PHYS 498 HW3 Solutions: 1. We have two equations: (1) (2)

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1 PHYS 498 HW3 Solutions: 1. We have two equations: (1) (2) Where Conc is the initial concentration of [B] or [SA] Since [B] = [SA], the second equation simplifies to: (3) Using equation (1) and (3), we get a quadratic equation, and can solve for [B-SA] a) When Conc = 1 µm, [B-SA] = µm, [B-SA]/Conc = b) When Conc = 10-9 M, [B-SA] = 9.99 x M, [B-SA]/Conc = c) When Conc = M, [B-SA] = 3.82 x M, [B-SA]/Conc = As the starting concentration decreases, the concentration of B-SA will decrease. What is a more useful quantity is [B-SA]/Conc, which tells us the fraction of [B-SA] over the total amount of proteins (the starting concentration). We plot the graph of [B-SA]/Conc below: We see that as the initial concentration decreases from 1e-08 to 1e-12, the fraction of B-SA slowly decreases. The fraction of B-SA starts decreasing very quickly as the initial concentration approaches the Keq, which is 1e-15. Below 1e-16, most of the biotin and streptavidin does not associate, and exist as individual proteins.

2 2. a) b) c) d) The value calculated from (b) is close to that in (d). The value from (b) and (c) are consistent with the value reported for the free energy of ATP, which is between 80 and 100 pn-nm 3. a) A gene chip or DNA microarray consists of many DNA spots attached to a solid surface Each DNA spot contains many DNAs of the same sequence from one gene (usually in picomolar amount). These DNA are the probes Each DNA will hybridize efficiently only with its complementary sequence. The complementary DNA is the target During cell division, we can determine which genes are turned on by probing the number of mrna transcribed for that particular gene. An alternative is to look at the number of proteins expressed by the gene, but it is difficult to accurately quantify very small amount of proteins The target mrnas is extracted from cell and reverse transcribed to complementary DNA (cdna). These cdna will be fluorescently tagged, and become the target for the gene chip The cdna will be added to all the DNA spots in the gene chip and allowed to hybridize with the probes. Unbound cdna are then washed, leaving only those that hybridize efficiently with the probe stuck on the surface By analyzing the fluorescence intensity of each DNA spot, the amount of fluorescent-cdna can be inferred. This will tell us the amount of mrna expressed and eventually we can determine which genes are being turned on b) Red: cancerous tissue, green: healthy tissue, yellow: both cancerous and healthy tissue. We want to decrease the gene expression of proteins that are overexpressed in cancerous tissues (oncogenes). We will therefore choose genes with red spots: gene 2, gene 8 and gene 15

3 c) We would increase the expression of genes that are suppressed in cancerous tissues (tumor suppressor genes). We will choose genes with green spots because these spots belong to genes normally present in healthy tissues, but absent in cancerous tissues. Thus, we will choose gene 4, gene 10 and gene The complementary sequence is: 5 TTT GAG CAA AGT CAC 3 The protein which is created from the two strands is completely different. The genetic code for any given protein is on one strand of the double helix, the coding strand. For the virus to have 2 genes overlap on the DNA, the sequence and its complement have to both code useful structures, and they have to evolve together. The proteins cannot be optimized independently Essay questions: 1. Describe the entire process of DNA replication: DNA is split into two single strands by helicases Each single strand is stabilized by single-stranded binding proteins (SSB) DNA polymerase III adds new nucleotides onto the leading strand. DNA is read from 3 to 5 and the new DNA is formed from 5 to 3 On the lagging strand, a short RNA primer is added by RNA primase DNA polymerase III then lays down new DNA and forms an Okazaki fragment DNA polymerase I replaces RNA primer with DNA DNA ligase links together two Okazaki fragments Thermodynamics review We have the free energy equation: ΔG = ΔH T ΔS In order for a reaction to be favorable (negative ΔG), we want ΔH to be negative (decrease in enthalpy) and ΔS to be positive (gain in entropy) Exergonic vs endergonic Exergonic: chemical reaction where the Gibbs free energy (ΔG) is negative (favorable reaction) Endergonic: chemical reaction where the Gibbs free energy (ΔG) is positive (unfavorable reaction) Exothermic vs endothermic Exothermic: chemical reaction that releases heat to the surrounding (ΔH is negative) Endothermic: chemical reaction that absorbs heat from the surrounding (ΔH is positive)

4 Anabolic vs catabolic reactions (Source: Two types of metabolic reactions take place in the cell: 'building up' (anabolism) and 'breaking down' (catabolism). 1 Anabolic reaction: make big molecules from small ones. They are endergonic, thus require energy input. Examples: amino acids form protein, sugar monomers form polysaccharide, and nucleotides form DNA Catabolic reaction: break down of big molecules into small ones. They are exergonic, releasing energy into the surrounding. Example: digestion of protein into amino acids, break down of DNA into amino acids, etc Chemical stability of DNA: Reactant: deoxy-nucleoside triphosphates (dntps) (sometimes just called nucleotide). These can be datp, dttp, dgtp or dctp Product: DNA Type of reaction: endergonic (product has higher free energy than reactant) Type of bond: phosphodiester bond/ covalent bond between two nucleotides DNA polymerase helps DNA polymerization in two ways: o It helps reduce the activation energy during polymerization (see Fig 1) o It uses energy from dntps to polymerize DNA. This energy is used overcome the reduced activation energy The phosphodiester bond formed is strong, which explains the stability of DNA o Even though the polymerization reaction is endergonic, which means that the reverse reaction is more favorable, the strong covalent bond makes it difficult to the DNA to depolymerize. It will take a long time for natural depolymerization of DNA to occur Since the polymerization reaction is endergonic, ΔG is positive. This is derived from positive ΔH and positive ΔS. The magnitude of ΔH is greater than T* ΔS such that eventually the ΔG is positive DNA forms double helices, and the free energy (ΔG) of formation of double stranded DNA is driven by the favorable enthalpy (negative ΔH, due to hydrogen bond interaction) and favorable entropy (positive ΔS). Entropy is gained because water lining single stranded DNA is freed when double stranded DNA is formed Fig 1. Free energy diagram of a condensation reaction with the help of enzyme. The final state has higher energy than the initial state. Enzymes like DNA polymerase and ribosome help lower activation energy. Source:

5 Chemical stability of protein: Reactant: amino acids Product: protein Type of reaction: endergonic (product has higher free energy than reactant) Type of bond: peptide bond/ covalent bond between two nucleotides Ribosome helps protein polymerization in two ways: o It helps reduce the activation energy during polymerization (see Fig 1) o It uses energy from GTPs to polymerize protein. This energy is used overcome the reduced activation energy The peptide bond formed is strong, which explains the stability of protein o Even though the polymerization reaction is endergonic, which means that the reverse reaction is more favorable, the strong covalent bond makes it difficult for protein to depolymerize. It will take a long time for natural depolymerization of protein to occur Since the polymerization reaction is endergonic, ΔG is positive. This is derived from positive ΔH and positive ΔS. The magnitude of ΔH is greater than T* ΔS such that eventually the ΔG is positive When protein folds, energetically favorable interactions are formed (hydrogen bond, ionic bond, Van der Waals interactions). This decreases the enthalpy, which decreases the free energy. However, as protein becomes more ordered when folded, entropy decreases, which increases the free energy. The free energy change of protein folding is therefore low due to the tradeoff between favorable enthalpy change and unfavorable entropy change. 2. Difference between DNA and RNA RNA: ribonucleic acid. DNA: deoxyribonucleic acid Structural: on the 2 carbon of the ribose ring, there is no hydroxyl group attached to DNA, while there is one hydroxyl group attached to RNA Structural: DNA and RNA are made up of 4 nucleotides. Three of them are the same: adenine, guanine and cytosine. Uracil in RNA is replaced by thymine in DNA. Thymine is simply uracil that is methylated. Methylation in effect protects DNA from nucleases (enzyme that breaks down DNA and RNA), thus allow longer lived DNA. Uracil readily pairs with adenine, but it can also pair with all other bases (guanine, cytosine, and also itself). Replacing uracil with thymine in DNA makes DNA replication more efficient by reducing the rate of mismatches. Structural: DNA is usually double stranded, and is made up of long chain of nucleotides. RNA mostly exists as single strand, and is usually shorter than DNA. RNA is more diverse structurally since short helices of RNA can be packed together to achieve an enzymatic function. This allows RNA to act as enzyme. Stability and function: DNA is chemically more stable than RNA, thus better for long term storage of information. RNA is used for short term storage of information. Function: RNA is more easily degraded than DNA, thus allowing regulations of proteins Types of RNA: messenger RNA (mrna): transfer genetic code from DNA to protein, allowing DNA to stay within the nucleus. Transfer RNA (trna): binds to amino acid and the codon of mrna, allows translation of genetic code from mrna to protein. Ribosomal RNA (rrna): makes up the ribosome, along with protein has catalytic activity and can form peptide bond