Biochem: Nucleotide Synth - DNA Repair

Question 1

What is pRpp? How is it involved in nucleotide synthesis reactions?

Answer 1

pRpp = phosphoribosyl pyrophosphate. It is formed during Step 1 of Purine Synthesis–Activation of the C1’ carbon–when ribose phosphate pyrophosphate kinase transfers a pyrophosphate from ATP to ribose-5’-phosphate. This activation reaction creates free energy available to do work of forming the N-glycosidic bond (Step 2 of purine synthesis)

Question 2

In what other reactions does pRpp participate? How is it that pRpp stands at a metabolic crossroads? What implications does this have for regulatory processes of purine synthesis?

Answer 2

the pRpp molecule (phosphoribosyl pyrophosphate) is not only used for the synthesis of purines, but also for the synthesis of (3) others: pyrimidines, and two amino acids Histidine and Tryptophan. Thus, pRpp has a number of different fates = “metabolic crosscroads”

Question 3

How does pRpp influence regulation of purine synthesis?

Answer 3

pRpp influences enzyme activity by feedforward (stimulatory) activity. (green arrow). Thus, increasing the concentration of pRpp can overcome other regulatory activity of purine synthesis, such as the feedback inhibition exerted by the end-product nucleotides, Adenine and Guanine.

Question 4

Interpret these graphs.

Answer 4

APRT = enzyme that converts IMP into either Adenoside or Guanine monophosphates (AMP & GMP).

Plotting APRT enzyme activity relative to [] of its two substrates (glutamine and pRpp) shows that the ultimate regulatory behavior is exerted through binding pRpp = sigmoidal curive. Guanine and adenine nucleotides individually shift the curve to the right. When added together (GXP & AXP), curve shifts more to right. However, increasing the [pRpp] can override the regulatory effects of GXP & AXP = feedforward activation

Question 5

What is the committed step in purine synthesis? Which enzyme is involved in this step and what products are formed?

Answer 5

The formation of the N-glycosidic bond (step 2) is the committed step in purine synthesis pathway. In this rxn, amidophosphoribosyl transferase enzyme transfers an amide nitrogen from side chain of glutamine to C1’ of pRpp, releasing the pyrophosphate and forming phosphoribosylamine.

Question 6

What disease is caused by HGPRT deficiency. Explain how deficiency of this enzyme influences de novo purine synthesis and subsequent effects on purine catabolism.

Answer 6

HGPRT (hypoxanthine/guanine phosphoribosyl transferase) = 1 of 2 enzymes involved in purine salvage pathways, which is process of cells recycling nitrogenous bases for making their own purines, rather than relying on getting them from diet).

Lesch Nyhan syndrome is a X-linked disease caused by HGPRT deficiency. Without HGPRT, pRpp levels increase, which causes an increase in purine synthesis (feedforward activation). Subsequently, there is an increase in purine catabolism and buildup of the end product–uric acid–leading to the symptoms of Lesch Nyhan syndrome.

Question 7

What are the clinical manifestations of Lesch Nyhan syndrome? Deficiency of what enzyme causes this disease?

Answer 7

Lesch Nyhan (LN) syndrome is an X-linked disease caused by deficiency of HGPRT enzyme. This is one of two enzymes involved in purine salvage pathways.

Deficiency of HGPRT → increased pRpp → increased purine synthesis → increased purine catabolism → higher uric acid levels

LN manifestations = gout-like sx (due to high uric acid), and neurological abnormalities with a predilection towards self mutilation, such as chewing off one’s fingers and lips (cause unknown)

Question 8

Why can uracil, and not cytidine, be used to treat orotic aciduria?

Answer 8

Phosphorylated compounds seldom make good drugs because they are highly charged and, thus, do not enter cells well. Uridine, on the other hand, can enter cells and be posphorylated to UMP through the pyrimidine salvage pathway. Whereas UMP can be used to synthesize CMP, the reverse is not true. So cytidine cannot be used as a source of UMP within cells.

Uracil alleviates symptoms caused BOTH by the lack of pyrimidines and by the accumulation of orotic acid.

Question 9

What does ribonucleotide reductase do?

Answer 9

An enzyme that reduces the 2’ carbon of ribose from -OH to -H, making deoxyribose. The reducing is accomplished by 2 thiol groups (S-H) on the enzyme that give up their H’s and form a disulfide bond (S-S). (Thus, the 2 thiols get oxidized, while ribonucleotide gets reduced!)

Question 10

Ribonucleoside v. Ribonucleotide

Answer 10

RibonucleoSide = base + carbohydrate without the phosphate

RibonucleoTide = base + carbohydrate + phosphate

Question 11

What is Hydroxyurea and what is its target?

Answer 11

Hydroxyurea is a cancer chemotherapeutic agent (anti-neoplastic) that is a bone marrow suppressant: suppresses WBC production and, thus, reduces inflammation.

It does one thing only, which is to quench the tyrosine radical that is essential to the functioning of the enzyme ribonucleotide reductase. If this enzyme doesn’t work, then can’t produce deoxyribonucleotides and cells can’t divide. And cells dividing the most = bone marrow

Question 12

What is the overall importance of folic acid derivatives to the synthesis of both purines and pyrimidines?

Answer 12

Folic acid = 1 carbon metabolism for deoxyribonucleotide and ribonucleotide synthesis, which is essential for DNA replication. Folid acid is the vitamin precursor (COENZYME) for:

• N¹⁰-formyl-THF, needed for two of the carbons in the purine ring (they come in as 1-carbon units)
• N₅,N₁₀-methylene tetrahydrofolate = 1 of 2 substrates used by enzyme, thymidylate synthase to convert U to T (pyrimidines). N₅,N₁₀-methylene THF gives a CH₃ to the C’5 of U, converting it to T, while N₅N₁₀ gets oxidized to dihydrofolate.

Question 13

What is a unique aspect of folic acid metabolism in the synthesis of thymidine?

Answer 13

N₅N₁₀ -methylene THF, a folic acid derivative, is used by thymidylate synthase to give a methylene to U (uracil), which converts it into T (thymidine). In the process of giving this methylene, N₅N₁₀ gets oxidized to dihydrofolate. This is the ONLY reaction in the human body where a folic acid derivative gets oxidized from tetrahydrofolate to dihydrofolate.

Thus, cells making large amounts of thymidine need the capacity to recycle dihydrofolate back to active tetrahydrofolate = enzyme dihydrofolate reductase

Question 14

Why is leucovorin used after methotrexate as an antidote to rescue patients, but is also used with fluorouracil to enhance fluorouracil’s mechanism of action?

Answer 14

All relates to their mechanisms of action. With fluorouracil, elevated amounts of THF will push the production of that tied-up intermediate, which ultimately leads to the suicide of thymidylate synthase.

But with methotrexate, the synthesis of amino acids and purines/pryimidines is being shut down, leading to cell death. Leucovorin comes in at the last minute and restores the THF pools, so cells can recover.

Question 15

What is fluorouracil and what is its target?

Answer 15

5’-fluorouracil is a suicide inhibitor used as an anti-cancer drug. It targets the thymidylate synthase reaction. The conversion of U to T ultimately reaches a state where thymidylate synthase, dUMP, and methyleneTHF are all bound up in a complex that can only be released with the extraction of the proton at C’5. But w/ fluorouracil replacing dUMP, the F at C’5 cannot be extracted, leaving the complex tied up and ultimately killing the enzyme’s activity. Cell can’t get Thymine, can’t get other dNTPs = apoptosis.

Side effects = bone marrow suppression, GI distress, and hair loss (cells in hair follicles can’t divide and die)

Question 16

What does dihydrofolate reductase do?

Answer 16

It’s an enzyme that uses NADPH to reduce dihydrofolate to tetrahydrofoalte. This is important in the thymidylate synthase reaction, where N₅N₁₀-THF gets oxidized to dihydrofolate in the process of converting U to T. **Note, that ONLY dividing cells are dependent on activity of dihydrofolate reductase. Thus, it is another target for cancer chemotherapy = methotrexate.

Question 17

How does methotrexate work? What is its target?

Answer 17

Methotrexate is a cancer chemotherapeutic drug that binds the enzyme dihydrofolate reductase (DHFR) with a 1,000-fold higher affinity than the enzyme binds to its substrate, dihydrofolate. Selectively inactivating this enzyme will deplete thymine production, shutting down DNA replication and, thus, cell division. (Remember, only dividing cells are dependent on the activity of DHFR)

Question 18

How are methotrexate and leucovorin used in cancer treatment?

Answer 18

They are used to treat childhood leukemias, by giving children a lethal dose of methotrexate, then rescuing the pt with leucovorin (methyl-THF). The methotrexate inhibits the recycling of DHF back to THF (by the activity of DHFR), thereby shutting down purine/pyrimidine synthesis, leading to cell death. NOT just the tumor, but all cells in body will die. Leucovorin comes in as the rescue, rapidly restoring the THF pools and normal cells can recover.

Question 19

What is leucovorin and how is it used?

Answer 19

Leucovorin is N⁵-methyl-THF. It is used as a rescue in childhood leukemias. After methotrexate has killed a lot of the dividing cells by shutting down THF recycling, leucovorin comes in and restores the THF pools. It is also used to enhance the activity of Fluorouracil. With elevated amounts of THF, more and more enzyme/methylene-THF/fluoruracil knots are made, inhibiting and killing the thymidylate synthase enzyme.

N⁵-methyl THF –> methionine (amino acid) + THF

Question 20

What is dihydropyrimidine dehydrogenase? How is it related to treatment with 5-fluorouracil?

Answer 20

Dihydropyrimidine dehydrogenase is an enzyme that works on pyrimidine catabolism (either U or T). 2-8% of the population has mild deficiency of this enzyme, and if you treat them w/ fluorouracil, you can kill them. Because of their inability to catabolize pyrimidine derivatives, they’ll get unusually high amounts of fluoro-deoxyuracil in cells = life-threatening toxicity.

Question 21

What function do the activity site and specificity site on Ribonucleotide Reductase serve?

Answer 21

The activity site = on/off switch for overall enzyme activity.

• ATP binding to this site = on
• dATP binding = off (b/c all other dNTPs have been produced in sufficient amounts)

The specificity site regulates substrate selection and assures balanced production of the four dNTPs. Important b/c imbalances in dNTP levels confuse DNA polymerase and can cause errors in replication

Question 22

Explain how the regulation of ribonucleotide reductase activity contributes to the clinical manifestations of adenosine deaminase deficiency.

Answer 22

ADA (adenosine deaminase deficiency) leads to SCIDS. Deficiency of the ADA enzyme that deaminates AMP to IMP leads to increased dAMP (b/c ADA only works on AMP and not dAMP). Both B and T lymphocytes have a lot of nucleotide kinases. These hyperactive lymphocyte kinases will then phosphorylate all that dAMP to dATP. And with increased dATP, ribonucleotide reductase will believe it has enough dNTPs and choke off production of C, G, and T = inhibited cell replication. Pt’s have affected ability to mount immune response involving B and T lymphocytes.

Question 23

What happens in ribonucleotide reductase when ATP binds to the specificity site?

Answer 23

binding of ATP to the specificity site stimulates reduction of either UDP or CDP by the active site. The dCDP is converted to dCTP, while UDP ultimately gets converted to dTTP.

dTTP then binds to the specificity site…

Question 24

What happens in ribonucleotide reductase when dTTP binds to the specificity site?

Answer 24

dTTP binding to the specificity site promotes binding of GDP to the active site and inhibits further reduction of UDP and CDP. The dGDP is converted to dGTP.

dGTP then goes on to occupy specificity site…

Question 25

What happens in ribonucleotide reductase when dGTP occupies the specificity site?

Answer 25

Once dGTP builds up, it occupies the specificity site, inhibiting further reduction of GDP and promoting reduction of ADP. The dADP gets converted to dATP, and when it builds up it binds to the activity site, inhibiting enzyme activity until the levels of the 4 dNTPs begin to fall again.

Question 26

During what phase of the cell cycle does ribonucletide reductase enzyme activity increase?

Answer 26

During S phase, when high levels of dNTPs are needed for replication.

Question 27

What are the two ways by which adenosine derivatives can be converted to inosine (ultimately going down pathway of purine degradation)?

Answer 27

(2) enzymes that deaminate = replace NH₄⁺ with C=O

• AMP deaminase deaminates AMP to IMP. It only works on AMP (ribonucleotide form) and NOT dAMP (deoxyribonucleotide form)
• So, dAMP can only be catabolized by first being dephosphorylated to deoxyadenosine and then being deaminated by adenosine deaminase.
  
  • deficiency of adenosine deaminase leads to buildup of dAMP and causes SCIDS

Question 28

What is SCIDS?

Answer 28

SCIDS = Severe combined immunodeficiency disorder. It is caused by a deficiency of the enzyme adenosine deaminase, which leads to increased levels of dATP that shut down ribonucletide reductase activity. Without the cell proliferation that requires ribonucleotide reductase, B and T lymphocytes can’t make enough of themselves to mount an adequate immune response. Therefore, pts with SCIDS have inability to fight off infection.

Question 29

What is the role of the hematopoeitic stem cell (HSC) in using a bone marrow transplant to treat a patient with adenosine deaminase deficiency?

Answer 29

HSCs give rise to all other blood cells, including lymphocytes and RBCs. ADA deficiency leads to SCIDS, which selectively kills B and T lymphocytes. So can treat with bone marrow transplantation. But:

• need to destroy pts normal HSC first (mainly to make room for new cells)
• MUST have matched donor (matched sibling = best)

Question 30

Why is Gout more prevalent in adult men relative to premenopausal women?

Answer 30

Uric Acid (end produce of purine catabolism) can take 3 forms. It’s conj. base form–sodium urate–is the predominant form in serum with a solubility of ~7.3-7.4 mg/dL. This is much more soluble than its acid form ~0.4 mg/dL.

Circulating levels of sodium urate in males ~7 mg/dL (almost at its solubility limit), whereas in premenopausal females ~6 mg/dL. This difference accounts for the 3:1 male:female ratio in gout.

Question 31

How do diet and alcohol consumption factor into gout pathophysiology?

Answer 31

Diet - foods rich in purines lead to increased levels of uric acid, include: shellfish, anchovies, and red meat

Alcohol - problematic in 2 ways:

• alcohol → dehydration → higher [uric acid] in serum
• ethanol can be metabolized to lactic acid, which blocks uric acid excretion of kidneys

Question 32

What is another name for gout? What is the overall pathophysiology for gout?

Answer 32

Gout, a.k.a. Crystalline arthropathy because crystals of uric acid deposit in the joint, causing extreme pain.

Problem of gout is ultimately that sodium ureate has low solubility (~7.3 mg/dL) in the serum and generally high [] in adult bodies. Thus, problem of gout is ultimately of solubility of uric acid. Crystals of uric acid crash out of the solution.

Question 33

How does the drug allopurinol reduce risk for gout flare-ups?

Answer 33

Allopurinol inactivates the enzyme xanthine oxidase, which is at the end of the purine degradation pathway, converting xanthine and hypoxanthine into uric acid. Thus, allopurinol decreases uric acid levels. (Both xanthine and hypoxanthine are still able to be eliminated from the body).

Question 34

Name a common characteristic between the way allopurinol and 5-fluorouracil inhibit their enzyme targets.

Answer 34

They are both substrate analogs, where the enzyme works on the drug up until a point where it can go no further (stuck), so that enzyme is inactivated.

• Allopurinol: xanthine oxidase works to add the 3 carbonyls it add to other purines. But is unable to add a C=O at the C’8 position of allopurinol, b/c there is a N substituted for the C.
• 5-fluorouracil: binds thymidulate synthase (and its other substrate, methylene-THF), keeping it tied up in complex, since the F at C’5 position cannot be extracted like the H typically can in uracil.

Question 35

What does the “repeat number” refer to in Huntington’s disease?

Answer 35

number of CAG trinucleotide repeats contained in gene encoding for Huntingtin protein. Normally, the CAG segment is repeated 10-35 times within the gene. In people with HD, repeat = 36 - 120+ times. People with 36 to 39 CAG repeats may or may not develop the signs and symptoms of Huntington disease, while people with 40 or more repeats almost always develop the disorder.

Question 36

Explain how genetic anticipation factors into Huntington disease pathophysiology.

Answer 36

HD is autosomal dominant inheritance. As the altered HTT gene is passed from one generation to the next, the size of the CAG trinucleotide repeat often increases in size. A larger number of repeats is usually associated with an earlier onset of signs and symptoms. This phenomenon is called anticipation.

People with the adult-onset form of HD typically have 40 - 50 CAG repeats in the HTT gene, while people with the juvenile form of the disorder tend to have more than 60 CAG repeats.

Individuals with 27-35 CAG repeats in the HTT gene do not develop HD, but are at risk of having children who will develop the disorder. As the gene is passed from parent to child, the size of the CAG repeat may lengthen into the range associated with HD (36+)

Question 37

What does CAG code for and how is this related to Huntington’s Disease?

Answer 37

CAG repeat codes for the amino acid Gluatmine (Gln). Glutamine repeats in a protein can cause aggregation, which is known to contribute to HD pathology.

Question 38

What are the (3) enzymatic activities that all DNA polymerases share? What are the (2) additional activities of DNA polymerase I?

Answer 38

All DNA polymerases:

• create complementary copies of a single-stranded template
• require a primer oligonucleotide with a free 3’ hydroxyl to extend
• new chain growth occurs only in 5’ to 3’ direction

In addition, DNA polymerase I:

• 3’ to 5’ exonuclease activity (edits out incorrectly added nucleotides)
• 5’ to 3’ exonuclease activity (removes damaged DNA or RNA primers

Question 39

Which special enzymatic activity of DNA polymerase I is responsible for removing RNA primers from Okazaki fragments? Which removes damaged DNA downstream of a nick in a DNA strand?

Answer 39

5’ to 3’ exonuclease activity (same direction as synthesis): removes damaged DNA and removes RNA primers. Then polymerase activity adds nucleotides back in.

3’ to 5’ exonuclease activity (opposite of synthesis direction): edits out incorrectly added nucleotides

Question 40

Exonuclase v. Endonuclease

Answer 40

First, remember nuclease is the opposite of polymerase. Nuclease is breaking down chains.

Exonuclease = coming in from the free end to chop away bases from that end; cleaving phosphodiester bonds

Endonuclease = internally breaking a phosphodiester bond. (v. Exonuclease, which only comes in from one end or the other)

Question 41

How does DNA polymerase I 5’ to 3’ Exonuclease activity work?

Answer 41

DNA polym I binds to single strand nicks and excises DNA in front of the nick with its 5’ to 3’ exonuclease activity. Then polymerase fills in behind it, adding in bases. *Same activity it uses to remove RNA primers. So can think of a nick as an Okazai fragment!

Question 42

Summarize the structure and function of telomerase. Why is it a good target for cancer chemotherapy?

Answer 42

Telomerase is a specialized enzyme that replicates telomeres. It is a protein/RNA complex that adds a repeat structure (5’-TTAGGG-3’) to the 3’ ends of each strand, creating G-rich-non-info sequences at chromosome termini. The RNA component of telomerase serves as the template for G-rich repeat. Using reverse transcriptase activity, telomerase uses the RNA template to add DNA bases.

Somatic cells often lack telomerase activity and only divide for several generations before telomeres fall below a critical length and cell division stops. However, immortal cells (such as tumors) typically reactivate telomerase activity, making telomerase a cancer chemo target.

Question 43

What are the (2) functions of G-rich telomeres?

Answer 43

1) solves the issue of replication, where chromosome ends would be shortened with each round of replication
2) provides G-quartet caps, that mask the ends of chromosomes, so they don’t look like damaged DNA fragments

Question 44

What is the Ames test used for? How does it work? How could you use it to test a theory that eating hot dogs increases your risk for cancer?

Answer 44

Used to identify chemicals that can cause damage to DNA, potentially creating a mutation (chemical mutagens) Way to identify potential carcinogens.

Test uses a His^- bacteria, which can only grow if provided with Histidine. Medium has no histidine. Then add chemical of interest, e.g. Nitrite (preservative in hot dogs known to cause mutations). If bacteria grow, shows capability of mutation.

Question 45

Why are rat liver extracts included in bacterial growth media of Ames test?

Answer 45

because certain mutagens need to be bioactivated (metabolized) by the liver in order to become mutagenic

Question 46

What is nitrite? Where is it found? How could it cause potential mutations?

Answer 46

Nitrite: the conjugate base of nitrous acids. Used as a preservative in meat products. e.g. hot dogs!

Nitrous acid can cause (2) different oxidative deaminations, which if not fixed, can cause a mutational change:

• Cytosine → Uracil (pair w/ A)
• Adenine → Hypoxanthine (pair w/ C)

Question 47

How can increased DNA repair activity lead to a glioblastoma (brain tumor) becoming resistant to Temozolomide?

Answer 47

Temozolomide is an alkylating that transfers methyl grps to guanine. This damages DNA in hopes that p53 will kill the rapidly dividing tumor cells. However, MGMT (methylguanine methyl transferase) is a cell’s natural defense mechanism that recognizes methylguanine and converts it back to G. Tumors can overexpress MGMT to protect themselves, thereby leading to temozolomide resistance.

Question 48

What is MGMT?

Answer 48

methylguanine methyl transferase. An enzyme used by cells to undo DNA damage done by alkylating guanine. Overexpression of MGMT can lead to resistance against Temozolomide (alkylating cancer chemo drug) in glioblastomas and other tumors.

Question 49

How can the base excision repair pathway contribute to repeat expansion in Huntington’s disease?

Answer 49

the CAG repeat forms a hairpin structure that interferes with recognition by the flap endonuclease. Hairpins full of C’s and G’s = very stable. If the flap is not removed and instead is joined to the remaining DNA by DNA ligase, an expansion of the trinucleotide repeat can occur.

Question 50

What does the base excision repair pathway do? What are the 4 general steps involved?

Answer 50

Removes and replaces oxidatively damaged bases.

1. DNA glycosylase OGG1 recognizes the damaged base (8-oxoguanine) and cleaves N-glycosidic bond leaving an abasic site in DNA
2. Abasic sites recognized by an endonuclease, APE1 endonuclease, which cleaves the phosphodiester bond upstread
3. Gap-filling polymerase fills in w/ correct bases
4. Flap endonuclease removes flap (displaced strand) and DNA ligase joins newly repaired strand w/ original one

Question 51

What is 8-oxoguanine? What pathway would work on it?

Answer 51

8-oxoguanine is oxidative damage to guanine where C=O is added to C’8 position of G. Base excision repair pathway would recognize this damaged base and work on repairing it. Beginning with recognition by DNA glycosylase OGG1.

Question 52

Base excision repair v. Nucleotide excision repair

Answer 52

base excision repair: recognizes 8-oxoguanine, uracil, and abasic sites

nucleotide excision repair: recognizes UV-induced damage (pyrimidine dimers) and bulky adducts (e.g. carcinogens from cigarette smoke)

Question 53

Why is essential that people with Xeroderma Pigmentosum (XP) wear sunscreen? Why would these people also be at an increased risk for smoking-associated lung cancer compared to the general population?

Answer 53

Pts with XP are deficient in the ~16 proteins involved in nucleotide excision repair. Nucleotide excision repair recognizes UV-induced and bulky adduct (e.g. from cigarette smoke) DNA damage and repairs. Thus, XP patients are highly sensitive to sunlight and have a 2000-fold increased rate of skin cancer relative to general population. For same reason, they have a higher risk of cigarette smoke-induced cancer.

Question 54

What is the mismatch repair (MMR) pathway and what is its link to colon cancer?

Answer 54

MMR repairs DNA mismatches done by DNA polymerase during replication. In humans, defects in mismatch repair proteins cause hereditary nonpolyposis colorectal cancer syndrome (Lynch syndrome)

Question 55

What is Lynch Syndrome?

Answer 55

nonpolyposis colorectal cancer. It is hereditary and can be caused by defects in mismatch repair (MMR) proteins. Recall that MMR repairs DNA mismatches done by DNA polymerase during replication.

Question 56

How does a property of one of HIV’s enzymes give rise to a high rate of drug resistance in treating AIDS patients?

Answer 56

reverse transcriptase lacks proofreading activity. Therefore, it is error prone with a high mutation rate. This high mutation rate contributes to drug resistance.

Question 57

Why is HIV reverse transcriptase a good target for drugs used to treat HIV infections?

Answer 57

HIV is a retrovirus that uses its own RNA genome to synthesize double-stranded DNA by way of HIV reverse transcriptase enzyme. (That DNA then gets incorporated into our DNA and is used to transcribe new viruses). This is a good target for tx, because–other than telomerase–we don’t have any other reverse transcriptase enzymes in our body!