP (green fluorescent protein) or two engineered versions of this gene containing an additional cis-regulatory element (crRL or crR12). The transcription of the wild-type GFP gene leads to the production of an mRNA containing a cis-acting element labeled X. Transcription of the engineered GFP genes leads to the production of mRNAs containing the cis- acting X and either crRL or crR12 in the 5’ portion of the mRNA. The synthesized mRNA molecules are diagrammed below.

1) To identify the function of X, scientists performed the following assay: purified small ribosomal subunits, large ribosomal subunits, 16S rRNA (ribosomal RNA), or Alanyl-tRNA were bound to a filter and the filter was then incubated with radioactive wild type GFP-mRNA molecules (with or without the cis-acting element X). The radioactivity associated with the filter was measured after removal of the unbound radioactive material. The results are given in the following table. Radioactive GFP mRNA

Small ribosomal subunit

Large ribosomal subunit

16s rRNA Ala-tRNA

X deleted 18 cpm 22 cpm 7 cpm 13 cpm X present 1500 cpm 20 cpm 2300 cpm 18 cpm

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Based on these data and your knowledge of translation in prokarytotes determine the function of X. Briefly explain. 2) The sequences of the 5’ end of crRL, crR12and WT-GFP mRNAs are:

A filter assay, similar to the one described above, was performed by adding radioactive crRL mRNA and crR12 mRNA to a filter with 16SrRNA bound to it. A second assay was performed to test the sensitivity of X to a RNAse that only cleaves single-stranded RNA. The results of both experiments are presented in the following table. Radioactive mRNA Filter bound 16S rRNA RNAse sensitivity of sequence X GFP mRNA 2300 cpm High crRL mRNA 22 cpm Low crR12 mRNA 9 cpm Low None 15 cpm Note: The RNase assay is performed in solution and in the absence of 16S rRNA Based on the sequence information and the experimental data, suggest a role for the cis-acting regulatory elements crRL and crR12. Briefly explain. 3) The level of GFP synthesis was measured in bacteria expressing GFP mRNA, crRL mRNA or crR12 mRNA. The following histogram shows the results of this experiment.

Level of GFP synthesis

0

50

100

150

200

250

WT GFP crRL crR12

Expressed mRNA

G FP

s yn

th e si

s

Based on the results of this experiment and the sequence information (see part 2), explain why the scientists describe crRL and crR12 as translation terminators. 4) The diagram below is a reminder of the sequence of the 5’-end of the crRL mRNA.

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Bacteria expressing a gene where the sequence of the crRL element was replaced by its complementary sequence had a normal level of GFP synthesis, whereas the level of GFP synthesis remained low when mutations were introduced in the underlined sequence only. Based on these information, draw the secondary structure adopted by the 5’ end of the crRL mRNA. Question 4 You have decided to study the role of SWI1 and SWI2 proteins in transcriptional enhancement via steroid receptors. You know that when a cell is exposed to glucocorticoids (steroid hormones) the intracellular binding of this hormone to its cognate receptor results in the transcriptional activation of a specific sub-set of genes. A promoter called the glucocorticoid response element characterizes these genes. Yeast cells are co-transformed with the following constructs:

– A yeast expression plasmid, pE, responsible for the constitutive expression of the gene encoding the glucocorticoid receptor. – A yeast reporter plasmid, pR, containing the gene encoding β-galactosidase under the control of the glucocorticoid response element.

a) Describe how the synthesis of β-galactosidase in the yeast cells is affected by the addition of glucocorticoid. Include in your description the variation in the level of transcriptional activity of the glucocorticoid receptor gene in both the absence and presence of hormone. b) The activity of β-galactosidase before and after addition of glucocorticoid was measured using three different strains of yeast each co-transformed with pE and pR. The strains are the following: – SWI+, is a wild-type strain producing active SWI1 and SWI2 proteins – swi1- and swi2- are two mutant strains producing inactive SWI1 and SWI2 proteins respectively. The results of the β-galactosidase activity measurements were the following:

Strain Glucocorticoid Activity (arbitrary units) SWI+ No 100 SWI+ Yes 1000 swi1- No 97 swi1- Yes 105 swi2- No 92 swi2- Yes 102

Based on the fact that proteins SWI1 and SWI2 are found in a multimolecular transcriptional complex, explain these results. c) If strains SWI+, swi1- and swi2- are only transformed by the yeast expression vector pE in which the β- galactosidase gene has replaced the glucocorticoid receptor gene the level of β-galactosidase activity is similar in each strain. Why is this control experiment critical to explaining the results presented in part (b)? Question 5: HP1 is a DNA binding protein that interacts with a specific sequence (TGCTTATTC). You want to analyze, by a Dnase I footprinting assay, the effect on DNA binding of the interaction between HP1 and its

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binding partner PA. In each assay, you combine a radiolabeled fragment of DNA that binds to HP1 and a specific combination of proteins. After incubation with DNase I, each reaction mixture was resolved by gel electrophoresis, and then exposed to film. The autoradiogram showing the results for each combination of proteins is diagrammed below.

Lane 1: DNA alone, no protein added in the nuclease assay. Lane 2: DNA + 5 ng (nanograms) of HP1 Lane 3: DNA + 5 ng of HP1 + 2ng of OCT1 (a protein known to interact with HP1) Lane 4: DNA + 5 ng of HP1 + 10 ng of OCT1 Lane 5: DNA + 5 ng HP1 + 5ng PA Lane 6: DNA + 5 ng PA

1) For each lane (1 to 6) give a brief interpretation of the results. 2) When DNA is mixed with PA and OCT1, the band pattern observed on the autoradiogram is identical to one of the pattern shown above. Which one? Explain. 3) Explain how DNA binding proteins can recognize a specific sequence without opening the DNA double helix. Question 6 IPTG is a structural analog of lactose commonly used to activate genes placed under the control of the Lac promoter. When added to a bacterial culture, IPTG diffuses freely inside the bacteria. 1) Bacteria, containing a functional lactose operon, were transformed by a plasmid that contains – LacI: the gene coding for the lac repressor under the control of a constitutive promoter – gfp: the gene coding for the green fluorescent protein GFP under the control of the lac promoter. The lac promoter is the promoter that normally controls the expression of the lac operon.

a) Could bacteria synthesize GFP in the absence of lactose in the culture medium? Briefly explain. b) Lactose is then added to the culture medium. Briefly explain the consequences (at the molecular level) of the addition of lactose on the synthesis of GFP when the culture medium contains glucose and when the culture medium does not contain glucose.

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2) The following diagram represents the amount of GFP synthesized per milligrams of bacteria (y-axis) over time when bacteria are grown in a medium without glucose and supplemented with either IPTG or with lactose.

a) Based on your knowledge of the lac operon, explain the difference in GFP synthesis when IPTG and lactose are used as inducers?

b) What is the main factor limiting the level of GFP synthesis when lactose is used as an inducer?