3B. Hint: Do you have enough information to calculate the Vmax?

3B.  Answer: Can’t predict. V_max = k₃E_o, but we do not know the k₃ for this substrate. Like the K_m, it need not be the same as for galactose-glucose (lactose) calculated in part A. Also the K_m and V_max are independent variables, determined by different aspects of the enzyme/substrate interaction. If the enzyme acts on a different substrate, and the K_m changes, the E_o (&  therefore the V_max) may or may not change.

 C2005/F2401 '06 -- Hints for Recitation Problems #4

1. What has changed, the Km, the Vmax, or both?

2. A. Remember that δGo's are additive.

    B1. How are ΔG and ΔG^o related?

    B2. What changes with concentration (of reactants and/or products)? ΔG ? ΔG^o ?

    B3. How is K_eq related to ΔG and/or ΔG^o ?

    B4. What determines direction? K_eq ? ΔG ? ΔG^o ? How can a reaction "go" in one direction, if the K_eq favors the "other" direction?

3. Having trouble with this problem? Remember you are starting with DHAP, not glucose!

    A. What pathways are involved here? Do you have the enzymes for Krebs etc.? In other words, what is DHAP being converted to? When you figure out the overall pathway, look carefully at the stoichiometry -- for every DHAP that goes through, what compounds are used up, and what compounds are synthesized (or regenerated)?

    B. Look at your handout (7-2) -- label the proper carbon in DHAP and trace it through the pathway. (What will be the end product of the pathway here?) What happens when you convert DHAP to glyceraldehyde -3-P ? Does the P move, or do you flip the molecule over -- so the "top" carbon in DHAP becomes the "bottom" carbon in glyceraldehyde -3-P?

    C. What is required for growth to occur?

C2005/F2401 '06 -- Answers to Recitation Problems #4

1.   V_max = X. V_max w/ cellobiose = the V_max in the absence of the inhibitor. In competitive inhibition., the V_max can always be reached by adding excess substrate.  In other words, the K_m has changed, but the V_max is the same.  With the inhibitor cellobiose present, the (apparent) K_m is increased -- the amount of lactose needed to reach ½ V_max is greater. So you can’t use the V at the old K_m value (measured w/o cellobiose) and assume that it = 1/2 V_max. 

2A. The ΔG^o for reaction (A)  is ___-61.5_____ .     How is this calculated? First add up (A), (B*) = reverse of (B), & (C) = same reactants and products as reaction 6.

    (A) Gald-3-P + 1/2 O₂ <--> 3-PGA    + H₂O                         ΔG^o = x kcal/mole

    (B*) NAD + H₂O <--> NADH₂ + 1/2 O₂                               ΔG^o = + 53 kcal/mole

    (C) 3-PGA + P_i <--> 1,3-diPGA                                             ΔG^o = + 10 kcal/mole

Sum = reaction 6: Gald-3-P + NAD + P_i <--> 1,3-diPGA + NADH₂         ΔG^o = +1.5 kcal/mole (value given for rxn 6)

Now solve for x:      53 + 10 + x = +1.5     x = -53 -10 + 1.5 = -61.5 kcal/mole = ΔG^o for reaction (A)

2B1. -1.3 kcal/mole:

     ΔG = ΔG^o + RT lnQ

        Q = (10^-6)(10^-5) / (10^-3)(10^-3)(10^-3) = 10^-11 / 10^-9 = 10^-2; RT = 0.6 kcal/deg-mole

     ΔG = +1.5 + 0.6 X 2.3 log (10^-2) = 1.5 + (1.38 X [-2]) = 1.5 - 2.76 = -1.26 = ~ -1.3 kcal/mole

Or, writing it all out in one step:

    ΔG = +1.5 + 0.6 X 2.3 log (10^-6)(10^-5) / (10^-3)(10^-3)(10^-3) = +1.5 + (0.6 X 2.3 log 10^-2), or

    ΔG = +1.5 + 0.6 X ln (10^-6)(10^-5) / (10^-3)(10^-3)(10^-3) = +1.5 + 0.6 X ln 10^-2

2B2. +1.5 kcal/mole. ΔG^o is a constant, given here as +1.5.

2B3. <1. Since ΔG^o = -RT ln K_eq, then since ΔG^o here is positive, and RT is positive, then ln K_eq must be negative, and so K_eq must be <1.

2B4. To the right, since ΔG is negative -- even though ΔG^o is positive (and so equilibrium lies to the left).  Remember it is  ΔG that determines direction, not ΔG^o . ΔG = ΔG^o + RT lnQ. In this example, the size of the negative RT lnQ term (-2.76) overrides the relatively small size of the positive ΔG^o term (+1.5), so that the overall  ΔG is negative, and reaction goes to the right. In other words, the relatively high concentrations of the starting materials, and the relatively low concentrations of the products, push the reaction to the right. This reaction is an example of how a reaction with a positive ΔG^o (corresponding to an unfavorable equilibrium constant) can occur in the "uphill" direction as part of a pathway. The reaction is pushed "uphill"  because the previous steps in the pathway provide substrate and the following steps remove the products, keeping the ratio of [products]/[substrates] at a low value.

3A. Having trouble with this problem? Remember you are starting with DHAP, not glucose!

    DHAP: (5)    -- you need to metabolize (ferment) one DHAP to get 2 ATP made.     ADP:   (10)   -- 2 used per DHAP metabolized (& per 2 ATP made).     Pi:         (5)    -- one used per DHAP.                                                             ATP:     (0)    -- not needed since you start with DHAP. You make ATP, but you don't use up any.     NAD:   (0.1)  -- catalytic amounts needed -- 1 NAD reduced and 1 NADH oxidized per DHAP metabolized (fermented).                                                           CoA:     (0)    -- not involved (needed only for Krebs)

Explanation: If you check out the stoichiometry of the pathway, you see that two ADP's are phosphorylated to ATP per DHAP broken down. No input of ATP is needed; you have bypassed the steps in glycolysis that use ATP by starting with DHAP instead of glucose. (When you start with glucose, you have to put in 2 ATP to "prime the pump." When you start with DHAP, you don't need to "spend" any ATP to get started.)  For each DHAP, one inorganic phosphate is picked up from solution, to make a di-phosphorylated intermediate. The two phosphates are then transferred in two separate steps to phosphorylate two ADP's to ATP. No CoA is involved -- it is needed for entry into the Krebs cycle, which is not occurring here. One NAD is needed for oxidation of each DHAP for step 6 of glycolysis. However the NAD reduced to NADH₂ in step 6 is recycled -- it is oxidized back to NAD in the last step of fermentation. So you only need a catalytic amount of NAD.

3B. Ethanol. The #1 carbon of DHAP (top as written on handout 7-2)  becomes #3 carbon  of glyceraldehyde-3 phosphate (bottom as written on handout). The phosphate group doesn't move -- you flip the DHAP molecule over and the hydroxyl on top becomes a carbonyl and the carbonyl at position 2 becomes a hydroxyl. The labeled carbon (now #3) becomes the methyl group of pyruvate. The pyruvate is subsequently decarboxylated to yield acetaldehyde and then reduced to ethanol in the ethanolic fermentation found in yeast. (If you forgot to flip the DHAP, then you would think that the radioactivity would be lost as CO₂.) Next best answer is lactate, if you forgot that yeast do ethanolic and not lactate fermentation. Third best answer is pyruvate, if you mistakenly added more than the minimum NAD to the reaction, such that fermentation is not necessary (though it would presumably take place anyway). Acetyl CoA would not be formed, as only the glycolytic and fermentation enzymes are present. The other compounds have no connection to this carbon.

3C. Yes. There is a net gain of ATP, and no net oxidation or reduction of NAD -- what is used is regenerated. Only one NAD is necessary in the glycolytic pathway converting DHAP to pyruvate, and that one NAD can be regenerated from NADH₂ in the final step of fermentation, the reduction of pyruvate to lactate (the usual product in E. coli).

Note:  E. coli can't really live on DHAP because phosphorylated sugars can't get transported into the cells.

C2005/F2401 '06 -- Key to Recitation Problems #5

Note: NADH₂ or NAD_red  is used as shorthand for reduced NAD (= NADH + H⁺⁾.

1.  E. coli can grow anaerobically in a minimal medium containing a mixture of glycerol and pyruvate, but not on glycerol or pyruvate alone.

ATP yield: You can't get any ATP from metabolizing pyruvate anaerobically, so the bacteria can't grow on pyruvate. One net ATP can be produced from metabolism of glycerol . One ATP is used up to produce DHAP -- it's used to phosphorylate glycerol to glycerol phosphate. 2 ATPs are produced in the reactions from DHAP to pyruvate. Net ATP generated = 2-1 =1. 

NAD/NADH₂ balance: Metabolizing one molecule of glycerol generates one molecule of pyruvate, and converts 2 molecules of NAD (oxidized) to 2 molecules of NADH₂ (reduced). Reducing the pyruvate to lactacte oxidizes only one of the two reduced NADH₂ molecules. If glycerol is the only carbon source, the 2nd NADH₂ cannot be oxidized, so the NAD/NADH₂ balance cannot be maintained --  the cell will run out of oxidized NAD. In the presence of excess pyruvate in the medium, the 2nd NADH₂ molecule can be oxidized back to NAD by reducing the added pyruvate to lactate. Thus NAD/NADH₂ balance will not be a problem in the presence of glycerol plus pyruvate.

2. A. Positive.  There are several ways to see this:     (i) By looking at the order in the electron transport chain: Reduced FAD feeds into the chain after reduced NAD -- that is, below it, on an energy scale. (See handout 8-2.) So NADH₂  should hand off hydrogens and/or electrons to FAD and not the other way around, since electrons go down the chain, not up it. In terms of affinity for electrons, the compounds "higher up" on the chain have a lower affinity for electrons, and tend to give them away to compounds further down the chain.  The compounds "lower down" on the chain have a higher affinity for electrons, and tend to grab electrons from the compounds above. (Oxygen, which is on the bottom, has the highest affinity for electrons -- it gets them in the end.)    

In terms of  ΔG^o: if NADH₂ is "higher up" on the chain, then the distance from it to oxygen is greater, and so the ΔG^o for reduction of NADH₂ must be more negative (than for FADH₂). If you want to formally add up the ΔG^o values for oxidation/reduction of NAD and FAD to get the overall ΔG^o for the reaction shown in the problem, see (iii).     (ii) By considering relative ATP yields -- Another way to get the relative ΔG^o values: Oxidation of FADH₂ drives production of only 2 ATP's while oxidation of NADH₂  drives production of 3 ATP's. So you must get more energy from oxidation of reduced NAD, so the ΔG^o for reduction of NADH₂ must be more negative.

    (iii) Adding up the ΔG^o values:                  NADH + H⁺ + 1/2 O₂ <--> NAD + H₂O   ΔG^o = -52 kc/mole or some large negative number if you don't remember exactly.

  FADH₂ + 1/2 O₂ <--> FAD + H₂O  ΔG^o = a much smaller negative number for the reasons given in (i) & (ii) -- since FAD enters the electron transport chain at a lower level, and since one derives less ATPs/electron pair from FADH₂ than from NADH + H⁺.

    Inverting the top reaction and changing the sign:

    NAD  <--> NADH + H⁺ + 1/2 O₂  ΔG^o = +52 kc/m or some large positive number

    FADH₂ + 1/2 O₂ <--> FAD + H₂O  ΔG^o = a much smaller negative number

    Net: NAD + FADH₂ <--> FAD + NADH + H⁺  ΔG^o =  a large positive number plus a much smaller negative number = highly positive.

    B. Less. In the bacteria, because there is so much enzyme present, all the NADH₂ will give up its electrons to FAD and then the FADH₂ will enter the electron transport chain "lower down" bypassing the first site of ATP formation. In humans, the NADH₂ will give up its electrons directly to the electron transport chain -- it will enter "higher up," before the first site of ATP synthesis. Therefore, in the bacteria, the electrons from  NADH₂ will only yield 2 ATP, compared to the 3 ATP/NADH₂ in humans. Since there are 10 NADH₂'s produced from the oxidation of one molecule of glucose, the difference in ATP yield is considerable. In bacteria, the yield is 10 X 2 = 20 ATP vs. 30 ATP in humans, a difference of 10 less in the bacteria.

3. A-1. pH will increase in the left. Hydrogen ions will be pumped across the membrane in the direction from what had been the inside or matrix side (left) to what had been the intramembrane space (right). The pH on the right will fall and the pH on the left will rise. (The hydrogen ion concentration will increase on the right and fall on the left.)

    A-2. Increase in the left. ATP will be synthesized from ADP + P_i as hydrogen ions flow back across the membrane from right to left through the ATP synthetase of the F_oF₁ complex in the membrane.

    A-3. NAD_ox will increase on the left. The reduced NAD gives up its electrons to the electron transport chain in the membrane, interacting with components on the inside or matrix side (left) of the membrane. (The electron chain is embedded asymmetrically in the membrane, so the protein complex that binds NAD faces the matrix side.)

    A-4. Neither. The reduced NAD (NAD_red) will decrease on the left as it gives up its electrons to the electron transport chain and is converted to NAD_ox in the membrane.

    A-5. Neither. Pyruvate can be metabolized two ways in a cell, but neither can occur here. Pyruvate can be metabolized by the entrance reaction to the Krebs Cycle, which requires CoA (Coenzyme A) to run, and CoA is not included (nor are the enzymes of the matrix). Pyruvate can also be metabolized by conversion to lactic acid in fermentation, but the enzymes to do this are found in the cytoplasm, and are not included here.

    B. pH, ATP, and NAD_ox or NADH_red. If there is no O₂ present to accept the electrons for the electron transport chain, then the electron transport chain will not be able to accept the electrons from NADr_ed. So compared to the situation above, there will be less NAD_ox  and more NAD_red on the left.  If there is no electron transport, no hydrogen ions will be pumped across the membrane, and no hydrogen ion gradient will be established. Therefore there will be no net flow of hydrogen ions back through the ATP synthetase and no ATP production (on the left). 

    C.  Phosphorylation of ADP to ATP & establishment of a pH difference between the two chambers will be affected.

The electron transport chain will continue to pump hydrogen ions from left to right across the membrane, but hydrogen ions will not accumulate in the right chamber since the pumped-out hydrogen ions will return to the left chamber through the tear in the membrane. Thus no pH gradient will be established. With no pH gradient, there will be no flow of hydrogen ions though the ATP synthetase, and no ATP will be produced.     NAD oxidation (oxidation of NAD_red to NAD_ox) will continue normally using the electron transport chain components in the membrane, so NAD_ox will increase on the left side as before. (You have uncoupled ox. phos. from electron transport.) Some of the NAD_ox generated on the left will diffuse into the right compartment through the tear, reducing the difference in NAD_ox between the two sides. However NAD oxidation will continue on the left side only.

C2005/F2401 '06 -- Recitation Problems #5 -- Hints.

1. (i). Consider the yields of ATP and the NAD_red/NAD_ox balance. What conditions must be met for the bacteria to grow?

    (ii). Not sure how to figure out the yields of ATP etc.? Look at the standard pathway and see where DHAP fits in. Use the equations shown to figure out what it takes to get the DHAP; use the pathway to figure out what happens to the DHAP.

2   A. Will FAD donate electrons/hydrogens to NAD, or vice versa? Consider where the two electron carriers normally feed into the electron transport chain.

     B. Where will the electrons/hydrogens of reduced NAD go in the bacteria? Will they follow their usual path?

3. (i). If you have a piece of inner membrane in its normal place, and you have the compounds listed, what processes will take place? Ox. phos? Electron transport? Krebs cycle? Fermentation? Anything else?

    (ii). In the apparatus, one side (left or right) is equivalent topologically to the mitochondrial matrix, and one side is equivalent topologically to the space between the mitochondrial membranes. Which is which? Once you've figured that out, where do electrons & protons normally go during electron transport, ox. phos., etc.? What chemical reactions involving NAD, ATP etc. normally occur on each side of the membrane?

C2005/F2401 '06 -- Answers to Recitation Problems #6

1. A.  A = T;  B.  AA = TT.   These are the proportions in the original double-stranded molecules and they will be the proportions found in the hydrolysate. AT does not equal TA because the strands are antiparallel.

Detailed Explanation: The conditions of hydrolysis are sufficiently harsh to break all weak bonds (although the weak bonds are not "hydrolyzed" -- they are not broken by addition of water across the bond). Therefore you have no base pairs, only single nucleotides and dinucleotides connected by covalent bonds. The DNA was broken at random spots, so you can't say where any individual molecule was broken, but overall, in the total hydrolysate, the proportions of bases and dinucleotides will be the same as the proportions in the original molecules. (See * below.) In the original double stranded molecules, A = T and AA = TT but AT is not = TA. Why? For every sequence of AA in one strand there is a TT in the other. For every 5'AT3' in one strand there is a 3'TA5' (or AT) in the other. So in the original molecule, AA == TT but AT = AT not TA (if everything is written 5' to 3').

* Note: You have many, many, molecules. Some are cut one way, some another. On average, the more times TT, for example, appears in the original sequence, the more often TT will turn up in the collection of dinucleotides, even though not every TT is preserved in the dinucleotides. If the strands of DNA were parallel, then you would find AT = TA. An experiment like this was actually done to confirm that the strands were antiparallel. (This was back in the ancient days, in the 60's, when there was no way to sequence DNA.)

2. A. G is a purine and a base.

    B. G* DNA is NOT degraded to bases. The sugar (deoxyribose) must remain attached to the G* if the radioactivity can be recycled to DNA but not RNA. If the sugar were removed, and the DNA was degraded to bases, the *G could be re-incorporated into both DNA and RNA. Some of the G* could  be attached to deoxyribose and incorporated into DNA; some could be attached to ribose and incorporated into RNA.  But this does not happen – radioactivity from G*-DNA breakdown does not end up in RNA. So DNA must be broken down to nucleosides or nucleoside monophosphates. Then the G*-deoxyribose or G*-deoxyribose-phosphate is incorporated only into DNA.

    C. Phosphate/G should be = 0; Moles of phosphate released should be 2. When the G*-DNA is broken down to mononucleotides, there is one phosphate per G*. The phosphate is not radioactive (but the G* is). Two phosphates must be added from ATP to convert the dG*MP to dG*TP, before the G* can be incorporated into DNA. The two added phosphates may be radioactive, but they are released when the dG* (really d*GMP) is added to the growing DNA chain. The only phosphate incorporated into the DNA is NOT radioactive.

    D. No colonies. Under the conditions described here, the DNA that is taken up by the recipient is degraded to nucleotides, so it cannot transfer any genetic information. So the recipients will all be unable to degrade compound X and will not form colonies.

C2005/F2401 '06 -- Hints for Recitation Problems #6

1. A & B. In the experiment described here, you have used conditions of hydrolysis (probably acid) that break most, but not all covalent bonds.  Will you break weak bonds?

    B. Consider the dinucleotide AT. What is the complementary dinucleotide in the "other" DNA strand? Be sure you have correctly labeled the 5' ends of the two dinucleotides.

2. A. This is not a trick question -- it's as simple as it looks. It's included just to be sure you know the terminology.

    B. Diagram out the experiment: Make a picture or flow chart of what happens to the starting G*, how you get G*-DNA etc. You need to be able to follow the G* through all the steps. Then consider what would happen in each of the possibilities given  -- if the DNA is degraded to bases (or nucleosides, or nucleotides), will radioactivity end up in both RNA and DNA? Why or why not?

    C. When the G*-DNA is broken down into nucleotides, what do you get? G*MP? G*DP? Something else? When you reuse the nucleotides, to make new DNA chains, what form do you use? G*MP? G*DP? Something else?

    D. What would it take to get colonies here? What is required for successful transformation?

C2005/F2401 '06 -- Recitation Problems #7 -- Answers

1. A. U could be at the 5' end, and synthesis should involve primase, pyrophosphatase, & DNA polymerase. An Okazaki fragment is a discontinuously made section of the lagging strand at a DNA replication fork. It is involved in DNA synthesis, but synthesis of an Okaziki fragment starts with an RNA primer. You need primer (& primase) to start the fragment, because DNA polymerase can not start strands; it can only add to pre-existing ones. The primer is made of RNA -- that's why you can have a U in it. The U has to be at or near the 5' end, because the rest of the fragment is made of deoxyXTP's and has T, not U. The enzyme primase, which is a special type of RNA polymerase different from regular RNA polymerase, is needed to make the primer. Then DNA polymerase can add to the 3' end of the primer, elongating it. You don't need ligase to make a single Okazaki fragment; you only need ligase to shown individual fragments into a complete strand. Pyrophosphatase is needed to ensure that polynucleotide chain growth is energetically favorable, that is, that the overall ΔG of polymerization is very negative.

    B. Either one, pyrophosphatase, DNA polymerase.  All nucleic acid (polynucleotide) chains grow 5' to 3' by addition to their 3' ends. Primase is no longer needed, but both pyrophosphatase and DNA polymerase are needed as in A.

2. A. The DNA should look like this:      

3' AGGTC...........TGCC 5'

5' TCCAG...........ACGG 3'

    B-1. The top strand. The new chain must grow 5' to 3 and left to right, antiparallel to the template, so template must be the top strand.

    B-2. U will be on the 5' end.  The product must contain U because it is RNA.

    B-3. The bottom strand. The sense strand is the one that is not the template, and is the same as the mRNA, except that it has T instead of U.

    C-1. The bottom strand. Picture should describe or show one replication fork, not 2 bidirectional forks. Molecule should be a Y shape, with Y horizontal, and opening of Y on the left. New (leading) strand is made continuously, complementary to the top template strand; it grows 5' to 3', antiparallel to the template, left to right. New (lagging) strand is made discontinuously, complementary to the bottom template strand.

    C-2. CTGGA. Template strand on bottom contains sequence 5' TCCAG. New complementary lagging strand will contain the sequence 3' AGGTC 5' = CTGGA. (This sequence will be on the 3' end of the Okazaki fragment, not at the beginning, 5' end, which would include RNA primer.)

    C-3. The first Okazaki fragment.   Each segment (Okazaki fragment) of the lagging strand is made right to left & left-most fragment is made first.         D-1. Neither.  The primer 5' AGGTC 3' would not hybridize to any of the sequences shown. It is parallel, not antiparallel, to the left end of the bottom strand. The primer 3' AGGTC 5' would hybridize to the left end of the bottom strand, but it could not act as primer to copy the DNA shown, because chain growth, if any, would occur in the wrong direction. Polymerase always adds to the 3' end of the primer, which is on the left of 3' AGGTC 5'.  There is no DNA here to act as template for growth to the left, and no 3' end to build on for growth of primer to the right.

    D-2. Neither strand.  All synthesis in PCR is continuous. Both strands are completely separated before replication, and the complement to each strand is made separately.

    E. Transcription should be different; replication should be the same.     During DNA replication, both strands serve as template, and the entire molecule is replicated. The direction of the fork determines which new strand is made continuously and which is made discontinuously, but the end result is the same, no matter which direction the fork starts from.     During transcription of any section of the DNA, only one strand serves as template. The newly made RNA strand must be antiparallel to the template (transcribed) strand, and the new RNA must grow 5' to 3'. Therefore the direction of transcription determines which strand of the DNA can serve as template. If you reverse the direction of transcription, you reverse which strand can be template.  Note that the terms "leading" and "lagging" strands apply only to replication, not to transcription.

C2005/F2401 '06 -- Recitation Problems #7 -- Hints

1. A-1 How does a DNA chain get started?

    A-2. What reaction does each of those enzymes catalyze? Is the corresponding function needed, and if so, what for?

    B-1. How is the newest nucleotide/base added on? Is it the same for the leading and lagging strands?

    B-2. How is this case different than the situation in A-2?

2. A. Are the two strands parallel or not? Note that the top strand has its 3' on the left, not its 5'.

    B-1. Which way does the new strand grow? from it's 3' or 5' end?

    B-2. Are you making RNA or DNA?

    B-3. How is the sense strand related to the mRNA?

    C. Which is the lagging strand? The one that is synthesized continuously or discontinuously? Remember all chains must grow from their 5' ends toward their 3' ends, and each new chain must be antiparallel to the template.

    D-1. Where will each primer hybridize, if at all? When the primer is extended, which way will the chain grow?

    D-2. When do you need discontinuous synthesis, and when can you avoid it?

    E. How many strands are used as template for replication? Transcription?

C2005/F2401 '06 -- Recitation Problems #8 -- Answers

1. A. The mRNA shown is translated and transcribed left to right (relative to the way it is written in the problem). Its 5’ end is on the left and its 3’ end on the right.

    B. Arg; 3’GCU5’ or 5’UCG3’; 57 or 56. (57 when it is made; 56 after met is removed. The initial met is usually removed after the protein is made.) The first codon translated is AUG and the last one translated is CGA. The start codon doubles as a "start" and as a met codon; the stop codon UGA is not translated.

    C.   The 5' end of the transcribed (template) strand corresponds to the right end of the RNA shown, and the RNA shown was made (transcribed) from its left to its right. Therefore the DNA of the template strand should look like this:

           Promotor …................... 3 ’ T G C T A C T C A T 5’; transcription left to right, OR