Post 1 Lec 3-5 : Discussion Board Answers PDF

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Bio 202.01 Spring 2020 Discussion Board Clarification Post 1 Lectures 3-5 Chapter 4: Molecular Diversity in Biology 1. Aldehydes

Which molecule shown above is an aldehyde? A. Molecule A *B. Molecule B C. Molecule C D. Molecule D Response: Although aldehydes have a carbonyl group at the end of the carbon skeleton, this does not mean that just because a carbonyl is attached to the last carbon in the chain, the structure is always an aldehyde. The carbon that is attached to the carbonyl must also be attached to a hydrogen or two hydrogens. If the carbonyl carbon is attached to a hydroxyl group (-OH) as well, then it is a carboxyl group, and not an aldehyde. This supports choice B as the best answer. As far as the structures presented in this question, molecule A is an alcohol, because it contains a hydroxyl group, (-OH). Molecule B is an aldehyde because as stated, it contains a carbonyl at the end of the carbon chain, and this carbon is bound to a hydrogen, as well as another carbon however the other carbon is not necessary for this to be an aldehyde. Molecule C is a ketone because the carbonyl group is within the carbon skeleton, and there is a carbon double-bonded to oxygen that is bound to two other carbons. Molecule D is a carboxylic acid because it contains a carboxyl group in which a carbon is both double bonded to an oxygen as well as being bound to an oxygen of a hydroxyl (-OH) group.

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Although it was not an answer choice, structure E can be identified as an amine, because it contains an amino (NH2) group. 2. Structure of Organic Molecules Organic molecules with only hydrogens and five carbon atoms can have different structures in all of the following ways except A. by varying the number of double bonds between carbon atoms. B. by varying the position of double bonds between carbon atoms. C. by forming a ring structure *D. by forming enantiomers. Response: This question tests your understanding of isomers with respect to the molecular structure of carbon. Carbon has four electrons in its valence shell that it can share either with the hydrogens or the other four carbon atoms described in the question. The carbon atoms can form double or single bonds in different positions like those found in structural and cis-trans isomers. The carbon skeleton can be straight, branching, or in a ring structure such as cyclopentane. This falsifies choices A – C. Enantiomers are mirror images of each other due to the presence of an asymmetric carbon. An asymmetric carbon is a carbon that is attached to four different atoms or molecules. The organic molecules described in the question contain only carbon and hydrogen and therefore cannot form enantiomers. See pages 61-62 of the Campbell Biology textbook for a detailed description of isomers.

3. Cis-Trans Isomers Which of the following statements correctly describes cis-trans isomers? A. They have an asymmetric carbon that makes them mirror images. B. They have different molecular formulas. C. Their atoms and bonds are arranged in different sequences. *D. They have variations in arrangement around a double bond. Response: An isomer is one of at least two compounds with identical molecular formulas, but their atoms arranged in different ways. Choice A can be falsified because it describes enantiomers not cistrans isomers. Choice B can be falsified because isomers of all types have the same molecular formula, this is part of what makes them isomers. Choice C can be falsified because the double bonds of cis-trans isomers are found between the same carbons in both isomers. Cis-trans isomers (aka geometric isomers) are isomers that have identical covalent bond arrangements but different spatial arrangements around the double bonds between two carbons. This identifies choice D as the best answer.

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4. Bonding Functional Groups What type of chemical bond joins a functional group to the carbon skeleton of a large molecule? *A. covalent bond B. hydrogen bond C. ionic bond D. disulfide bond Response: This question tests your understanding of different bond types given the circumstances of functional groups binding to a carbon skeleton. Prof. Erster explained in the Chapter 4 lecture that carbon can share its four valence electrons with other atoms to fill its valence shell. This sharing of electrons describes a covalent bond. This supports choice A. Carbon cannot participate in hydrogen bonding because it is not electronegative enough to form a polar covalent bond, which is essential for hydrogen bonding, falsifying choice B. Ionic bonds are unlikely to form between a carbon skeleton and a functional group because carbon is not an ion, and electrons would be stripped not shared, making the carbon skeleton less stable. This falsifies choice C. Disulfide bonds, as the name implies, are formed between two sulfhydryls which do not include carbon, falsifying choice D and confirming choice A as the best answer. Chapter 5: Macromolecules – Carbohydrates, Lipids, and Nucleic Acids 5. Glucose and Fructose Examine the two molecules in the figure below. Which of the following statements is false?

A. Glucose and fructose have the same empirical formula B. Both glucose and fructose can exist in linear form and in ring form C. Glucose is an aldose, and fructose is a ketose *D. Glucose and fructose are geometric isomers Response: The answer to this question is D, glucose and fructose are geometric isomers. This is the only statement that is not true. Glucose and fructose both have the same empirical formula which is CH2O. This is the empirical formula for carbohydrates in general. Glucose and fructose can both also exist in both linear and ring form. The ring form is formed by one of the oxygens linked to the carbon backbone (carbon 5) attacking the carbonyl carbon, creating a bond between this carbon and the oxygen. This oxygen (from carbon 5) loses the hydrogen it was bound to and this COPYRIGHT 2020 BIOLOGY ONLINE

hydrogen binds to the oxygen that was originally part of the carbonyl group. Glucose is an aldose because it is a sugar that contains an aldehyde group while fructose is a ketose because it contains a ketone group, making choice C true. Statement D is false because glucose and fructose are not geometric isomers. Geometric isomers have their atoms covalently attached in the same way, the only difference is the spatial arrangement of the atoms, such as cis and trans orientation around a double bond. In glucose and fructose, there is no double bond, and the covalent arrangement of the atoms actually differs, making them structural isomers. Understanding the differences between different sugars is important as these differences will affect the nature of the polysaccharides that they form, as well as their interaction with the body. Fructose taste sweeter than glucose due to differences in the way it interacts with the body. This is advantageous in the food industry as less fructose can be added to a food to give the same amount of sweetness. 6. Glucose Polymer The molecular formula for glucose is C6H12O6. What would be the molecular formula for a linear polymer made by linking ten glucose molecules together by dehydration reactions? A. C60H120O60 *B. C60H102O51 C. C60H100O50 D. C60H111O51 Response: When two glucose molecules are joined by dehydration reactions, a molecule of water is formed. One glucose molecule loses a hydroxyl group and the other loses a hydrogen atom. The glucose molecules are then linked via a covalent bond called a glycosidic linkage. For every glycosidic linkage two hydrogen atoms and one oxygen atom are removed. Linking ten glucose molecules this way requires nine glycosidic bonds and thus nine molecules of water are produced. Using the molecular formula for glucose provided in the question you know that 10 molecules of glucose is C60H120O60. This falsifies choice A because no molecules of water have been lost. If you then subtract the eighteen hydrogen and nine oxygen atoms needed to form the glycosidic bonds you get C60H102O51 – this supports choice B. Choice C can be falsified because an extra molecule of water has been lost. This would be correct if you were looking for the molecular formula for a linear polymer of eleven glucose molecules or a ring structure of ten. Choice D describes only one molecule of hydrogen lost for every water molecule produced. This falsifies choice D and confirms choice B as the best answer. Clarification of Confusion Specific to the Spring 2020 Discussions: Some posts to this question demonstrated difficulty falsifying the answer choices and/or contained confusion regarding dehydration reactions. The main points of confusion are clarified below. 1. Dehydration reactions involve the loss of water. Hydrolysis adds water to break bonds and dehydration reactions remove water to make bonds. This question asks specifically about the dehydration reactions involved in synthesis of a polysaccharide. COPYRIGHT 2020 BIOLOGY ONLINE

2. For questions that involve a calculation like this one, answer choices cannot be falsified by simply saying they do not match your calculation. This is not explaining why an answer is wrong in the context of the question. It is attempting to confirm the answer you think is correct. This can lead to answering incorrectly if you have a mistake in your original calculation. Suggestion: Practice falsifying answers that involve calculations by considering the circumstances under which that answer would be correct or what that answer is actually describing. Questions to Consider: 1. What would be the molecular formula for a polymer formed by linking 8 glyceraldehyde molecules together? 2. How many molecules of water would be formed by linking 11 glucose molecules together? 7. Glycosidic Linkages The enzyme amylase can break glycosidic linkages between glucose monomers only if the monomers are the α (alpha) form. Which of the following could amylase break down? A. Cellulose B. Chitin *C. Glycogen D. Collagen Response: Glycosidic linkages are found between glucose monomers. In order to answer this question, you need to identify which polymers contain glucose monomers. The only carbohydrates on the list are choices A-cellulose, B-Chitin, and C-Glycogen. Cellulose consists of β-glycosidic linkages which you can infer from the question cannot be broken down by amylase. Chitin is a glucosederived polysaccharide found in fungus and exoskeletons. Monomeric units of chitin are linked via beta-glycosidic linkages like those found in cellulose. This falsifies choice B and identifies choice C, glycogen as the best answer. Animals use glycogen as storage molecules for glucose. Choice D, collagen, is false because collagen is a protein and contains peptide bonds, not glycosidic linkages. 8. Identify the Molecular Formula A molecule with the formula C18H36O2 is probably a(n) A. Nucleic Acid *B. Fatty Acid C. Carbohydrate D. Protein Response: A molecule with this formula is probably a fatty acid B. It helps to understand some basic indicators of each type of molecule in order to answer this question. The absence of nitrogen falsifies choices A and D. The even number of carbons and presence of 2 oxygen instead of 18 falsifies choice C and confirms choice B as the best answer. COPYRIGHT 2020 BIOLOGY ONLINE

9. Carbon The chemical bonds present in a molecule contribute to the properties of the molecule. An unusual property of carbon is that it can form multiple bonds. Which of the following statements about carbon bonding is false? A. A carbon-to-carbon cis double bond is the type found in nature and is associated with cardiovascular health. B. A carbon-to-carbon trans double bond that is made artificially in food processing is associated with poor cardiovascular health. C. A carbon-to-carbon double bond in the cis configuration creates a bend in the hydrocarbon chain. *D. Saturated fats are those that have a carbon-to-carbon double bond and are associated with good health. Response: This question highlights the role of carbon bonding in the structure of fats by testing the ability to identify a false statement. Carbon to carbon cis double bonds are found in unsaturated fats which are known to be more beneficial to cardiovascular health because they are less likely to form arterial plaques due to the kink the bonds form in the hydrocarbon chain. This falsifies choices A and C because they are true statements. Choice B can be falsified because it correctly describes trans fats which are made through the artificial process of hydrogenation. Saturated fats are detrimental to cardiovascular health because they do not have a carbon-tocarbon double bond in their hydrocarbon chains. This is why saturated fats are able to pack tightly together and form the arterial plaques that lead to atherosclerosis. This identifies choice D as the best answer because it is a false statement. 10. Phospholipid A phospholipid is composed of *A. one glycerol molecule linked to one phosphate head group and two fatty acids. B. one fatty acid molecule linked to one glycerol molecule and two phosphate groups. C. one glycerol molecule linked to three phosphate groups. D. one fatty acid molecule linked to three glycerol molecules. Response: The best answer to this question is choice (A). A phospholipid is composed of one glycerol molecule linked to one phosphate group and two fatty acid chains. Phospholipids are amphipathic meaning they are both hydrophilic and hydrophobic. The two fatty acid tails are hydrophobic, and the phosphate group comprises the hydrophilic (water loving) head. A phospholipid bilayer contains the hydrophobic tails on the interior, away from the aqueous cytoplasm and extracellular fluid, and the hydrophilic heads on the exterior in contact with the cytoplasm and the extracellular fluid. 11. Nucleic Acids Nucleic acids have a definite polarity, or directionality. Stated another way, one end of the molecule is different from the other end. How are these ends described? COPYRIGHT 2020 BIOLOGY ONLINE

A. One end has one phosphate group; the other end has two phosphate groups. B. One end contains a nitrogenous base; the other end lacks it. C. One end has a hydroxyl group on the 2 carbon; the other end has a hydrogen atom on the 2 carbon. *D. One end has an unlinked 3 carbon; the other end has an unlinked 5 carbon. Response: Choice (D) is the best answer to this question. The directionality of nucleic acids is described by the unlinked 3 carbon (3’) and unlinked 5 carbon (5’) ends. You learned that the functional group found on the 3’ end is an unlinked hydroxyl and a single phosphate group is found on the 5’ end thus ruling out the other answer choices. All of the other 5’ and 3’ carbons present in the rest of the molecule participate in phosphodiester bonds. Only the 5’ and 3’ carbons on the ends are unlinked. 12. DNA Double Helix What kind of chemical bond is found between the complementary paired nitrogenous bases of the DNA double helix? A. covalent bonds *B. hydrogen bonds C. N-glycosidic bonds D. phosphodiester bonds Response: If the double helix was held together by covalent bonds, it would be very difficult to separate the DNA in order to replicate or transcribe genes. Even if breaking the bonds was possible, they would have to be reformed, creating a major energy burden on the cell. This falsifies choice A. Hydrogen bonds are not very strong however with the large amount of hydrogen bonds forming between the bases of two DNA strands, the cumulative effect of all these bonds makes the interaction quite stable. The instability of hydrogen bonds does however allow the DNA double helix to be pulled apart, which is required for DNA replication as well as transcription. The fact that the double helix must be stable enough to preserve the DNA but at the same time flexible in coming apart, make hydrogen bonds perfect for holding the two strands together. This supports choice B. N-glycosidic bonds are found in DNA however, they are the type of bond that attaches the purine or pyrimidine to the pentose sugar, not the bond between base pairs. You should know that Nglycosidic bonds are a type of covalent bond. This falsifies choice C. Phosphodiester bonds are covalent bonds formed between the 3’ and 5’ carbons of the sugars in neighboring nucleotides in the nucleic acid backbone. This falsifies choice D and confirms choice B as the best answer. The fact that hydrogen bonds hold together the double helix is essential to the nature of DNA. If you take a look at pages 85-86 of the Campbell Biology textbook, including Figure 5.23 you can see that adenine always pairs with thymine via 2 hydrogen bonds and cytosine pairs with guanine via 3 hydrogen bonds. This interaction allows DNA to be stable enough to preserve genetic information, allows DNA to act as a template for its own replication as well as giving it the flexibility it need to be used for the transcription of genes. COPYRIGHT 2020 BIOLOGY ONLINE...