BioChem final Exam

Question 1

The bacterium E. coli requires simple organic molecules for growth and energy-it is therefore a:

photoautotroph
lithotroph
chemoheterotroph
photoheterotroph
chemoautotroph

Answer 1

Chemoheterotroph

Question 2

It is believed that life evolved with RNA as the genetic material, but RNA has been replaced by DNA in all current cellular life. Which feature of DNA probably accounts for this?

DNA is more stable than RNA

Only DNA can form the genetic material of viruses

DNA is a nucleic acid while RNA is not

DNA can direct its own replication while RNA cannot

Answer 2

DNA is more stable than RNA

Question 3

ATP (adenosine triphosphate) can be termed the universal energy currency in life because it:

• traps energy released from the oxidation of different classes of food molecules
  • enables cells to disobey the first law of thermodynamics
  enables cells to disobey the second law of thermodynamicsreleases energy as heat when it undergoes hydrolysis

Answer 3

traps energy released from the oxidation of different classes of food molecules

Question 4

Which of the following groups of elements represent the most common elements in biological systems?

Carbon, nitrogen, sodium, and hydrogen
Carbon, sodium, hydrogen, and oxygen
Carbon, sodium, potassium, and nitrogen
Carbon, nitrogen, oxygen, and hydrogen

Answer 4

c,h,o,n

Question 5

What functional groups are present on this molecule?

HO-CH2-CH2-COOH

Hydroxyl and aldehyde
Hydroxyl and ester
Hydroxyl and carboxylic acid
Ether and aldehyde
Hydroxyl and ketone

Answer 5

Hydroxyl and carboxylic Acid

Question 6

catalysts do not alter deltaG°

True
False

Answer 6

True, what alters delta G is : changes in temperature, enthalpy (ΔH), and entropy (ΔS), as well as the reaction quotient (Q) or equilibrium constant (K)

Question 7

Binding of chiral biomolecules is not stereospecific.

True or False

Answer 7

False, binding of chiral biomolecules is stereospecific.

Question 8

Solutes that are hydrophobic are those that…

dissolve poorly in water because they contain polar covalent bonds.

have been reduced by addition of hydrogen atoms.

dissolve easily in water because they contain polar covalent bonds.

dissolve poorly in water because they contain non-polar covalent bonds.

dissolve easily in water because they contain non-polar covalent bonds.

Answer 8

dissolve poorly in water because they contain non-polar covalent bonds.

Question 9

Which one of the following is equal to the pKa of a weak acid?

The pH of a solution containing equal amounts of the acid and its conjugate base

Its relative molecular mass

The equilibrium concentration of its conjugate base

The pKb of its conjugate base

Answer 9

The pH of a solution containing equal amounts of the acid and its conjugate base

Question 10

Substances, such as salt, that readily dissociate in water are said to be water loving, or __________, while those that do not are called water fearing, or __________. Some molecules, such as proteins and nucleic acids, have both of these properties and are called __________.

amphipatic, hydrophobic, hydrophilic
hydrophobic, hydrophilic, amphipatic
amphipatic, hydrophilic, hydrophobic
hydrophilic, hydrophobic, amphipatic

Answer 10

philic, phobic, amp

Question 11

Water surrounding non polar molecules has lower entropy than bulk “free” water.

True
False

Answer 11

True

Question 12

The pKa of NH4+ is 9.25. What is the highest pH that a buffer system composed of NH4+ and NH3 will buffer at effectively?

13.25
11.25
12.25
10.25

Answer 12

10.25

Question 13

A 1 M solution of NaCl is hypertonic to a 1M solution of glucose.
True
False

Answer 13

TRUE

Question 14

The hydrophobic effect can make important energetic contributions to:

enzyme-substrate interactions.

folded protein structures.

All of the answers are correct.

oil droplets coming together.

membrane structure.

Answer 14

All of the answers are correct

Question 15

Of the 20 amino acids, only ___________ is not optically active. The reason is that its side chain ________.

Proline; forms a covalent bond with the amino group.

Glycine; has a hydrogen atom in its R group.

Glycine; is unbranched.

Alanine; is a simple methyl group.

Lysine; contains only nitrogen.

Answer 15

Glycine; has a H atom in its R group

Question 16

Titration of valine by a strong base, for example NaOH, reveals two PK’s. The titration reaction occurring at PK2 (PK2=9.62) is:
E. -NH2 + OH-→ -NH- + H2O
C. –COO- + -NH2+ –→ –COOH + NH2
D. -NH3+ + OH- → -NH2 + H2O
A. –COOH + OH-→ –COO- + H2O
B. –COOH + -NH2 → –COO- + NH2+

Answer 16

D

Question 17

For amino acids with neutral R groups, at any pH below the PI of the amino acid, the population of amino acids in solution will have:

A net positive charge.
No net charge.
A net negative charge.
Positive and negative charges in equal concentration.
No charged groups.

Answer 17

net positive charge

Question 18

A dipeptide has 2 amino acids and two peptide bonds.
True
False

Answer 18

False, 1 peptide bond

Question 19

Protein separation techniques are often based on the following properties except…

Charge of the protein.
Specific binding properties of the protein.
Solubility of the protein.
Viscosity of the protein.

Answer 19

viscosity

Question 20

By adding SDS (sodium dodecyl sulfate) during the electrophoresis of proteins, it is possible to…

determine an enzyme’s specific activity.

separate proteins exclusively on the basis of molecular weight.

determine the amino acid composition of the protein.

preserve a protein’s native structure and biological activity.

determine a protein’s isoelectric point.

Answer 20

separate proteins exclusively on the basis of molecular weight.

Question 21

Peptide bond is…

rigid with partial double bond character.
Considered to be a covalent bond.
All answers are correct.
A dehydration synthesis.
A condensation product of two amino acids.

Answer 21

all answers are correct

Question 22

processing of mRNA includes?

Answer 22

splicing out introns and rejoining any.

adding a 5 cap ( a 7-methylguanosine: protects RNA from nucleases and forms a binding site for ribosome). 2’OH groups.

adding a 3 poly (A) tail

degradation.

Question 23

about the UP element

Answer 23

• left of e coli promoter
• upstream of the -35 region
• rich in AT bases
• : Enhances transcription by providing an additional binding site for the alpha subunit of RNA pol in strong promoters

Question 24

spacer regions are between

Answer 24

Up & -35 and -10 & +1

Question 25

the alpha subunit

Answer 25

Involved in enzyme assembly. Interact with regulatory proteins and UP elements of promoters. Help positioning the enzyme on DNA.

Question 26

beta subunit

Answer 26

Catalytic center: responsible for RNA chain elongation. Binds ribonucleoside triphosphates (NTPs) and contributes to phosphodiester bond formation.

Question 27

B' Subunit

Answer 27

Binds the DNA template during transcription. Forms part of the catalytic site with the beta subunit. Involved in DNA unwinding and maintaining the open complex.

Question 28

Omega Subunit

Answer 28

Helps in folding and stabilizing the B’ subunit. Not essential for transcription but enhances enzyme assembly and stability

Question 29

Sigma subunit

Answer 29

promoter recognition. It specially binds to the -35 and -10 promoter regions. It also facilitates the formation of the closed and open complexes and ensures transcription initiates at the correct site. It is released after transcription initiation, leaving the core enzyme to continue elongation. RNA pol lack proofreading 3’  5’ exonuclease active site. Thus, the error rate for transcription is higher than that of DNA replication.

Question 30

Describe the steps of transcription initiation and elongation.

Answer 30

- Promoter Recognition (closed complex formation. - DNA unwinding (open complex formation - Transcription initiation - Promoter Clearance - Elongation - Termination

Question 31

Describe the structure and role of the following processes that lead to the maturation of the mRNA in Eukaryotes: 5’ cap, 3’ poly A, and splicing

Answer 31

A newly synthesized transcript is called a primary transcript. The primary transcript for a eukaryotic mRNA typically contains sequences encompassing one gene, although the sequences encoding the polypeptide may not be contiguous. Noncoding tracts that break up the coding region of the transcript are called introns, and the coding segments are called exons.

Question 32

the fundamental rules of replication

Answer 32

- Replication is semi- conservative - replication begins at an origin and proceeds bidirectionally - synthesis of new DNA occurs in the 5 --> 3 direction and is semidiscontinuous

Question 33

What is Cairns' experiments

Answer 33

DNA was radiolabeled by growing cells in 3H (tritium); DNA was isolated and spread under a photographic emulsion. - showed circulur DNAs with an extra loop - showed that both strands are replicated simultaneously - Two replication forks, so bidirectional replication.

Question 34

Inman's experiments

Answer 34

- Denatured bacteriophage DNA at A=T rich regions -> "bubbles" - bubbles were mapped - showed that loops always initiate at a unique point which they named the ORIGIN

Question 35

Synthesis proceeds in which direction?

Answer 35

5 - 3 (3' OH) - The leading strand is made continuously as the replication fork advances * The lagging strand is made discontinuously in short pieces (Okazaki fragments) that are later joined together

Question 36

Exo and endonucleases

Answer 36

* Exonucleases―cleave bonds that remove nucleotides from the ends of DNA * Endonucleases cleave bonds within a DNA sequence

Question 37

The five DNA pol in E. Coli

Answer 37

DNA pol I: Abundant but not ideal for replication. – Rate (600 nucleotides/min) is slower than observed for replication fork movement – Has low processivity – Its primary function is in clean-up DNA Pol III: the principal replication pol II, IV, and V: Involved in DNA repair

Question 38

DNA pol I unique 5 - 3 exonuclease activity

Answer 38

* In addition to the 3’5’-exonuclease activity * Moves ahead of the enzyme, hydrolyzes things in its path * Does nick translation―a strand break moves along with enzyme * This activity and the polymerase activity are in the Klenow fragment―a distinct domain that can be separated by protease treatment

Question 39

The enzymes/proteins for DNA replication

Answer 39

– Helicases (use ATP to separate DNA strands) – Topoisomerases (relieve the stress caused by helicases separating strands) – DNA-binding proteins to stabilize separated strands – Primases to make RNA primers – DNA ligases to seal nicks

Question 40

Initiation of DNA replication

Answer 40

oriC is 245 bp highly conserved sequence elements - binding sites for initiator binding protein - AT rich, DNA unwinding element Dna A: bind at R and I (DNA proteins are ATPases 8 Dna A bind to the r and i sites DNA wraps around the complex forming a positive supercoil) - DnaB helicase continues initiation ONLY INITIATION IS REGULATED

Question 41

DnaB helicase

Answer 41

* DnaB hexamer structure is opened by DnaC * Once DnaB loaded, DnaC dissociates * DnaB migrates along ssDNA 5’3’ and unwinds the helix * All replication proteins bind to DnaB either directly or indirectly

Question 42

Tell me about initiation being regulated

Answer 42

* After replication, oriC adenine is hemimethylated (on template/parental strand only) * Hemimethylated oriC sequences interact with plasma membrane So regulation of cell cycle due to 1. DnaA binding 2. Hemimethylation 3. Membrane association

Question 43

Elongation phase of replication

Answer 43

* Primase (DnaG) makes RNA primer * DNA Pol III adds nucleotides * Gyrase (a topoisomerase) move ahead of replication complex relieving stress of unwinding Synthesis of two antiparallel strands! * One 5’  3’ and one 3’  5’ * But replication only in 5’  3’ direction! * DNA Pol III – Dissociates – Binds upstream to start new Okazaki fragment * DNA Pol I – Removes RNA primer – fills in the gap * DNA ligase seals the nick

Question 44

Termination phase of dna replication

Answer 44

Replication forks meet at terminus region – with 20-bp sequences Ter (TerA-TerF) – Ter sites are found near each other but in opposite directions – Create a site that replication forks can pass but cannot leave – Final few hundred bp copied by unknown mechanism

Question 45

What is the AMES test?

Answer 45

a screen for human carcinogens

Question 46

What is the physiological role of 3’→5’ (proofreading) activity?

Answer 46

The 3′→5′ exonuclease activity of DNA polymerase serves as a proofreading mechanism to maintain the accuracy of DNA replication. This activity allows the polymerase to detect and remove incorrectly incorporated nucleotides, preventing mutations and ensuring the fidelity of genetic information.

Question 47

What is the physiological role and key functions of 5’→3’ (exonuclease) activity?

Answer 47

The 5′→3′ exonuclease activity of DNA polymerase plays a crucial role in DNA repair and replication by removing RNA primers, damaged nucleotides, and other unwanted DNA segments. This function is primarily associated with DNA polymerase I in E. coli and is critical for ensuring accurate and efficient DNA synthesis. Key functions: 1. Removal of RNA primers in DNA replication. 2. Repair of Damanged DNA (excision repair) 3. Removal of Mismatched or incorrectly incorporated nucleotides 4. processing of Okazaki fragments

Question 48

function for nucleotides

Answer 48

– Energy for metabolism (ATP) – Enzyme cofactors (NAD+) – Signal transduction (cAMP)

Question 49

function for nucleic acids

Answer 49

– Storage of genetic info (DNA – only known function) – Transmission of genetic info (mRNA) – Processing of genetic information (ribozymes) – Protein synthesis (tRNA and rRNA)

Question 50

about Nitrogenous bases

Answer 50

* Derivatives of pyrimidine or purine * Planar or almost planar structures * Absorb UV light around 250–270 nm * All are good H-bond donors and acceptors * Neutral molecules at pH 7

Question 51

about the phosphate group

Answer 51

* Negatively charged at neutral pH (pKa near 0!) * Typically attached to 5’ position – Nucleic acids are built using 5’-triphosphates * ATP, GTP, TTP, CTP – 5’ phosphate attaches to pentose 3’ OH – Creates ‘sugar-phosphate backbone’ – Backbone hydrophilic * May be attached to other positions

Question 52

Why is RNA hydrolyzed in alkaline conditions but not DNA?

Answer 52

* RNA is unstable under alkaline conditions * Hydrolysis is also catalyzed by enzymes (RNase) * RNase enzymes are abundant around us: – S-RNase in plants prevents inbreeding – RNase P is a ribozyme (enzyme made of RNA) that processes tRNA precursors – Dicer is an enzyme that cleaves double-stranded RNA into oligonucleotides * protection from viral genomes * RNA interference technology. Why not DNA is due to the difference on their sugar backbones. DNA's deoxyribose sugar lacks a hydroxyl group at the 2' position, making it less reactive and less susceptible to base-catalyzed hydrolysis.

Question 53

complementarity of DNA strands

Answer 53

* Two chains differ in sequence (sequence is read from 5’ to 3’) * Two chains are complementary * Two chains run antiparallel * G=C versus A=T

Question 54

conformation around N-Glysocidic bond

Answer 54

* Relatively free rotation can occur around the C1'-N-glycosidic bond * But rotation limited by crowding (steric constraints) to either: – Angle near 0 corresponds to syn conformation – Angle near 180 corresponds to anti conformation * Purines can be either, pyrimadines usually anti * Anti conformation is found in normal B-DNA

Question 55

the 3 forms of DNA

Answer 55

* A form – low water environments – not found in vivo * B form – normal conformation in cells – Watson and Crick structure * Z form – – found occasionally in vivo – Left-handed double helix! – Sequence-dependent.

Question 56

RNA, single stranded structure

Answer 56

Single stranded RNA adopts a right handed helix. - longer than 3x the protein length.

Question 57

DNA Denaturation

Answer 57

- Covalent bonds remain intact – Genetic code remains intact * Hydrogen bonds are broken – Two strands separate * Base stacking is lost – UV absorbance increases (Hyperchromic effect) Free nucleotides > ssDNA > dsDNA Denaturation can be induced by high temperature, or change in pH denaturation may be reversible: annealing.

Question 58

Factors affecting DNA denaturation

Answer 58

* The midpoint of melting (Tm) depends on base composition. – High CG increases Tm * Tm depends on DNA length – Longer DNA has higher Tm – Important for short DNA * Tm depends on pH and ionic strength – High salt increases Tm AT rich regions melt at a lower temperature than GC-rich regions

Question 59

What are the molecular mechanisms of radiation-induced mutagenesis?

Answer 59

1.Deamination - very slow reactions - large number of residues - net effect significant - repair mechanisms exist (DNA C -> U events are repaired because if not, CG pairs would eventually disappear. 2. Depurination - N-Glycosidic bond is hydrolyzed - Significant for purines: 10k purines lost/day in a mammalian cell. - AP sites vulnerable to degradation via isomerization of sugar - Cells have mechanisms to correct most of these modifications. 3. UV lights - induces dimerization of pyrimidines; this may be the main mechanism for skin cancers. - Other forms of ionizing radiation wreck havoc too (radioactive materials, X-Rays) - Oxidative damage the most potent DNA threat! 4. Ionizing radiation (X-rays and gamma rays) causes ring opening and strand breaking. Difficult to fix - Cells can repair some of these modifications, but others cause mutations. Accumulation of mutations is linked to aging and carcinogenesis.

Question 60

functions of the DNAs

Answer 60

RNA is essential in interpreting and executing DNA’s instructions. Messenger RNA (mRNA): * Function: Transcribes genetic information from DNA and transports it to the ribosomes, where it serves as a template for protein synthesis. * Process: mRNA is transcribed from a gene on the DNA, carrying the genetic code in a form that ribosomes can read to synthesize proteins. Ribosomal RNA (rRNA): * Function: Combines with proteins to form ribosomes, which are the sites of protein synthesis. * Process: rRNA provides structural support and catalyzes peptide bond formation, linking amino acids together in a protein chain. Transfer RNA (tRNA): * Function: Delivers specific amino acids to ribosomes during protein synthesis. * Process: Each tRNA molecule matches an amino acid with the appropriate mRNA codon by using its anticodon region, ensuring the correct sequence of amino acids in the protein. Small Nuclear RNA (snRNA): * Function: Involved in RNA processing, particularly in the splicing of pre-mRNA (removal of introns and joining of exons). * Process: snRNA forms complexes called spliceosomes, which help edit the mRNA before it is translated. MicroRNA (miRNA) and Small Interfering RNA (siRNA): * Function: Regulate gene expression by binding to mRNA and either blocking its translation or causing its degradation. * Process: miRNAs and siRNAs play roles in gene silencing, influencing various cellular functions and pathways, including development and defense against viral infections. Long Non-Coding RNA (lncRNA): * Function: Regulates gene expression, often at the transcriptional level. * Process: lncRNAs can modulate chromatin structure, recruit transcription factors, or act as molecular scaffolds, impacting gene regulation, genome stability, and cell differentiation

Question 61

Is DNA soluble in water? Are free bases soluble in water?

Answer 61

Yes because: Hydrophilic phosphate backbone Hydrogen bonding Double Helix Structure No Because Hydrophobic nature of nitrogenous bases Lack of phosphate group: Phosphate introduces the negative charge but they don't have that. H Bond capacity

Question 62

Base Pairing

Answer 62

A, T (Purines) form 2 H bonds C,T,U (pyrimidines) form 3 H bonds stackign interactions support DNA structure.

Question 63

Parameters that influence Tm

Answer 63

1. DNA Length: the longer the higher Tm Salt concentration. 2. Higher salt concentrations stabilize DNA double helix by shielding the - charges on the phosphate backbone, increasing the Tm 3. DNA concentration: slightly increase Tm 4. pH: extreme conditions can affect the stability of the hydrogen bonds between base pairs and, therefore, affect the Tm

Question 64

Enzymes

Answer 64

- Most are globular proteins - They are selective - They need co factors (cations) and coenzymes (derivatives of vitamins). - They do not affect equilibrium (Delta G and Keq), they increase reaction rates by decreasing Delta G.

Question 65

Steps in catalysis

Answer 65

1. Substrate approaches active site. 2. Enzyme - substrate complex forms. 3. Substrate transformed into products. 4. Products released. 5. Enzyme recycled. !!!

Question 66

Enzyme classes

Answer 66

Oxireductases Transferases Hydrolases Lyases Isomerases Ligases

Question 67

What are the 2 categories of inhibitors?

Answer 67

Irreversible and reversible

Question 68

competitive inhibitors

Answer 68

- looks similar to the substrate - competed with the substrate for binding to the enzyme. - Binds to active site non covalently - Does not affect catalysis - V max unchanged, Km increased

Question 69

Non competitive inhibition

Answer 69

- does not resemble the substrate - does not compete with the substrate for the active site. Binds to a regulatory site on the enzyme. Inhibits both substrate binding and catalysis. - decrease V max, Km unchanged - This inhibition cannot be outcompeted