An enzymatic reaction converts starch to simple sugars using a fungal enzyme that is very pH sensitive and works efficiently ?f and only if the pH of the reaction environment is ?.20. Your statistician, Mr. Biswadeep Pai decides to use a pH meter that is known to give normally distributed replicate measurements. Suppose that ?0 independent replicate measurements of pH yield the following values: {?. ?8, ?. ?7, ?. ?6, ?. ?5, ?. ?1, ?. ?7, ?. ??, ?. ?8, ?. ?9, ?. ??}.

(a) What conclusion can Biswadeep draw at the ? = ?. ?0 level of significance?

(b) What conclusion can Biswadeep draw at the ? = ?. ?5 level of significance?

(c) Is "?f and only if" in the above problem statement a typo?

Many cancer drugs have been developed that inhibit or activate enzyme function.

a. Geftinib is a tyrosine kinase inhibitor that has been shown to specifically target EGFR. EGFR is a receptor tyrosine kinase (RTK) and can become constitutively active in a cancer cell. Before we continue, explain how a RTKs function to transmit a signal (EGF) from the outside of the cell across the membrane to the inside of the cell. Be specific, what changes occur on the outside of the cell when ligand is bound? What changes occur on the inside of the cell that pass that information from the receptor to the first downstream target? I don’t need the entire signal transduction cascade. Rubric (4): correct explanation with the effect of ligand binding by the receptor on outside of the cell (2) and effect of ligand binding on inside of the cell (2).

b. Geftinib has been shown to occupy the ATP binding site of the tyrosine kinase domain of the EGFR, therefore we can hypothesize that it acts as a(n) __________________ inhibitor. (2pts)

c. Let’s think about a specific mutation in the EGFR (L858R).   The Kd for ATP binding to the tyrosine kinase domain of the EGFR (L858R) mutation is 148mM while the Kd of Geftinib for the L858R mutation of the EGFR is 2.4nM. Which binds more tightly (ATP or Geftinib) to EGFR (L858R) and how do these numbers confirm your conclusion in “b”? Rubric (4): correct analysis of binding (2) and correct explanation that relates analysis to conclusion in “b” (2).

d. Researchers determined proteins downstream of activated EGFR that were affected by Geftinib. We know some of these proteins! In the experiment, cancer cells with constitutively active EGFR are treated with or without Geftinib. In these experiments, Raf was downregulated in the cells treated with Geftinib when compared to untreated. Why does this make sense?   Make sure you substantiate your answer with concepts covered this week! Rubric (4): correct correlation is made between specific effect of the drug (1) and obvious application of concepts taught this week is applied to the answer, and they are correct (3).

A colorimetric LDH assay provides a method to measure the enzyme activity of LDH. NADH (MW 633.4 g/mol) absorbs light at 340 nm while NAD+ does not show any absorption at that wavelength. The Beer-Lambert Law relates the absorbance of a solute to its concentration: A = ε*c*L A is absorbance, ε is the molar extinction coefficient, c is concentration, and L is the path length of the solution in the spectrophotometer (usually in cm). In the lab you have available NADH. You dissolve 4.8 mg in 50 ml of buffer and measure an absorbance of 0.2363 at 340 nm in a 0.5 cm cuvette. Subsequently in a 1 cm cuvette, you mix 25 µl of 20 mM NADH, 25 µl pyruvate 0.2 M , 400 µl of TRIS buffer (pH=7.8), and 50 µl of a solution containing lactate dehydrogenase at a concentration of 2 mg/l. You place the cuvette in a spectrophotometer and follow the kinetics of absorbance at 340 nm. You measure a rate of decrease of absorbance (ΔA) of 0.50 min-1 .

a) Determine the absolute amount of NADH consumed after 20 seconds in the reaction (in nmol)

b) Determine the specific activity of LDH as NAD+ produced per LDH in the unit of time (in µmol/mg/min)

Both glyceraldehyde 3-phosphate and dihydroxyacetone phosphate are substrates for the enzyme triose phosphate isomerase. In a study by Knowles et al the k_M for glyceraldehyde 3-phosphate was determined to be 0.47mM with a Vmax of about 410 uM/min. Whereas, for the substrate dihydroxyacetone phosphate the k_M was determined to be 0.97 mM and the Vmax was about 41 uM/min. (Biochem J 129 (2) 1972:301-310). Triose phosphate isomerase has a relatively weak affinity for _______

A. glyceraldehyde 3-phosphate because the k_M is lower in comparison

B. dihydroxyacetone phosphate because the k_M is higher in comparison

C. glyceraldehyde 3-phosphate because the Vmax is higher in comparison

D. dihydroxyacetone phosphate because the Vmax is lower in comparison

Consider the following paragraph.  

Superoxide dismutase is an enzyme that detoxifies a free radical called superoxide.a This radical is responsible for causing injury during reperfusion after a heart attack, and is also thought to play a role in Parkinson’s Disease.b Depending on what organism the protein is found in, it occurs either as a dimer or tetramer of identical subunits.c The subunits are mostly a-helical with some β-sheet near the C-terminus.d A metal binding site for manganese, iron, nickel or copper/zinc is composed of residues from both the N-terminus and the C-terminus.e These residues are: His-26, His-81, Asp-167, His-171.f

Write the letter of the sentence(s) represent primary structure, secondary structure, tertiary structure, and quaternary structure in the above paragraph? You may or may not use all the letters. Explain your choices.

Primary structure =

Secondary structure =

Tertiary structure =

Quaternary structure =

You have worked on YFE1 (your favorite enzyme!) for awhile and know it is a serine/threonine kinase important in a crucial signal transduction pathway in the cell (not related to metabolism). You have developed in vitro and in vivo assays for its activity. You have the genetic tools to overexpress the protein and to produce mutants of the protein that show decreased function while still expressing the protein as well as RNAi to silence the gene. You know that eliminating the gene function leads to impaired cells that do not survive well because the pathway it is in is very important in the cell. Another researcher just published results showing that knocking out the YFE1 gene in mice leads to pups that survive less than 10 days. They found during day 1, the pups developed hyperglycemia which they believe was the cause of their early deaths. There has been no previous evidence YFE1 played a role in glucose metabolism; none of your work has ever shown a link to the pathway, but you acknowledge that you have never done experiments that directly test this. Given these new results, you want to see if you can figure out if YFE1 could be involved in glucose metabolism. Write a hypothesis. Outline the experiments you would conduct to test your hypothesis. Define the cells you will examine if that is your choice of experiments. (in particular, the tissue source). You may do any in vitro or in vivo assay – no whole animal studies. [Don't just say you are going to run a microarray. Tell me what cells you would use as control and experimental. Don't just say you are going to run a Northern blot - tell me what sample you are going to run it on and against what gene or sequence your probe will be. However, you should not write a Method section description of the experiment.]

Consider a continuous culture of Bacillus subtilis bacteria, which are used to produce the enzyme amylase during the growth phase. The relevant parameters for growth of this bacteria are μ_MAX = 0.88 hr^-1, Y_X/S = 0.5 gdw/gr substrate, S₀ = 4.0 gr substrate/L, and the Monod constant K_S = 0.3 gr/L, while the protein yield during growth phase is Y_P/X (aka product parameter α) = 0.12 g protein/gdw. The chemostat is operated at a dilution rate that is 60% of the maximum growth rate. Using this information, what is the steady state rate of product formation in the chemostat culture, in units of grams protein/L-hour?

1. The enzyme that unwinds the DNA double helix is called
   1. DNA gyrase                c. helicase
   2. primase                       d. ligase
2. RNA polymerase has
   1. 5’- 3’ exonucleases
   2. 3’- 5’ exonucleases
   3. both exonucleases
   4. no exonucleases

3. In the synthesis of RNA the addition proceeds by    
   1. attack of the 3’OH to the a phosphate of the incoming NTP
   2. attack of the 3’OH to the a phosphate of the incoming dNTP
   3. attack of the 5’OH to the a phosphate of the incoming NTP
   4. attack of the 5’OH to the a phosphate of the incoming dNTP                     

4.    The RNA transcript is identical to the
   1. nontemplate strand of DNA
   2. coding strand of DNA but T is replaced by U
   3. coding strand of DNA but U is replaced by T
   4. template strand

5.   The specific binding sites for RNA polymerase on DNA are called
   1. core subunits              
   2. transcripts
   3. introns
   4. promoters

The restriction enzyme ApaI recognises and cleaves the sequence GGGCC^C (^ indicates where the DNA is cleaved).

You previously determined the size of the product amplified in the PCR. which is the region between the bold section

TGGGCTAGGTGTAGGGGTCCTGAGTTCCGGGCTTTGCTACCCAGCTCTTGACTTCTGT

TTCCCGATTTTA AATGAGCAGTTTGGACTAAGCCATTTTTAAGGAGAGCGATGGGGAGG

GCTTCCCCCTTAGCACAAGGGCAGCCCTGGCCCTGGCTGAAGCCCAACCCCAACCTC

CAAGACTGTGAGAGGATGGGGACTCATCCCTGGAGGAGGTGCCCCTCCTGGTATTGAT

AAAGAATGCCCTGGGGAGGGGGCATCACAGGCTATTTGAACCAGCCCTGGGACCTTG

GCCACCTCAGTGTCACTGGGTAGGGGGAACTCCTGGTCCCTTGGGTATATGGAAGGTA

TCAGCAGAAAGCCAGCACTGGCAGGGACTCTTTGGTACAATACCCAGCATGCATGCTG

TGCCAGGGGCTGACAAGGGTGCTGTCCTTGGCTTCCCCATTTTGGAGTGGTCACTTG

CCTCTACTCCAGCCCCAGAAGTGGAAACTGAGATGATGTGTGGAGGAGAGAGCCAGC

GTTCATGTTGGGAATCTTGAGGCTCCTTTCCAGCTCTCAGATTCTGTGATGCTCAAAGG

GTGAGCTCTGTGGGCCCAGGACGCATGGTAGATGGAGCTTAGTCTTTCTGGTATCCAG

CTGGGAGCCAAGCACAGAACACGCATCAGTGTTTATCAAATGACTGAGGAAATGAATGA

ATGAATGTCTCCATCTCAACCCTCAGCCTGGTCCCTCCTTTTTTCCCTGCAGTTGGTAC

AGATGGCATTGTCCCAGTCTGTTCCCTTCTCGGCCACAGAGCTTCTCCTGGCCTCTGC

CATCTTCTGCCTGGTATTCTGGGTGCTCAAGGGTTTGAGGCCTCGGGTCCCCAAAGGC

CTGAAAAGTCCACCAGAGCCATGGGGCTGGCCCTTGCTCGGGCATGTGCTGACCCTG

GGGAAGAACCCGCACCTGGCACTGTCAAGGATGAGCCAGCGCTACGGGGACGTCCT

GCAGATCCGCATTGGCTCCACGCCCGTGCTGGTGCTGAGCCGCCTGGACACCATCCG

GCAGGCCCTG

Now, identify the ApaI site within this region. If you digest the PCR product with ApaI, how many fragments of DNA would you get, and what would be the sizes of the resulting DNA fragments (in base pairs)?

QUESTION 29 1. In Avery, MacLeod and McCarty's experiment, what enzyme was present in the mixture of R & S Strain that when injected into the mice did not result in death?

a. Protease b. All the enzymes were present c. DNASE d. RNASE

QUESTION 31

1. Frederick Griffith's experiments resulted in which of the following novel findings?