F24 Biology 210/212 Exam 2 Key PDF

Name____________________________ Key UMS number______________________ Circle one: 210 or 212 Lab Section ______...

Name____________________________ Key UMS number______________________ Circle one: 210 or 212 Lab Section ______ Biology 210/212 Exam 2 – November 4, 2024 Instructions: 1) TWO points for correctly filling in the information at the top of every page. If you are not in a lab, write “none” where it asks for the lab section. If you don’t know the lab section, day and time or TA will suffice. 2) This exam is worth 100 points. The point value for each question is in parentheses. Fo 3) You have 50 minutes to complete this exam. If you get stuck on a question, move on to an easier question r and return to the more difficult questions later. (Do the easy ones first!) Fa 1. Consider step 1 of glycolysis, where ll glucose is phosphorylated by hexokinase. 20 i) For this reaction, the creation this 24 phosphate bond has a ∆G°= 12.8 kJ/mol. Is this an energetically favorable Bi reaction? (2 points) o No. 21 ii) In the cell, the cell couples the hydrolysis of the phosphate bond in ATP to the creation of 0/ glucose-6-phosphate. (See reaction above). We know that the creation of ATP from ADP (ADP + 21 Po —> ATP) has a ∆G° = 30.6 kJ/mol. Given this information, can you: a) calculate ∆G° for step 1 of glycoloysis and b) explain how this coupling affects the ∆G° of the reaction (compared to part i)? 2 Show your work for full credit. (5 points) st a)ATP —> ADP + P deltaG°= -30.6 kJ/mol; creation of the phosphate bond is deltaG°=12.8 kJ/mol ud -30.6 kJ/mol + 12.8 kJ/mol = - 17.8 kJ/mol (must get math correct to get full credit). b) Coupling the en unfavorable creation of the phosphate bond to the favorable hydrolysis of ATP makes this a favorable ts 2. You decide to isolate the hexokinase gene from Drosophila, where the genome sequence is rxn (deltaG° on known. The map below shows the hexokinase gene. The striped area indicates where the hexokinase gene is located and sequences flanking the hexokinase gene are shown in is now negative). ly enlargements..N o sh a rin g i) Circle the primers that could be used to amplify the hexokinase gene from Drosophila DNA. (6 w points) ith 5’-GCGTAGTGTAGTACA-3’ 5’- AAAAAAAAAAAAAAA -3’ ou tp 5’-CATAGGCCTGTACTA-3’ 5’-GTATCCGGACATGAT-3’ er m 5’-TCAGGTAGATGCGTA-3’ 5’-AGTCCATCTACGCAT-3’ is s 5’-TGGTACTTTCATCGC-3’ 5’-ACAATGAAAGTACGC-3’ io ii) For the primer that hybridizes to the top strand, write out the next seven bases that will be n. added by Taq polymerase (the DNA polymerase used in PCR) to the growing DNA strand. Please label the 5’ and 3’ ends. (4 points) 5' - CTTTCAT - 3' ©Linda Huang, University of Massachusetts Boston, 2024, All Rights Reserved. This is for UMass Boston Biology 210/212 students only. Sharing without permission is not allowed.  Name____________________________ Key page 2 of 5 3. You are studying an antiport, Pump2, that pumps one molecule of magnesium (Mg2+) ouside the cell for two potassium ions (K+) that are moved into the cell. Pump2 uses energy from ATP to move molecules across the membrane. Experiment 1: You make an artificial membrane vesicle that has Pump2 as the only protein on the membrane. Pump2 is oriented on this artificial membrane vesicle as it is oriented in a cell, and works Fo just like it would in the cell, moving Mg2+ out and K+ in. At the beginning of your experiment, the r concentration of Mg2+, K+, and ATP are each at 100 µM both inside and outside of the vesicle. Fa ll Under these conditions (2 points each, 4 points total): 20 i) Which ion(s) will be at a higher concentration inside the vesicle after 1 minute? _______ K+ 24 Bi ii) Which ion(s) will be at a higher concentration outside of the vesicle after 1 minute? _______ Mg++ o 21 iii) Will a charge difference across the membrane be established in 1 minute under these conditions? 0/ Explain. (4 points) 21 No. Because one Mg2+ moves in for every 2 K+, this movement is electrically neutral, since 2 you are exchanging two positive charges in for two positive charges out. st ud Experiment 2: You make artificial membrane vesicles that have the potassium leak channel, Pot3, as en the only protein on the membrane. At the beginning of your experiment, you have 100 µM of Mg2+, ts K+, and ATP outside the vesicle and only 100 µM ATP on the inside of the vesicle. on ly iv) You wait two minutes and reexamine the solution inside the vesicle. Do you predict it will contain.N K+? Explain. (4 points) o Yes, the electrochemical gradient of K+ will cause it to move in through Pot3. There is both a sh concentration difference, with much much more K+ outside, as well as a large charge difference a across the membrane, with an excess of positive charges outside of the vesicle at the start of the rin experiment. g Experiment 3: You now make artificial membrane vesicles that have both Pump2 and Pot3, all oriented w as they would be on a cell membrane. At the beginning of this experiment, both the inside and outside ith of the vesicle will contain 100 µM each of Mg2+, K+, and ATP. ou tp v) Given these conditions, which ion(s) will be at a higher concentration outside of the vesicle after 1 er minute? (2 points) Mg2+ m is vi) After 1 minute, which ion(s) will be closest to its equilibrium potential? Explain. (5 points) s io K+. The membrane is permeable to K+ because it contains a leak channel, so K+ will flow down n. it's electrochemical gradient every time the channel opens. There will be a concentration gradient that is caused by Pump2 movement of ions and there will eventually be a charge difference across the membrane as K+ moves down its electrochemical gradient. The membrane is not permable to Mg2+. ©Linda Huang, University of Massachusetts Boston, 2024, All Rights Reserved. This is for UMass Boston Biology 210/212 students only. Sharing without permission is not allowed.  Name____________________________ Key page 3 of 5 AGA UUA AGC AGG UUG AGU GCA CGA GGA CUA CCA UCA ACA GUA GCC CGC GGC AUA CUC CCC UCC ACC GUC UAA GCG CGG GAC AAC UGC GAA CAA GGG CAC AUC CUG AAA UUC CCG UCG ACG UAC GUG UAG GCU CGU GAU AAU UGU GAG CAG GGU CAU AUU CUU AAG AUG UUU CCU UCU ACU UGG UAU GUU UGA A R D N C E Q G H I L K M F P S T W Y V stop Fo 4. Your friend wants to express GFP (Green Fluorescent Protein) under the control of the promoter from the lac operon. They are happy because they find Xba1 and EcoRV sites that cut before the start r Fa of transcription for the LacZ gene (see A in Figure below). They cut out this piece of DNA and put it into their plasmid (see B of the Figure below), which they also cut with XbaI and EcoRV. (Recall that ll the CAP activator binds to the CAP binding site and the Lac repressor binds the operator.) 20 24 i) When they cut the plasmid (Fig B) with Bi Xba1 and EcoRV, how o many linear pieces will they create? (2 points) 21 0/ 21 2 2 st ii) Can they find this Xba1-EcoRV fragment from the lac operon (Fig A) in a cDNA library? Explain ud briefly. (3 points) en No, cDNA libraries are made from mRNA and do not contain regions that are not transcribed, ts such as the regulatory region of the LacZ gene. on ly.N iii) What enzyme will they need to use to join the nicks in the sugar phosphate backbone so that they can create the new plasmid with the Xba1-EcoRV portion from the lac operon promoter and GFP? (2 o points) sh DNA ligase iv) Your friend puts the successfully created plasmid into bacteria and adds media that does NOT arin contain either glucose or lactose. Do you think that GFP will be produced? Explain. (7 points) g Yes. The operator sequence that binds to the lac repressor is not included in the Xba1-EcoRV w fragment, so the promoter will not respond to lactose. Because there is no glucose, cAMP levels will ith be high and CAP will bind to the CAP binding site and recruit RNA polymerase to transcribe ou the GFP gene under these conditions. tp er v) The enlargement from the Figure above shows part of the sequence of the LacZ gene. This part of m the gene encodes the peptide SCRKML (peptide is written from N-terminus to C-terminus). Given is this information, write out the sequence of the portion of the RNA that is made from the LacZ gene s (label 5’ and 3’ ends) below and indicate how it encodes that peptide (7 points). io n. 5' = G AGC UGC CGU AAG AUG CUU AG =3' n term - S C R K M L - c term Must show the translation explicitly to get full credit, with peptide shown in diagram. ©Linda Huang, University of Massachusetts Boston, 2024, All Rights Reserved. This is for UMass Boston Biology 210/212 students only. Sharing without permission is not allowed.  Name____________________________ Key page 4 of 5 5. You have discovered an enzyme called Glp2 that is important for fat cell function. You isolate the DNA from the regulatory region of Glp2 and place this DNA upstream of the gene for green fluorescent protein (GFP). You put this into mice and find GFP produced in the fat cells and not the blood cells, as expected (experiment 1). You discover there are two regions important for proper cell type expression, labelled A and B in the Figure. These two regions bind different gene regulatory proteins: DBP1 binds to sequence A and DBP2 binds to sequence B. To determine the function of these proteins (DBP1 and DBP2), you create additional pieces of Fo recombinant DNA that have different combinations of sequence A or B; if the piece of DNA is missing the binding site, it has been r removed and will no longer bind Fa the protein. You put these new ll pieces of DNA into a different 20 set of mice and look at fat and blood cell expression. (There are 24 a total of 4 mouse strains – one Bi strain per construct. Results are o shown on the Figure on the right.) 21 0/ From the data in the Figure, 21 answer the following questions: 2 i) Which binding site binds to a gene activator? (3 points)_____________________ B st ud ii) Which binding site binds to a gene repressor? (3 points) ____________________- A en iii) From this data, do you predict that you can find DBP1 protein in fat cells? Explain your data, citing ts the appropriate experiments that support your claim. (8 points) on I would accept any answer that discussed scenarios consistent with the data. Answers must ly explicitly state what protein is binding in which cell, and what experiments allow you to draw.N that conclusion. The simplest most straightforward answer is: o No - Experiment 4 shows that DBP2, the activator, is present in both fat cells and blood cells, sh as you can see expression in both cell types when site B is present. Experiment 3 a rin suggests that DBP1, which binds to site A, may be a repressor, but does not tell us what g cell type DBFP1 is expressed in. Experiment 1 shows no expression in blood, even in w the presence of DBP2 (which we saw in experiment 4), and suggests that when both DBP1 ith and DBP2 are present in blood, the repressor will turn off expression in blood. Thus, it is unlikely ou that the repressor is present in fat cells, as experiment 1 shows expression in fat cells and tp experiment 4 shows that DBP2 is present in fat cells; if DBP1 were present in fat, there would not be er fat cell expression in experiment 1. m Explanations that do not cite and interpret experiments did not receive full credit. is 6. Circle the amino acid(s) below that would be favored for use in an a-helix that spans the lipid s bilayer. (8 points) io n. valine histidine aspartic acid arginine alanine lysine isoleucine leucine ©Linda Huang, University of Massachusetts Boston, 2024, All Rights Reserved. This is for UMass Boston Biology 210/212 students only. Sharing without permission is not allowed.  Name____________________________ Key page 5 of 5 7. For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once. (2 point each; 10 points total) 50 1492 polypeptides 1842 peripheral 2001 ribosome 20,000 Fo lipid 99 terminase amino acid rFa 2023 vacuole endosome 1 ll 1 million nucleus 1776 nucleotide 20 24 The sequence of the human genome was first published in _____________________. 2001 Sequencing Bi the genome relied on our knowledge of replication and DNA polymerase function, to create a o 21 situation where the addition of a dideoxy-_____________________ would cause a stop in 0/ nucleotide 21 2 replication. We now know humans have about _____________________ 20,000 protein coding genes. st ud Surprisingly, only about _____________________ 1 percent of the human genome codes for protein en ts coding exons. In a eukaryotic cell, the genome can be found in the _____________________. nucleus on ly 8. Write true or false on the line following each sentence. If a statement is false, cross out the.N incorrect word or phrase in the sentence and write the correct word on top of the incorrect word. o true a) Human cells add cholesterol to their membranes to make them stiffer. _______________ sh cyclical a rin b) The citric acid cycle is a linear pathway that takes place inside the mitochondria and creates g NADH molecules for energy. _________________ false w ith c) When there is not enough oxygen, the pyruvate produced from glycolysis does not enter the ou tp mitochondria and is converted to lactic acid. _____________ true er amphipathic m d) Detergents are hydrophilic molecules that are used for the study of membrane proteins because is s false they can interact with the charged and non-polar parts of lipids. _______________ io n. e) Cloning vectors need an origin of replication so that the DNA will be replicated by the cell. true _______________ ©Linda Huang, University of Massachusetts Boston, 2024, All Rights Reserved. This is for UMass Boston Biology 210/212 students only. Sharing without permission is not allowed.