Biochem Exam IV

Question 1

What are the three steps of fatty acid synthesis?

Answer 1

1) Acetyl CoA is transferred from the mitochondria to the cytosol in the form of citrate

2) Acetyl CoA is then converted to malonyl CoA and both become attached to an acyl carrier protein to produce acetyl ACP and malonyl ACP

3) Fatty acid elongation then occurs, and this happens 2 carbons at a time

Question 2

How is acetyl-CoA translocated from the mitochondria to the cytosol?

Answer 2

1) Acetyl-CoA combines with oxaloacetate to form citrate in the mitochondria

2) Citrate is transferred from the mitochondria to the cytosol

3) ATP-Citrate lyase cleaves citrate to produce acetyl CoA and oxaloacetate from CoA-SH

Citrate + ATP + CoASH + H2O –> Acetyl CoA + ADP + Pi + Oxaloacetate

This process requires oxaloacetate to be present in the mitochondria

Question 3

How is oxaloacetate generated in the mitochondria to allow for fatty acid synthesis?

Answer 3

1) Malate dehydrogenase reduces oxaloacetate in the cytoplasm to malate

Oxaloacetate + NADH + H+ –> Malate + NAD+

2) Malic enzyme oxidatively decarboxylates malate into pyruvate (COO- attached to CH2 gets removed)

Malate + NADP+ –> Pyruvate + NADPH+ + CO2

3) Pyruvate can freely return to the mitochondria

4) Pyruvate carboxylase in the mitochondria carboxylates pyruvate back into oxaloacetate

Pyruvate + CO2 + ATP + H2O –> Oxaloacetate + ADP + Pi + 2H+

Question 4

How is Malonyl CoA formed in the cytoplasm? What is the overall reaction and what are the two steps?

Answer 4

Acetyl CoA Carboxylase catalyzes the two step reaction where acetyl CoA is converted to malonyl CoA. The overall reaction is

Acetyl-CoA + ATP + HCO3(-) –> Malonyl-CoA + ADP + Pi + H+

Malonyl CoA is carboxylated using HCO3(-) (not CO2) and the two steps are

1) Biotin-enzyme + ATP + HCO3(-) –> CO2-biotin (carboxybiotin) + ADP + Pi + H+

2) CO2-biotin + Acetyl CoA –> Malonyl CoA + Biotin-enzyme

Question 5

What is the structure of the acyl carrier protein and what is it used for?

Answer 5

An acetyl transacylase catalyzes the transfer of an acyl carrier protein (ACP) to Acetyl CoA forming Acetyl ACP (contains an acyl group), and a malonyl transacylase catalyzes the transfer of an ACP to Malonyl CoA to form Malonyl ACP. These are used in fatty acid synthesis

ACP is made up of a single polypeptide and consists of a B-Mercaptoethylamine unit with a SH reactive group at the end and this is attached to the pantothenate unit. These units make up the phosphopantetheine group and this group is found on Coenzyme A and ACP with slight differences

Question 6

What are the steps of fatty acid elongation?

Answer 6

1) Condensation where B-Ketoacyl synthase catalyzes the formation of Acetoacetyl ACP (a B-ketoacyl ACP) from Acetyl ACP and Malonyl ACP releasing ACP and CO2

(Acetoacyl ACP consists of a B carbon (ketone) attached to an a carbon which is attached to a C=O connected to S-ACP)

Acetyl ACP + Malonyl ACP –> Acetoacetyl ACP + ACP + CO2

2) Reduction where B-ketoacyl reductase reduces Acetoacetyl ACP into D-3-hydroxybutyryl ACP

Acetoacetyl ACP + NADPH –> D-3-Hydroxybutyryl ACP + NADP+

3) Dehydration where 3-hydroxyacyl dehydratase forms Crotonyl ACP from D-3-hydroxybutyryl ACP

D-3-Hydroxybutyryl ACP –> Crotonyl ACP + H2O

4) Reduction where enoyl reductase (inhibited by Triclosan an antibacterial agent) reduces Crotonyl ACP into Butyryl ACP (an acyl ACP)

Crotonyl ACP + NADPH –> Butyryl ACP + NADP+

5) Elongation where B-Ketoacyl synthase catalyzes the formation of C6-B-ketoacyl ACP (A B-ketoacyl ACP) from Butyryl ACP and Malonyl ACP via a condensation reaction. Malonyl ACP can continuously be added via condensation to form a chain where 2 carbons are added at a time.

Butyryl ACP + Malonyl ACP –> C6-B-ketoacyl ACP + ACP + CO2

Question 7

How is fatty acid synthesis regulated? What enzyme is targeted?

Answer 7

Acetyl CoA Carboxylase can be regulated allosterically. It is stimulated by citrate and inhibited by Palmitoyl (C16) CoA.

Acetyl CoA Carboxylase is also inactive when phosphorylated (fatty acid synthesis inhibited) and active when unphosphorylated (fatty acid synthesis stimulated) and it can be regulated covalently.

1) Insulin can stimulate enzyme by activating protein phosphatase 2A which dephosphorylates enzyme

2) Glucagon and Epinephrine can inhibit enzyme by activating AMP which activates a Protein Kinase and phosphorylates enzyme (not PKA)

Citrate can be used to partially activate enzyme even when it is phosphorylated (inactive) via allosteric regulation

Question 8

What is the structure/function of fatty acids and how are they formed?

Answer 8

Free fatty acid chains are hydrocarbon chains that terminate with a carboxylic acid group, and the first carbon after the carboxylic acid carbon is the a-carbon atom

Fatty acids are stored in the form of triacylglycerols and these are formed via the esterification of three fatty acid chains to a glycerol molecule

Lipases are used to hydrolyze a fatty acid chain and remove it from the triacylglycerol and form a free fatty acid and diacylglycerol. Another lipase can then produce a monoacylglycerol from a diacylglycerol, and these processes allow fatty acids to be metabolized

Question 9

What lipases are used to metabolize fatty acids and how are they regulated?

Answer 9

Adipocyte Triglyceride Lipase (ATGL) hydrolyzes triacylglycerols to diacylglycerols releasing a free fatty acid. This is regulated by CGI-58

Hormone Sensitive Lipase (HSL) hydrolyzes diacylglycerols to monoacylglycerols releasing a free fatty acid. This is regulated by Perilipin A

Monoacylglycerol lipase (MGL) hydrolyzes monoacylglycerols to produce a free fatty acid and glycerol

CGI-58 and Perilipin A are regulated via phosphorylation and are phosphorylated/activated by PKA. When phosphorylated, CGI-58 can bind to ATGL and Perilipin A can bind to HSL. When this binding occurs, ATGL and HSL can then be phosphorylated/activated and this activates lipolysis. HSL can also be partially activated after phosphorylation by PKA. Glucagon and Epinephrine activate this PKA pathway and stimulates lipolysis

Insulin can inhibit lipolysis by activating phosphodiesterase (PDE) via Protein Kinase B (Akt), and this converts cAMP to 5’-AMP interfering with the PKA pathway preventing the phosphorylation/activation of Perlipin A, CGI-58, and HSL

Question 10

How is glycerol metabolized during fatty acid degradation?

Answer 10

Glycerol is converted into D-Glyceraldehyde-3-phosphate

1) Glycerol kinase produces L-Glycerol-3-phosphate from glycerol

2) Glycerol phosphate dehydrogenase oxidizes glycerol-3-phosphate into dihydroxyacetone phosphate (DHAP)

3) Triose phosphate isomerase isomerizes DHAP into D-Glyceraldehyde-3-phosphate

Question 11

How are fatty acids transported into the mitochondria for fatty acid oxidation?

Answer 11

The fatty acid is first acylated into carnitine

1) Acyl-CoA synthetase forms acyl-CoA from a fatty acid and coenzyme A in two steps.

i. First the fatty acid reacts with ATP to form acyl adenylate releasing PPi which quickly dissociates into two inorganic phosphate groups making the reaction irreversible (this expends the equivalent of 2 ATP since 2 high energy phosphate bonds are removed)

ii. The sulfhydrl (SH) group of CoA then attacks the acyl adenylate to form acyl-CoA and AMP

2) Carnitine acyltransferase I transfers a CoA group from the sulfur atom of acyl CoA to the hydroxyl group of carnitine to form acyl-carnitine and acetyl-CoA

3) Translocase shuttles the acyl carnitine across the inner mitochondrial membrane

4) In the mitochondrial matrix carnitine acyltransferase II re-forms the acyl-CoA for fatty acid oxidation along with carnitine which moves out of the mitochondria

Question 12

What are the steps of the fatty acid oxidation of saturated and even numbered fatty acid chains?

Answer 12

1) Acyl CoA dehydrogenase oxidizes acyl CoA into enoyl CoA with a trans double bond between C-2 and C-3. FADH2 is produced

2) Enoyl CoA hydratase hydrates enoyl CoA into L-3-hydroxyacyl CoA and this removes the double bond and H2O is used

3) 3-hydroxyacyl CoA dehydrogenase oxidizes 3-hydroxyacyl CoA into 3-ketoacyl CoA and this produces NADH + H+

4) B-ketothiolase cleaves 3-ketoacyl CoA via thiolysis which involves the addition of an HS-CoA molecule. One acetyl CoA and one acyl CoA are formed (the acyl-CoA is shortened by two carbon atoms)

Question 13

What are the steps of fatty acid oxidation of for an unsaturated fatty acid chain?

Answer 13

1) Acyl CoA dehydrogenase, enol CoA hydratase, and 3-hydroxyacyl CoA dehydrogenase are used and the first three steps are the same as for unsaturated fatty acid digestion until a double bond between C3 and C4 is present

2) When a double bond between C3 and C4 is present, cis-3-enoyl CoA isomerase moves the double bond to C2 and C3 forming trans-2-enoyl CoA

3) Trans-2-enoyl CoA can by hydrated by enoyl CoA hydratase and can continue in the B-oxidation cycle

Question 14

How much ATP is produced from the fatty acid oxidation of palmitic acid (C16)?

Answer 14

For each B-oxidation cycle, an acyl CoA
1) is shortened by 2 carbons
2) produces one FADH2 (1.5 ATP)
3) produces one NADH+H+ (2.5 ATP_
4) produces one Acetyl CoA which produces 10 ATP via the citric acid cycle since 3 NADH+H+, 1 FADH2, and 1 GTP is formed

Seven cycles are needed to completely oxidize palmitic acid, so 8 acetyl CoA are produced in addition to 7 FADH2 and 7 NADH+H+

Net production is 106 ATP since activation of palmitic acid to palmitoyl CoA “consumes” 2 ATP. This is equivalent to 6.625 ATP per carbon and one molecule of glucose forms 5.333 ATP per carbon

Question 15

Meselson and Stahl Experiment

Answer 15

1) A CsCl gradient was used to separate different densities of labeled DNA

2) 14N and 15N was used to label two different DNA strands

3) E. Coli was initially incubated in a medium with heavy nitrogen (15N) which served as the only source of nitrogen for the cells

4) Then the cells were transferred into a medium containing only light nitrogen (14N)

5) The samples were then taken after a period of time and DNA was isolated and separated on a CsCl gradient

6) A single band that fell between the light and heavy nitrogen were observed as replication was occurring and after 1 generation or cycle of complete replication, a thick band was observed in between the 14 and 15N

7) A second round of replication occurred in the presence of only light nitrogen and after the second round was completed two distinct bands of DNA were observed. One with 14N and one with combined 14 and 15N.

8) This points to semi conservative replication rather than conservative or fragmented since two distinct bands are observed

Question 16

What are the components and steps of DNA replication in prokaryotes?

Answer 16

1) DNA replication starts at an origin of replication (oriC) which is made up of 245 base pairs, three A-T rich binding sites made up of 13 base pairs, and 5 DnaA binding sites for recognition

2) DnaA binds to oriC (one of the binding sites) and recruits DnaB

3) DnaB complex (helicase) opens/unwinds the double helix at AT rich 13-bp sites and this requires ATP

4) Single-strand binding proteins bind to single-stranded DNA to prevent the reformation of the helix

5) Primase, an RNA polymerase, synthesizes a ~10 bp RNA primer complementary to the template strand which starts DNA synthesis

6) Type II Topoisomerases uncoil dsDNA by cutting, unwinding, and rejoining it and this requires ATP. Type I Topoisomerases uncoil ssDNA by cutting, unwinding, rejoining

7) DNA replication is catalyzed by DNA polymerases and this requires dNTPs, a primer, a template, and Mg(2+) as a co-factor. Strand synthesis occurs in a 5’ –> 3’ direction. DNA Polymerase III replicates DNA in a 5’ –> 3’ direction but it has a 3’ –> 5’ exonuclease function where it can repair DNA if an incorrect base pair is added. As the leading strand is synthesized in a 5’ –> 3’ direction, a loop is formed in the lagging strand and Okazaki fragments are synthesized (also in a 5’ –> 3’ direction)

8) DNA Polymerase I has a 5’ –> 3’ polymerase function (replication), a 5’ –> 3’ exonuclease function for primer removal, and a 3’ –> 5’ exonuclease function for backwards repair

9) DNA Ligase seals “nicks” that connect DNA Okazaki fragments

Question 17

What are the components and steps of transcription in prokaryotes?

Answer 17

1) Catalyzed by RNA polymerase and requires a divalent cation cofactor. Also requires a template where an RNA strand synthesized is complementary to the “template strand” of DNA. This is extremely similar to the DNA coding strand except that thymine is replaced with uracil. Process requires RNTP’s and it is synthesized in the 5’ –> 3’ direction

2) Initiation - facilitated by the sigma subunit in RNA polymerase. This subunit recognizes and binds to a TATA (Pribnow) box ~10 base pairs upstream from the transcriptional start site and another sequence ~35 bp upstream from the start site

3) Rho dependent termination - uses a Rho protein that binds to a phosphorylated 5’ end of the mRNA sequence. The rho protein then moves up the RNA polymerase using ATP and dislodges it from the template terminating synthesis

4) Rho independent termination - a transcribed GC rich stop signal in mRNA forms a hairpin loop with H bonds and this is followed by 4 or more uracils. This destabilizes the RNA Polymerase complex which then disassociates from the template strand

Question 18

What does the lac operon consist of?

Answer 18

The lac operon encodes three genes, LacZ which encodes for B-Galactosidase (metabolizes lactose), LacY which encodes for Permease (allows lactose to enter the cell), and LacA which encodes for Transacetylase (helps metabolize lactose). These three genes are regulated by the promoter and operator

These genes are downstream from an operator region, where a repressor binds to preventing RNAP assembly/progression, and a promoter region, which contains the -10 and -35 promoter regions and allows for RNAP binding

Further downstream is the i gene that encodes for the repressor protein, and this has its own promoter region without an operator. This means that the assembly of repressor proteins is constant and transcription is turned off

Question 19

How does lactose regulate the lac operon?

Answer 19

When lactose enters the cell, it can be converted into allolactose and this binds to the repressor protein forming a repressor-inducer complex. This complex does not bind DNA, and this means the lac operon is activated

Question 20

How does glucose regulate the lac operon?

Answer 20

To activate the lac operon, a protein called the Catabolite Activator Protein (CAP) binds to cAMP to form cAMP-CAP complexes that recruit RNA polymerase to the promoter near the -35 site

When glucose is abundant, cAMP levels are low and since there are few cAMP-CAP complexes, the lac operon has lower levels of transcription. This is catabolite repression by glucose

Therefore the lac operon has the highest level of activity when glucose levels are low and lactose levels are high, second highest level when lactose and glucose are both present, second lowest level when lactose and glucose are both absent, and lowest level when lactose is low and glucose is high

Question 21

What are characteristics of the mRNA genetic code?

Answer 21

The code has to be non-overlapping, it has to have no punctuation and is read sequentially and continuously, it has directionality and is read in a 5’ –> 3’ direction, and it is degenerate where amino acids are encoded by more than one codon so you cannot tell the RNA sequence from an amino acid sequence but you can tell the amino acid sequence from RNA

Question 22

What is tRNA and what is its role in translation?

Answer 22

tRNA is the adapter molecule between the codon and a specified amino acid. It is made up of 73-93 ribonucleotides and is in an “L-shaped” and “Cloverleaf” pattern. Made up of nucleic acids (A, U, C, G) and some derivatives.

The 5’ terminus is phosphorylated and the 3’-CCA terminus serves as the acceptor stem where an amino acid attaches to. There are several loops and one of the loops is the anticodon loop which binds to mRNA

Question 23

What are aminoacyl-tRNA synthetases and what is their role in translation?

Answer 23

They pair the correct amino acid with the correct tRNA based on its anticodon.

This enzyme catalyzes the formation of aminoacyl-adenylate via the attachment of a high-energy bond from ATP to the amino acid. It then catalyzes the transfer of an aminoacyl group to a specific tRNA matching the anticodon to the amino acid

Question 24

What are ribosomes and what is their role in translation?

Answer 24

Ribosomes are a ribonucleic protein that contain both protein and RNA. It consists of a large 50S subunit with 34 different proteins and 2 RNA molecules and a small 30S subunit with 21 different proteins and one RNA molecule. Translation 5’ –> 3’ on the mRNA. Proteins are also synthesized in an amino to carboxyl (N to C) terminus direction.

Question 25

What are the components and steps of PCR?

Answer 25

The components are a Taq polymerase (heat stable), primers, template DNA, nucleotides, and sometimes cofactors

1) Primers are generated based on the flanking sequence of a target sequence. First excess primers are added and the tube is heated to 95 deg C to separate strands

2) The reaction is cooled to ~60 deg C to anneal the primers

3) The reaction is heated to room temp ~72 deg C to synthesize new DNA and this forms two new long strands

4) A second cycle begins forming two short strands from the newly synthesized strands that act as templates. As cycles continue, short strands containing the target DNA are amplified exponentially (millions within a few hours)

Question 26

What are the steps of RT-PCR?

Answer 26

This process combines reverse transcription (where DNA is formed from RNA, generally viral) and PCR

1) Reverse transcriptase is used to synthesize a single strand of DNA using RNA as a template. For a eukaryotic source of mRNA, the primer that is used is an Oligo(T) primer complementary to a Poly A tail

2) Oligo(G) is attached to the 3’ end of ss cDNA, and this process does not require a template

3) An Oligo(C) primer targets the OligoG sequence and DNA polymerase and dNTPs can then be used to synthesize the double stranded DNA

4) If the source of the RNA sequence in known, any primer that is complementary to the RNA sequence can be used, not just an Oligo(T) primer that targets a Poly A tail

Question 27

What are qPCR and RT-qPCR and what are the steps?

Answer 27

qPCR is quantitive PCR or Real-time PCR and it allows you to amplify and detect changes in a copy number (monitoring PCR), and RT-qPCR combines reverse transcription and qPCR.

A fluorophore (signal) and quencher (inhibits fluorescence) are present on either sides of a TaqMan Probe that binds to a template of a target sequence. The TaqMan Probe binds to the sequence along with the primers, and when the taq polymerase targets the probe, the probe is broken and the fluorophore is released. This means that everytime you synthesize a strand, a fluorophore is released and the fluorescent activity over time can be monitored. This process was used to detect COVID

Question 28

Machado-Joseph Disease

Answer 28

Neurodegenerative disorder in parts of the the brain called the hind brain. Characterized by progressive clumsiness in limbs. Caused by a mutation in the ATXN3 gene where excess CAG repeats are present and a mutated protein results

Question 29

Huntingtons Disease

Answer 29

Neurodegenerative disorder in parts of the brain called the striatum leading to cognitive/motor impairment. Caused by CAG repeats in huntington gene resulting in a mutated protein

Question 30

Amyotrophic Lateral Sclerosis

Answer 30

Neurodegenerative disorder in parts of brain called the primary motor cortex. Leads to motor impairment where cognitive function is still intact. Caused by a hexanucleotide repeat expansion (GGGGCC) in the C9orf72 gene