Answers to Quiz 1.

1. Compared to S34, the radioactive isotope S35 has a. a different atomic number b. one more neutron c. one more proton d. one more electron e. a positive charge ANSWER: b. 2. T or F (please circle) A cation is an atom that has fewer electrons than does a neutral atom of the same element. ANSWER: T 3. Knowing the valences of C, O, N, S, and Al, and H; which of the following molecules is not likely to exist? a. c. b. d. e. ANSWER: both a and b are not likely to exist (could select either one). 4. T or F (please circle) Prokaryotic organisms are cells that lack a plasma membrane but do contain DNA for replication. ANSWER: F 5. Name the three recognized Domains into which 5 Kingdoms are currently classified? _____________________________ _____________________________ _____________________________ ANSWER: Bacteria, Archaea, and Eukarya. 6. Who was awarded the Nobel Prize in Medicine for the discovery of DNA? a. Leeuwenhoek b. Darwin c. Fleming d. Crick e. Hooke f. Van der Waal ANSWER: d. 7. Of the following elements, which is least prevalent in our body? a. Ca b. N c. Cl d. Fe e. P ANSWER: d. 8. If you traveled to South America and had to perform an experiment as close to room temperature as possible, what degrees farenheit would this approximate? a. 65 b. 68 c. 72 d. 74 e. 76 ANSWER: C x 1.8 + 32 = F RT in C is 20. Solve for F ANSWER: b. 9. Given the following information, how much MgCl2 would you weight out if you wanted to make 400 ul of a 2M solution? Mg = 12 electrons Cl = 17 electrons For calculation purposes assume that mass number = atomic weight ANSWER: number of electrons equal number of protons. Protons + Neutrons = mass number that approximates the atomic weight (given). THerefore MgCl2 would be 24 in atomic weight (Mg) + 68 in atomic weight (for 2 atoms of Cl). Therefore FW = 92 g (2)(92g/L)(1L/1000ml)(1ml/1000ul)(400 ul) = 0.0736 g or about 7.4 mg 10. Oxygen has a valence of a. 1 b. 2 c. 3 d. 6 e. 8 ANSWER: b. 11. The _____________ is the smallest unit of matter while the __________________ is the smallest unit of life. ANSWER: atom, cell Answers to Quiz 2.
1. A solution with a hydroxide ion concentration of 0.0001 a. is very acidic b. has a pH of 10-4 M c. has a pH of 4 d. has a pH of 10-10 M e. has a pH of 10 ANSWER: e. 2. Draw out at least three carbon skeletons for the molecular formula C6H14 (i.e. different isomers) ANSWER: see problems two and three on page 67 of your text book for similar scenarios. Looking for structural isomers of different branching patterns. 3. Which of the following macromolecules are formed from glycosidic linkages and dehydration reactions? a. hemoglobin b. sucrose c. phospholipids d. DNA e. testosterone ANSWER: b. 4. Which of the following organelles would you find in your dog but not in your pine tree? a. nuclear envelope b. lysosomes c. centrosomes d. mitochondria e. central vacuole ANSWER: b. 5. T or F The nuclear membrane is actually a double phospholipid bilayer because it consists of two membranes partitioned by the nuclear lamina. ANSWER: T 6. Why are mitochondria and chloroplasts considered semi-autonomous organelles? ANSWER: They have their own DNA and ribosomes but cannot live outside the confines of a cell. 7. Which of the following organelles functions to synthesize or make lipids? a. the golgi apparatus b. thylakoids c. mitochondria d. smooth endoplasmic reticulum e. the nuceolus ANSWER: d. 8. Draw out the defined functional groups of a naturally occurring amino acid? Label all functional groups. ANSWER: See purple text box on page 78 of your text book. 9. Circle the aldehyde group in the following organic molecule: ANSWER: Last three atoms on the right. 10. Which of the following regularly spaced H-bond is found in the secondary structure of a protein? a. Van der Waals b. disulfide bridge c. subunits d. alpha helix e. riboses ANSWER: d. 11. Which of the following purine-pyrimidine pair is correct for DNA-DNA complementary base pairing? a. A-G b. G-T c. A-U d. C-G e. T-C ANSWER: d. Answers to Quiz 3.
1. If you placed your red blood cell (normal osmolarity = 300 mosm) in a solution of 180 mM of NaCl, then you would expect your cell to a. shrink because it is in a hypertonic environment b. lyse because it is in a hypotonic environment c. become turgid because it is in a hypertonic environment d. become plasmolyzed because it is in a hypertonic environment Answer: a 2. T or F ? (Please circle) Cellular respiration is an exergonic reaction because it has a negative ?G, is a spontaneous process, and energy will be made available to do work. Answer: T 3. Which of the following is reduced by enzymes to speed the rate of reactions? a. the chemical bonds b. the free energy c. the activation energy d. the products e. the co-factors Answer: c. 4. An inhibitor that blocks the active site of an enzyme is referred to as a. a co-factor b. a noncompetitive inhibitor c. a competitive inhibitor d. an induced fit e. feedback inhibition Answer: c. 5. If you discovered a new life form that used a similar electron transport chain as that of humans, how many total ATP could be produced in a metabolic pathway that yielded 3 NADH and 6 FADH2? a. 9 ATP b. 18 ATP c. 21 ATP d. 24 ATP e. 30 ATP ANSWER: c. 6. If a noncompetitive inhibitor blocked the activity of co-enzyme A, so that acetyl CoA could not enter the Kreb’s Cycle.....what would be the net ATP yield from 2 molecules of glucose in the presence of oxygen? a. 2 ATP b. 8 ATP c. 16 ATP d. 28 ATP e. 36 ATP ANSWER: d. 7. In the Chemiosmotic Hypothesis, H ions flow down the concentration gradient established by the inner mitochondrial membrane (IMM) and activate what enzyme to produce ATP? a. phosphofructose kinase b. ATP synthase c. dehydrogenase d. cytochrome c e. substrate level phosphorylation ANSWER: b. 8. You come across your young nephew that has a large gum ball caught in his throat and cannot breathe oxygen. How many net ATP will he produce from 8 molecules of glucose if he cannot undergo oxidative phosphorylation and the pyruvate shuttle pump is unoperational? a. 4 ATP b. 8 ATP c. 16 ATP d. 32 ATP e. 48 ATP ANSWER: c. 9. During cellular respiration, there are two ways to synthesize ATP via phosphorylation: 1) Ninety percent of ATP is made in the mitochondria via the electron transport chain and is called _____________________________ phosphorylation. 2) Ten percent of ATP is made by _____________________________ phosphorylation that occurs in the cytosol and in the mitochondria but not via the chemiosmotic hypothesis. ANSWER: oxidative, substrate-level 10. T or F ? (please circle) Enzymes can react with specific substrate at a rate of 1000 times per second and each time emerge in their original form (never consumed or used up). ANSWER: T 11. What are the three types of energy important to most organisms? 1. _________________________ _________________________ 2. _________________________ _________________________ 3. _________________________ _________________________ ANSWER: Potential energy, chemical energy, kinetic energy Answers to Quiz 4.
1. Which of the following scientists discovered bacteria transformation? a. Griffin b. Crick c. Danieli d. Avery e. Morgan ANSWER: a. 2. How many nucleotide bases are there for every helical turn of DNA? a. 2 b. 5 c. 10 d. 4 e. 20 ANSWER: c. 3. Using the provided mRNA codon table, if a protein had the peptide sequence phe-pro-lys, which of the following could be the proper 5' to 3' DNA template? (hint: the table provides 5' to 3' RNA sequence and remember DNA is antiparallel) a. uuc ccu aaa b. uuu cca aag c. ttt agg gaa d. aaa ttc ggg e. more than one of the above ANSWER: c. 4. Which of the following joins the Okazaki fragments together in the lagging strand during DNA replication? a. primase b. ligase c. RNA primer d. DNA polymerase I e. DNA polymerase II ANSWER: b. 5. Which of the following types of RNA would you never find in the nucleus? a. mRNA b. tRNA c. pre mRNA or primary transcript d. snRNA e. SRP RNA ANSWERS: b. or e. 6. T or F (please circle) The synthesis of RNA under the direction of DNA is called RNA processing whereas the synthesis of a protein under the direction of mRNA is called transcription. ANSWER: F 7. Using the provided mRNA codon table, what is the produced peptide (small protein) for this dna sequence? 3' taa tac tct agg tcg cat act cca gtg 5' WORK AREA - To work this problem....first start by calculating the 5' to 3' nucleotide sequence of mRNA complementary to the DNA sequence above (remember anti-parallel) = ____________________________________________________ ANSWER: auu aug aga ucc agc gua uga ggu cac Now use the codon table to give the protein, amino acid sequence = ________________________________________________________________________ ANSWER: met arg ser ser val a. met-arg-ser-ser-val b. arg-met-arg-ser-thr-gly-his-arg c. met-his-gly-thr-leu-ser-ser-arg d. met-phe-gly-ser-ala e. met-leu-pro-pro-arg-his-gly-arg ANswer: a. 8. Which of the following is not involved in transcription of DNA to the RNA primary transcript? a. RNA primer b. RNA polymerase II c. TF = transcription factor d. TATA box e. promoter ANSWER: a. 9. During translation of mRNA to protein a. the start codon met is positioned at the A site. b. a transcription factor (TF) must bind to the large ribosomal subunit to initiate the first assembly of the ribosom. c. a release factor hydrolyzes the growing peptide by blocking the E site or exit. d. the growing peptide chain is moved from the P site to the A site when the peptide bond is formed. e. the small ribosomal subunit uses ATP and catalyzes the formation of the peptide bond ANSWER: d. 10. T or F (please circle) During the process of RNA splicing, the primary transcript may be shortened by the sliceosome that removes intervening non-coding portions called exons. ANSWER: F 11. Which set of scientists creatively used radioactive labeling and a generic waring blender to determine for the first time that dna, and not protein, was the transforming material? a. Watson and Crick b. Hershey and Chase c. Avery and Chargaff d. Wilkins and Franklin e. Meselson and Stahl ANSWER: b. Answers to Quiz 5.
1. T or F (please circle) In the two chambered heart of the fishes, both the ventricle and atria of the heart contain relatively deoxygenated blood. ANSWER: T 2. Which of the following animals has no mixing of oxygenated and deoxygenated blood within the heart due to having a completely separated ventricle chambers? a. the earthworm (annelids) b. the sea star (echinoderms) c. sharks d. turtles e. one of those Gainesville reptiles ANSWER: e. 3. Blood contained in the left atria a. will pass through the tricuspid valve to reach the right ventricle b. will be squeezed out to the right ventricle during the cardiac period called systole c. would have just left the pulmonary artery d. might be picked up as the fourth sound on the stethoscope during the contractile period of the atria e. will have just passed through the aortic semilunar valve ANSWER: d. 4. If you have mitral stenosis (hardening of the mitral/bicuspid valve) you would hear irregularities during which sound of a phonocardiogram? a. S1 b. S2 c. S3 d. S4 ANSWER: a. 5. Which of the following animals use a tracheal system to breath? a. clams b. crustaceans c. frogs d. turtles e. grass hoppers ANSWER: e. 6. Which of the following vessels possess back wash valves? a. arterioles b. venules c. metaarterioles d. capillaries e. lymph nodes ANSWER: b. 7. If you centrifuged a blood sample, which of the following items would likely be at the lowest level of the test tube towards the bottom? a. platelets b. neutrophils c. unbound oxygen d. fibrin e. water ANSWER: b. 8. Which of the following cellular elements is beneficial for clotting after vessel damage? a. stem cells b. monocytes c. macrophages d. erythrocytes e. platelets ANSWER: e. 9. What is the name of the electrical signal that is the basis for all your brain thoughts on this evaluation? a. the potassium channel b. neurotransmitters c. a hyperpolarization d. the synapse e. the action potential ANSWER: e. 10. According to Boyle’s Law which is a true statement with regards to respiration? a. At exhalation the diaphragm relaxes to increase the volume in the lung. b. When the external intercostals contract, air moves in due to larger lung volume. c. At inhalation, the diaphragm moves up to decrease lung volume and force air in. d. At inhalation, air moves into aveoli as they enlarge due to relaxation of the diaphragm. e. When the rib muscles fall down with gravity and move in, we take an inspiration (inhale) because the lung volume has increased. ANSWER: b. 11. During the repolarization phase of an action potential, which of the following events help drive the membrane potential back down towards a resting state? a. absolute refraction b. opening of delayed K channels c. activation of voltage-dependent Na channels d. operation of the Na/K ATPase pump e. Ca channels ANSWER: b.

Answers to Exam 1. Compositions Written (unedited) by Your Colleagues as Good Example Responses -

Answers to Exam 2. Compositions Written (unedited) by Your Colleagues as Good Example Responses -

Essay 1: By Aline Rackley

Below the molecule of ATP drawn out, with the three phosphate groups, ribose, and adenine labeled correctly.

DNA also has the functional group adenine. Also, proteins contain phosphate functional groups.

Hydrolysis: ATP ? ADP + Pi with delta G= -7.3

The hydrolysis of ATP is coupled to other reactions to allow them to proceed spontaneously. When ATP is hydrolyzed, a phosphate group is removed from this molecule which involves the breaking of bonds. This breaking of bonds releases energy which is used to spontaneously start other reactions. For example, during the Calvin Cycle in photosynthesis, a 6C sugar with a phosphate group is hydrolyzed to release a carbon for glucose requiring the energy of ATP.

Essay 1: By Natalie Nadeau

Below correctly drawn molecule.

Other molecules we have studied include carbohydrates, which can have a pentose structure like ribose. There are also hexose sugars like glucose, fructose, and galactose found in the foods we eat. Another molecule is an amino acid, which is a subunit of a protein. These contain amine groups, which have nitrogenous bases like adenine. Adenine is also a component of nucleic acids and is one of the building blocks of DNA.

When ATP is hydrolyzed, one of the phosphates is removed and transformed into an inorganic phosphate. This inorganic phosphate can be coupled to an intermediate, and with this inorganic phosphate’s presence, reactions can proceed spontaneously and produce an exergonic – delta G.

Essay 2:

The membranes of cells are selectively semi-permeable, meaning that only certain molecules/compounds/etc can enter the cell without help of energy (active transport) or facilitated diffusion. For example; water, sugar and ions cannot permeate the cell. For this reason, we have the Na/K pump, which pumps 3 Na+ out of the cell (against its concentration gradient) for every 2 K+ pumped into the cell (also against its concentration gradient). Because there are 3/2 as many + charged ions outside the cell, the inside of the cell becomes extremely negative, with high potential energy due to the high voltage difference. This is vital and worth the high cost in energy because it allows for symport—allowing otherwise impermeable molecules to enter the cell. ie: Na+/glucose symport: Na+ is built up outside the cell and when it finally goes down its concentration gradient, glucose gets a “free ride.”

Essay 3: By Kate Harrison

The main idea of the Chemiosmotic Hypothesis is that electron flow couples energy production. This means that the state of electrons plays a significant role in metabolic reactions, based on potential energy (unstable, excited state) and the release of energy (stable, ground state). In animals, cellular respiration produces ATP in the mitochondria. Hydrogen accumulates in the inner mitochondrial membrane and it acts as an electron donor in redox reactions. In plants, photosynthesis and the Calvin cycle act as a reverse electron flow because ATP is used for the production of glucose, which goes on to cellular respiration. Photosynthesis occurs in the thylakoids, while the Calvin cycle (“dark reaction”) occurs in the stroma. Plants are autotrophs that produce their energy using light, with O2 as a waste product. Animals, heterotrophs, use this oxygen for cellular respiration, and they also consume the plants as well as other animals. In cellular respiration, the last step is the electron transport chain, which uses coenzymes to transfer electrons to hydrogen carrier NAD+. Gradients also are related to electron flow; the voltage difference between the inside and outside of a cell affects the molecules that move in and out, because the inside is more negative. The Chemiosmotic Hypothesis explains many elements of energy production and provides a better understanding of reactions in both plants and animals.

Essay 3: By Sheri Rahman

Peter Mitchell’s Chemiosmotic Hypothesis explained how ATP production is coupled to electron transport. The hypothesis states that NADH allows an electron from hydrogen to move down the electron transport chain in the inner mitochondrial membrane and it then accumulates, causing the membrane of the mitochondria to become acidic. The hydrogens then move down their concentration gradient to the mitochondrial matrix with the help of an ATP synthase pump. It is different between plants and animals because of the spatial organization of the electron transport chain and the source of the electrons. In plants, the electron transport chain resides in the thylakoid membranes of the chloroplasts, while in animals, it is in the inner mitochondrial matrix. Also, plants get their electrons from light energy, while animals get them from the hydrogen in glucose. The purpose of the hypothesis is to explain how hydrogen concentration and electron transport affects production of ATP and the way in which everything is coupled together.

Answers to Exam 3. Compositions Written (unedited) by Your Colleagues as Good Example Responses -

Essay 1

Hershey and Chase used radioactivity to answer the question of whether protein or DNA was the genetic material. They labeled the DNA in a T2 phage with radioactive phosphate, and labeled the protein coat of the T2 phage with radioactive sulfur. The T2 would infect E. coli. The T2 phages were then chopped off of the E. coli with a blender and the substance was centrifuged. As E. coli is significantly larger than the virus, the two would separate. It was found that the E. coli was labeled with radioactive phosphate and that DNA was therefore the genetic material. The negative control was labeling the protein with radioactive sulfur. Had protein been the genetic material, the E. coli would have been labeled with radioactive sulfer.

Years later, Meselson and Stahl set out to determine which of the following models accurately demonstrated DNA replication: conservative, semi-conservative or mosaic. Meselson and Stahl cultured E. coli in radioactive 15-N and then placed it in a 14-N environment. After 20 minutes (one replication cycle), the E. coli was centrifuged. If the conservative model were to be true, two layers would have appeared- the heavier parent DNA labeled with 15-N and the lighter daughter copies labeled with 14-N. This was not the case. Instead there was one thick band containing 14-N and 15-N, leaving the semiconservative model and mosaic model as palpable options. After another 20 minutes (another replication cycle), the E. coli was centrifuged again. With the mosaic model, Meselson and Stahl would have found one bold layer containing 14-N and 15-N again. Instead, they found one layer of 14-N and 15-N and another layer with just 14-N, proving the semiconservative model of DNA replication (the 14-N E. coli used the 14-N daughter strand from the first replication as template. 14-N + 15-N still had one parent strand and a 3rd generation daughter strand). By knowing what the other possible outcomes would have meant,Meselson and Stahl had an appropriate negative control.

Essay 2

The first notable group of scientists that used radioactivity to an experimental advantage were Alfred Hershey and Martha Chase. In the 1950s, Hershey and Chase were trying to decipher if the DNA or proteins were the genetic material. To figure this out, they used T2 bacteriophage, a virus that infects E. coli. To perform their experiment, they labeled the protein coat of the T2 with radioactive sulfur (35-S) and the DNA of the phage with radioactive phosphorus (32-P) They then allowed the T2 virus to infect the e. coli and after this sheared off the head of the T2 using a warring blender. The resultant mixture was then centrifuged, and since the bacteria was longer than the T2 that meant it would be heavier and therefore form a pellet. Once the mixtures had been centrifuged they looked at the results. They found that the radioactivity of the phosphorous had travelled to the pellet. That meant that DNA had been injected into the e.coli, and therefore was the genetic material. Since the radioactive sulfur was not found in the pellet, that meant that the protein had not been injected into the e. coli, so this can be viewed as alternate result that negated other standing theories. The other group of scientists that used radioactivity as an experimental advantage was Meselson and Stahl. Meselson and Stahl were trying to figure out which of the existing theories of how DNA replicated was actually correct. At the time the three possible theories were the conservative, semiconservative, and dispersive theories. To determine which theory was valid, they cultured several generations of e. coli bacteria in a medium that had the heavy radioactive nitrogen (15-N). They then transferred the bacteria into medium that had the lighter 14-N. From this they took 2 DNA samples. Because replication occurs in 20 minutes, they took the first sample at 20 minutes to see the first generation DNA and then took another sample at 40 minutes to see the second generation of DNA. Therefore, they were using staggered timing and differentiated centrifugation to determine which one was the valid theory of replication. For the sample taken at 20 minutes, their results showed a hybrid bond of DNA which automatically disproved the conservative theory of replication. The sample taken at 40 minutes showed a light band (14-N) and a hybrid band of DNA. This proved the semiconservative theory of replication and disproved the dispersive model.

Essay 3

Two great advances concerning the knowledge of DNA were discovered advantageously using radioactive labeling. The first such experiment was designed by Alfred Hershey and Martha Chase to put to rest the long standing debate over what actually was the genetic medium: protein or DNA? Hershey and Chase radioactively labeled T2 bacteriaphages; the protein component with radioactive sulfur and the DNA component with radioactive phosphorous. Then, they let the T2 phages react with E. coli bacteria, using a warring blender to shear off the protein heads of the phages and centrifuging the resultant mixture of bacteria with injected T2 DNA. After analysis of the supernatant and recognizing which radioactivity- if any- were a part of the new DNA mix- they found radioactive phosphorous in the bacteria (with DNA). This experiment provided evidence that the genetic material was in fact DNA.

The second group of scientists who utilized radioactivity with a classic experimental design was Meselson and Stahl. They were seeking to discover which method of replication DNA used: conservative, dispersive, or semiconservative. They used two types of radioactive Nitrogen, 14-N and 15-N, allowed these separate mixtures of radioactively labeled DNA to replicate at varying times, and centrifuged the mixture. In conservative theory, there is one parent strand and one daughter strand which after one replication would result in two bands, a 14-N band (which is lighter), and a 15-N band. This was not the case, as after one replication there was only one band where the amount of 14N = 15-N. This eliminated the conservative theory. After two replications, the dispersive theory, in which the resulting strands are a mixture, or mosaic of daughter and parent, there would be only one band, where 14-N =15-N. This was not the case, and there was one band of 10-N and one band of 15-N =14-N. This eliminated the dispersive theory and favored the conservative theory, which the parent strand is a template for the daughter, so a double helix would include a parent and daughter.

Essay 4

One of the notable groups was Hershey and Chase. They set to determine the transformed material was really DNA or the protein capsule. To do so they used the T2 bacteriophage that infected E. coli. With the bacteriophage, the DNA inside the phage was labeled with 32-P (radioactive) and the protein coat was labeled with radioactive 35-S. The T2 phage was allowed to infect E. coli. The heads of the bacteriophage were sheared off using a waring blender, and the supernatant was centrifuged. In the centrifuged solution, 32-P was found, indicating that the transforming material was DNA. This discovery helped prove Oswald Avery’s initial hypothesis that DNA was the transforming material, not the protein capsule. This also supported Griffin’s earlier experiment as it again showed that the transforming material was the DNA.

Another notable group was Meselson and Stahl, who proved Watson and Crick’s theory of semi-conservative replication. In their experiment they labeled one half of the DNA double helix with radioactive 14-N and the other with 15-N. By labeling these strands separately they could distinguish between the original strand and the new strands as 14-N is lighter than 15-N. In their experiment, they allowed for the labeled DNA to replicate one round, which revealed one band of DNA (post-centrifuge), which eliminated the conservative hypothesis meaning that at the end result of replication, there would be 1 parent strand and a completely new daughter strand.

They then allowed the DNA to replicate again. The DNA was centrifuged and the resultant mixture showed one 14-N band and one 15-N band. This proved semi-conservative replication because the original strand was preserved as well as creating the new daughter strands. It also disproved dispersive replication; if it was dispersive there would be only one 14N, 15N band.

Essay 5

Hershey and Chase used radioactive isotopes of sulfur and phosphorous to help them test whether protein or DNA was responsible for genetic material. The two noticed that the T2 that infected the e. coli bacteria could rapidly reproduce itself and then reprogram the cell to infect other cells. They injected a radioactive isotope of sulfur into the protein in one batch of T2 and a radioactive isotope of phosphorous into the DNA of a second batch of T2. Hershey and Chase then allowed the T2 to infect cells. When tested, they discovered the radioactive isotope of phosphorous in the DNA had made its way inside the infected cell and was multiplying. This gave them proof that DNA was responsible for genetic material. They used protein as the other test because at the time of the experiment, protein was thought to contain genetic material.

Watson and Crick used radioactivity in their experiment as an advantage by using the X-ray crystallograph made by Franklin to help them assemble their DNA model. Using the X-ray picture, they were able to determine that the structure was 2 nm wide, made 1 helical turn every 3.4 nm, and bases .34 nm apart, and was right handed. Using the 2 nm width, they determined the structure was a double helix, not the previously believed triple strand. Also using the width and known width of the bases, they were abel to determine A and T must be paired while C and G were paired. The 3.4 nm for every 1 turn and the distance between each base pair allowed Watson and Crick to say there were ten base pairs for every helical turn. Using all the information deducted from Franklin’s X-ray, Watson and Crick were able to publish their first paper, which contained the structural model of DNA, which is still accepted today.