Biochemistry & Genetics Important Concepts Q&as Dr Kumar Ponnusamy & Jegathambigai RN, Biochemistry & Genetics, International Medical University (IMU) / AIMST University School of Medicine, Malaysia

“Dr Kumar Ponnusamy & Ms Jegathambigai RN, Biochemistry & Genetics, AIMST University School of Medicine, Malaysia” Important Concepts in Biochemistry & Biomedical Genetics, USMLE PreparatoryMBBS Curriculum

“The Authors Acknowledges All the Biochemistry Text Books, On-Line Resources” Inborn Errors of AA Metabolism AA to Specilized Products Purine & Pyrimidine Heme Synthesis & Degradation Bilirubin Metabolism, Porphyrias, Biomedical Genetics: Single Gene Disorders, Chromosomal Abnormalities, Population Genetics, Genetic Diagnosis, Presursors of Heme, Purine and Pyrimidine Biosynthesis. SCID, Adenosine Deaminase Deficiency. Coenzymes required for catecholamine, heme, purine and primidine synthesis & degradation pathways. Rate-limiting Step of Heme Biosynthetic pathway and Purine & Pyrimidine pathways. Allosteric Inhibition Lead toxicity Lesch-Nyhan Syndrome, cause Crigler-Najjar Syndrome, cause Dubin-Johnson Syndrome Neonatal Jaundice, causes Vitamin B6 deficiency Hereditary Spherocytosis, cause Isoniazid and Vitamin B6 Nutritional Deficiency & Anemias Bilirubin Conjugation, enzyme catalaysed and site of conjugation & deconjugation. Albinism, enzyme deficiency, signs and symptoms. Phenylketoneuria, enzyme deficiency, Signs & Symptoms. Alkaptonuria, enzyme deficiency. Cystationuria, causes and clinical signs & symptoms. Catecholamine Synthesis & Coenzymes involved Melanin Synthesis Gout, causes

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Allopurinol, mode of action. PCT, Erythropoietic Porphyria , enzyme deficiencies. HGPRT, function consequences. UDP-GT, Function. Lens dislocation THF deficiency & Methionine metabolism Mode of action of 5-FU and Methotrexate as an anticancer agent. Mode of action of methotrexate as an imuuno suppressor. Mode of action of Mycophenolic acid as a an immunosuppressor. Folate anti-metabolite as an Antibiotic (sulfa drugs) Obstructive Jaundice. What are the precursors require for Heme Biosynthesis?. Glycine + Scccinyl CoA. What is the regulatory and rate-limiting step in the biosynthesis of heme? Reaction catalyzed by the enzyme δ-Amino Levulinic Acid (δ-ALA). What is the Allosteric inhibitor of heme Biosynthesis? Heme. What vitamin co-enzyme is required in the first step catalyzed by the enzyme δ-ALA. Vitamin B6 / Pyridoxal Phosphate. Name an anti-tuberculosis antibiotic?. Isoniazid. Patient on TB therapy with Isoniazid, which vitamin one must supplement? Vitamin B6 / Pyridoxal phosphate. Which one of the common environmental heavy metal toxin leads to anemia? Lead. Name the enzymes inhibited by lead in the heme biosynthetic pathway?.

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ALA Dehydrase & Ferrochelatase. What is the regulatory and rate-limiting step in the biosynthesis of heme? Reaction catalyzed by the enzyme δ-Amino Levulinic Acid (δ-ALA). What is Bilirubin? Bilirubin is a degradation product of heme. What is meant by jaundice? Accumulation of the yellow color pigment bilirubin in the circulation, sclera and skin. Name the 2 types of Bilirubins? Conjgated Bilirubin = Bilirubin diglucuronide = Direct Bilirubin. Unconjugated Bilirubin = Bilirubin ( No diglucuronide) = Indirect bilirubin. Name the 3 types of jaundice? Hemolytic jaundice Obstructive jaundice. Infective hepatitis. Name the straw yellow colored pigment excreted in the urine> Urobilinogen. Name the brown colored pigment excreted in the stools?. Stergobilinogen. In which type of jaundice one can expect clay-colored stool and WHY?. Obstrutive jaundice. Obstruction of bile duct by galll stone or tumor prevent the flow of bile to the deodenum. Vit B6 Deficiency, iron Deficiency & Lead Poisoning leads to __________ anemia. Microcytic anemia. Vit B6 Deficiency & Lead Poisoning leads to __________ in bone marrow.

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Ringed sideroblasts. Urobilinogen Decarboxylase deficiency leads to ______. Porphyria Cutanea Tarda (PCT). Proptoporphyria is caused by the deficiency of the enzyme ________. Ferrochelatase. PCT is inherited as an autosomal dominant trait, caused by the congenital deficiency of the enzme ________. Uroporphyrinogen Decarboxylase (UROD). What is meant by Dubin-Johnson Syndrome (DJS)?. Dubin-Johnson Syndrome (DJS) is an autosomal recessive disorder which causes an increase of conjugated-bilirubin without elevation of liver ALT & AST. DJS is associated with a defect in the ability of hepatocytes to secrete conjugated-bilirubin into the bile. In which cells the Bilirubin is synthesized?. Macrophages. What is the organ in which conjugation of bilirubin occurs?. Liver. The enzyme ____________ converts bilirubin into Conjugated bilirubin in liver. UDP-Glucuronyl Transferase (UDP-GT). What laboratory test will be helpful to identify the type of bilirubin? Van Den Berg’s Diazotised sulfanilic acid test. What is the cause of neonatal jaundice? Congenital deficiency of UDP-Glucuronyl Transferase (UDP-GT) causes Neonatal Jaundice.

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What type of bilirubin is found to be elevated in the case of Hemolytic Jaundice? Unconjugated bilirubin (Bilirubin). What type of bilirubin is found to be elevated in the case of Obstructive Jaundice? Conjugated Bilirubin (Bilirubin Diglucuronide) / Direct Bilirubin. What type of bilirubin is found to be elevated in the case of hepatocellular jaundice?. Both Conjugated & unconjugated Bilirubin. What type of bilirubin is found to be elevated in the case of Neonatal jaundice?. Indirect bilirubin. Name the causes of hemolytic crisis. v Episode of hemolysis in G-6-PDH deficiency. v Sickle cell crisis. v Rh disease of newborn. What is the cause of Crigler-Najjar Syndrome?. Bilirubin conjugation is low because of genetic or functional deficiency of the Glucuronyl transferase” system, unconjugated & total bilirubin increase.

“UDP-

What are the causes of Bile Duct Occlusion?. Occlusion of the bile duct (gallstone, primary biliary cirrhosis, pancreatic cancer) prevents conjugated bilirubin leaving the liver.

Inborn Errors of Amino Acid Metabolism

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1. Which one of the following statements concerning a one-week-old male infant with undetected classic phenylketonuria is correct? A. Tyrosine is a nonessential amino acid for the infant. B. High levels of phenylpyruvate appear in his urine.√ C. Therapy must begin with the first year of life. D. A diet devoid of phenylalanine should be initiated immediately. E. When the infant reaches adulthood, it is recommended that diettherapy be discontinued. Answer: B. Phenyllactate, phenylacetate, and phenylpyruvate which are not normally produced in significant amounts in the presence of functional phenylalanine hydroxylase, are elevated in PKU, tyrosine cannot be synthesized from phenylalanine and, hence becomes essential and must be supplied in the diet. Treatment must begin during the first 7-10 days of life to prevent mental retardation. Discontinuance of the phenylalanine-restricted diet before 8 years of age is associated with poor performance on IQ tests. Adult PKU patients show deterioration of attention and speed of mental processing after discontinuation of the diet. Life-long restriction of dietary phenylalanine is, therefore, recommended. 2. A new test is developed that can non-radioactively “label” compounds in the human body. As a physician with a back-ground In the new field of metabolomics, you assess a 21-year-old with classical phenylketonuria (PKU). Phenylalanine is fed with a label in the phenyl ring. In the urine, in which of the following compounds would you expect to find the greatest amount of label?. A. B. C. D. E.

Tyrosine. Tryptophan. Epinephrine. Phenylketone.√ Acetate.

Answer: D. Phenylketonutria (PKU) is a defect in phenylalanine hydroxylase, resulting in a block in the conversion of phenylalanine to tyrosine. Phenylalanine accumulates in in both disorders and to converted to phenylketones. Tyrosine is the product whose formation is blocked, and epinephrine, a product of tyrosine, would not be made or “labelled’. Acetate and tryptophan are very far downstream from tyrosine.

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Phenylketoneuria (PKU) The coenzyme Tetrahydro biopterin is required for the bioconversion of Phenylalanine into Tyrosine.

Congenital deficiency of the enzyme “Phenylalnine hydroxylase” hydroxylase” leads to PKU, which is characterized by the ↑sed excretion of phenylpuruvate, phenylpuruvate, phenylacetate & phenyllactate in Urine (Phenyl (Phenyl keto acids). acids).

Deficiency

3. A 56-year-old man with a history of genetic disease undergoes hip replacement surgery for arthritis. During the operation the surgeons notes a dark pigmentation (ochronosis) in the man’s cartilage. His ochronotic arthritis is most likely caused by oxidation and polymerization of excess tissue: A. B. C. D. E.

Homogentisic acid.√ Orotic acid. Methylmalonic acid. Uric acid. Ascorbic acid.

Answer: A. Adults with alcaptonuria show a high prevalence of ocronotic arthritis due to deficiency of homogentisate oxidase. 4. A four-year-old boy of a first-degree consanguineous couple was noted by the parents to have darkening of the urine to an almost block color when it was left standing. He had a normal sibling, and there were no other medical problems. Childhood growth and development were normal. Which of the following is most likely to be elevated in this patient? A. Methylmalonate. B. Homogentisate.√ C. Phenylpyuvate. D. α-Ketoisovalerate. E. Homocystine.

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Answer: B. Alkaptonuria is a rare metabolic disease involving a deficiency in homogentisic acid oxidase, and the subsequent accumulation of homogentisic acid in the urine, which turns dark on standing. The elevation of methylmalonate (due to methylmalonyl CoA mutase deficiency), phenylpyruvate due to phenylalanine hydroxylase deficiency), α-ketoisovalerate (due to branched-chain α-keto acid dehydrogenase deficiency), and homocystine (due to cystathionine synthase deficiency) are inconsistent with a healthy child with darkening of the urine. 5. A-56-year-old man with a history of genetic disease undergoes hip replacement surgery for arthritis. During the operation the surgeon notes a dark pigmentation (ochronosis) in the man’s cartilage. His ochronotic arthritis is most likely caused by oxidation and polymerization of excess tissue: A. B. C. D. E.

homogentisic acid.√ orotic acid. methylmalonic acid. uric acid. ascorbic acid.

Answer: A. Adults with alcaptonuria show a high prevalence of ochronotic arthritis due to deficiency of homogentisate oxidase.

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Alkaptonuria Homogentisic Acid Formation HO

Transamination NH3 +

O

CH2 CHCO 2 -

CH2 CCO2-

Tyrosine

?

HO

Deficient in Alkaptonuria

p-HydroxyphenylPyruvate

O2

p-HydroxyphenylHydroxyphenyl-

Cleavage of aromatic ring

Homogentisate dioxygenase

XO

2

Fumarate + Acetoacetate

OH

pyruvate dioxygenase (ascorbateascorbate-dep.) dep.) CO2 CH2 CO2-

OH

↑ Homogentisate

7. A 7-day-old girl has had a seizure. The mother explains that the baby has been vomiting and having difficulty feeding for the past 2 days. There is also a strange, sweet smell to her diapers. Physical examination is unremarkable, except for indications of dehydration. Serum test results show normal levels of glucose and ammonia. Uninalysis reveals the presence of α-keto-isocaproate and α-keto-isovalerate. Elevation of this patients liver would reveal deficiency of which of the following enzymes? A. Alaninine aminotransferase. B. Glutamine synthetase. C. Cystathionine β-synthase. D. Carbamoyl phosphate synthetase. E. α-Keto acid dehydrogenase.√ Correct Answer: E: The infant appears to be suffering from maple syrup urine disease (MSUD). This id consistent all the symptoms and is strongly supported both by the subjective, “sweet” smell of the baby’s urine and the findings of α-keto acids in her urine. This condition is produced by inherited deficiency of α-keto acid dehydrogenase, a key enzyme in degradation of the branched-chain amino acids.

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8. A 7-year-old girl has seizure. The mother explains that the baby has been vomiting and having difficulty feeding for the past 2 days. There is also a strange sweet smell to her diapers. Physical examination is unremarkable, except for indications of dehydration. Serum test results show normal levels of glucose and ammonia. Urinalysis reveals the presence of α-keto-isocaproate and α-keto-isovalerate. Evaluation of this patients liver would reveal deficiency of which of the following enzymes? A. B. C. D. E.

Alanine aminotransferase (ALT). Glutamine synthetase. Cystathionine β-synthase. Carbamoyl phosphate synthetase. α-Keto acid dehydrogenase.√

Ans: E. The infant appears to be suffering from MSUD. This is consistent with all symptoms and is strongly supported by both by the subjective, “sweet’ smell of the baby’s urine and he findings of αketoacids in her urine. This condition is produced by inherited deficiency of α-keto acid dehydrogenase, a key enzyme in degradation of the BCCAs. None of the other enzyme deficiencies listed can Account for this unique set of symptoms. Some cases of ALT deficiency have been reported but only in association with viral infections and there were no symptoms arrtibutable to reduced ALT activity. Glutamine synthetase deficiency is rare, autosomal recessive disorder that has produced severe malformations of the brain, multiple organ failure, and neonatal death in confirmed cases. Cystathionine-β-synthase deficiency is key to homocysteinuria. Deficiency of carbamoyl phosphate synthetase produces hyperammonemia, which is not observed in this case.

Degradation of Leucine, Valine & Isoleucine Maple Syrup Urine Disease (MSUD) Deficiency of Branched Chain α-Keto Acid Dehydrogenase Complex Leads to MSUD. MSUD.

☻?

Leucine, Leucine, Valine & Isoleucine are Branched Chain Amino Acids (BCAA).

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9. A 11-year-old-girl being evaluated for a possible learning disorder. Recent testing indicates her IQ is 73. She reports having problems seeing in class and frequently finds it necessary to squint as she reads. Physical examination reveals lens dislocation (downward orientation); long, thin fingers and toes (arachnodactylyl); high arched palate; and mild scoliosis. A series of radiographs indicates generalized osteoporosis. An elevation of this patient’s blood and urine would most likely reveal an elevated level of which of the following compounds? A. Methionine.√ B. Cyseine C. Lactate D. Cystathionine E. Urea. Correct Answer: A. The constellation of symptoms exhibited by this patient is characteristic of homocystinuria. The impairment of her cognitive function could be attributed to many conditions, but the key findings are ectopia lentis with downward lens dislocation and osteoporosis In a female of this age. Homocysteinuria is produced by inherited deficiency of one of the enzymes in the pathway of Met conversion to Cys. The most Common form is cystathionine β-synthase deficiency, which results in accumulation of all upstream components of the pathway, including homocysteine, which is responsible for the toxic effects, and Met, which becomes elevated in the blood. Cystathionine and cysteine, which are both down stream of the block in the pathway Caused by cystathionine β-synthase deficiency, would be decreased. Metabolic pathways for lactate and urea are not involved in this mechanism. 10. A 49-year-old man with a rare recessive condition is at high risk for deep vein thrombosis and stroke and has had replacement of ectopic lenses. He has a normal hematocrit and no evidence of megaloblastic anemia. Amino acid analysis of this patient’s plasma would most likely reveal an abnormally elevated level of, A. B. C. D. E.

lysine. Leucine. methionine.√ ornithine. cysteine.

Answer: C. Only methionine is degraded via the Homocysteine / Cystathionine pathway and would be elevated in the plasma of a cystathionine synthase-deficient patient via activation of homocysteine methyltransferase by excess substrate.

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11. A 49-year-old man with a rare recessive condition is at high risk for deep vein thrombosis and stroke and has had replacement of ectopic lenses. He has a normal hematocrit and no evidence of megaloblastic anemia. A mutation in the gene encoding which of the following is most likely to cause this disease? A. B. C. D. E.

Cystathionine synthase.√ Homocysteine methyltransferase Fibrillin. Lysyl oxidase. Branched chain α-keto acid dehydrogenase.

Answer: A. Homocysteine, the substrate for the enzyme, accumulates increasing the risk of deep vein thrombosis and disrupting the normal crosslinking of fibrillin. Defficiency of homocysteine methyltransferase would cause homocysteinuria, but would also predispose to megaloblastic anemia.

Dislocated Lens in Homocystinuria

♫?

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A

Degradation & Resynthesis of Methionine

Enzyme Deficiency in Homocystinuria TransTrans-Sulfuration Pathway TransTrans-Sulfuration pathway is analagous to transamination for AAs. AAs. Two RXNs, RXNs, both use pyridoxal phosphate (Vit (Vit B6) as a cofactor (as with transamination). transamination).

☻ Cystathionuria Cystathinase

Homocystinuria

B6

PLP NH 3

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12. A 49-year-old man with a rare recessive condition is at high risk deep vein thrombosis and stroke and has had replacement of ectopic lenses. He has a normal hematocrit and no evidence of megaloblastic anemia. Amino acid analysis of this patient’s plasma would most likely reveal an abnormally elevated levels of, A. B. C. D. E.

lysine. leucine. methionine.√ ornithine. cysteine.

Answer: C. Only methionine is degraded via the homocysteine / cystathioninemia pathway and would be elevated in the plasma of cystathionine synthase-deficient patient via activation of homocysteine methyltransferase by excess substrate. 13. In the synthesis of cysteine, the carbon atoms (nearest precursor) are provided by which of the following? A. B. C. D. E.

Aspartic acid. Methionine. Oxaloacetic acid. Serine.√ Homocysteine.

Answer: D. This question refers to the precursor of the carbon atoms of cysteine that are provided by serine. Homocysteine, a direct precursor, only supplies sulfur. These two molecules form cystathionine. Methionine is the precursor to homocysteine. Aspartate and oxaloacetate are involved in transamination and do not directly affect cysteine synthesis. Which of the following amino acids is a precursor to cysteine? A. B. C. D. E.

Threonine. Methionine.√ Glutamine. Lysine. Alanine.

Ans: B.

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Heme Matabolism: 14. Which of the following statement about heme and iron metabolism is correct? A. Iron is strored in the liver as transferrin. B. Iron (as Fe2+) is inserted into protoporpyrin IX in the last step of heme synthesis C. δ-ALA synthase catalyzes the regulated and rate-limiting step in heme biosynthesis.√ D. The major route for bilirubin excretion is via the urine. E. The iron produced by heme degradation is excreted in the feces. Answer: C. The first rate-limiting step in heme biosynthesis involves the condensation of glycine and succinyl CoA to form Delta-ALA. Iron is stored as ferritin & transported in the blood in transferrin. As Fe2+, it is inserted into protoporpyrin IX to form heme. When heme is degraded to form bilirubin (which is excreted mainly via the intestine), iron is returned to the body’s iron stores and is not excreted. Bleeding is the only significant means by which Iron is lost from the body 15. A 44-year-old woman is brought to the emergency department doubled over with abdominal pain. Her husband states that the pain began several hours earlier, comes in waves, but has not really subsided completely even for brief periods since then. Oral antacids have not helped the pain at all. Her discomfort is not relieved by defecation. A stool sample is light gray or claycolored. Physical examination shows right upper quadrant abdominal pain. Her sclerae are slightly yellow in color. A sonogram shows a 2-cm mass in the region of the bile duct. Testing of her serum would be expected to reveal elevated levels of which of the following? A. Albumin-bound bilirubin. B. Porphobilinogen. C. Free, unconjugated bilirubin. D. Conjugated bilirubin / Bilirubin Diglucuronide / Direct Bilirubin. E. Biliverdin. Ans: D. 16. A woman 7 months pregnant with her first child develops anemia. Laboratory evaluation indicates an increased mean cell volume (MCV), hypersegmented neutrophils, and altered morphology of several other cell types. The most likely underlying cause of this woman’s anemia i s, A. B. C. D. E.

Folate deficiency.√ Iron deficiency. Glucose-6-phosphate dehydrogenase deficiency. Cyanocobalamin (B12) deficiency. Lead poisoning.

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Answer: A. Pregnant woman with megaloblastic anemia and elevated serum homocysteine strongly suggests folate deficiency. Iron deficiency presents as microcytic, hypochromic anemia and would not elevate homocyateine. B12 deficiency is not most likely in this presentation. 17. A 23-year-old, single, unemployed woman in her eighth month of pregnancy is seen in a volunteer-staffed obstetrics clinic. Her first child, born at home and exclusively breast-fed, had prolonged diarrhea and died from an intracranial hemorrhage at 1 month of age. To help prevent a similar problem in this pregnancy, the resident gives her a free prescription for a vitamin and advises her to take one 20-mg tablet each day. He also informs her that the infant should receive an injection of this vitamin soon after birth. The vitamin prescribed is required as a coenzyme by which of the following enzymes? A. B. C. D. E.

δ-Aminolevulinate synthase.√ L-GlutamyI carboxylase. Homocysteine methyltransferase. Prolyl hydroxylase. Thrombin.

Ans: A. 18. A 36-year-old Greek man with viral pneumonia has a self-limiting episode of hemolysis. Over the next week, he has an increased rate of reticulocytosis. Which of the following compounds serves as a precursor to heme in the reticulocytes? A. B. C. D. E.

α-Ketoglutarate Fumarate Isocitrate Oxaloacetate Succinyl-CoA√

Ans: E. 19. A 23-year-old, single, unemployed woman in her eighth month of pregnancy is seen in a volunteer-staffed obstetrics clinic. Her first child, born at home and exclusively breast-fed, had prolonged diarrhea and died from an intracranial hemorrhage at 1 month of age. To help prevent a similar problem in this pregnancy, the resident gives her a free prescription for a vitamin and advises her to take one 20-mg tablet each day. He also informs her that the infant should receive an injection of this vitamin soon after birth. The vitamin prescribed is required as a coenzyme by which of the following enzymes? (A). δ-Aminolevulinate synthase.√ (B). L-GlutamyI carboxylase (C). Homocysteine methyltransferase (D). Prolyl hydroxylase (E). Thrombin. Ans: A. 16

PLP

20. A 62-year-old man being treated for tuberculosis develops a microcytic, hypochromic anemia. Ferritin levels are increased, and marked sideroblastosis is present. A decrease in which of the following enzyme activities is most directly responsible for the anemia in this man? A. B. C. D. E.

Cytochrome oxidase. Cytochrome P450 oxidase. Pyruvate kinase. δ-aminolevulinate synthase.√ Lysyl oxidase.

Ans: D. δ-aminolevulinate synthase (δ-ALA Synthase). Sideroblastic anemia in a person being treated for Tuberculosis (with isoniazid) is most likely due to vitamin B6 deficiency. δ -Aminolevulinate synthase, the first enzyme in heme synthesis, requires vitamin B6 (pyridoxine).

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Pathway of Porphyrin Synthesis: Formation of Porphobilinogen Isoniazid = PLP Antagonist

PLP

PLP = Pyridoxal phosphate

Pathway of Porphyrin Synthesis: Formation of Heme

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Formation of Specialized Products From Amino Acids: 21. Which of the metabolites below is a precursor of tyrosine. A. B. C. D. E.

L-Dihydroxyphenylalanine (dopa). Dopamine. Norepinephrine. Epinephrine. Phenylalanine.√

Ans: E.

22. A 48-year-old man developed abdominal colic, muscle pain, and fatigue. Following a 3-week hospitalization, acute intermittent porphyria was initially diagnosed based on a high level of urinary δ-aminolevulinic acid. Subsequent analysis of the patient’s circulating RBCs revealed that 70% contained elevated levels of zinc protoporphyrin, and the diagnosis was corrected. The correct diagnosis is most likely to be, A. B. C. D. E.

Protoporphyria. Congenital erythropoietic porphyria. Lead poisoning.√ Barbiturate addiction. Iron deficiency.

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Ans: C. Lead poisoning. Lead inhibits both ferrocheatase (increasing the zinc Protoporphyrin) and ALA dehydrase (increasing δ-ALA). 23. Which of the following porphyrins gives stools their characteristic brown color? A. B. C. D. E.

Biliverdin. Urobilinogen. Heme. Stercobilin.√ Urobilin.

Ans: D. Stercobilin.

Bilirubin Excretion

Bilirubin diglucuronide 2 glucuronate

liver

Bacterial enzyme

Intrahepatic urobilinogen cycle

Bilirubin 8H

Bacterial enzyme

kidneys

intestines

Urobilinogen

Urobilin

Urine

Stercobilin

Feces

kidneys

Bacterial enzymes

Stercobilinogen

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Purine & Pyrimidine Metabolism 24. Pregnant women frequently suffer from folate deficiencies. A deficiency of folate would decrease the production of: A. creatine phosphate from creatine. B. all of the pyrimidines required for RNA synthesis. C. the thymine nucleotide required for DNA synthesis.√ D. phosphatidylcholimne from diacylglycerol and CDP-choline.

Answer: C. The only pyrimidine that requires folate for its synthesis is thymine (dUMP → dTMP). Folate is required for incorporation of carbons 2 and 8 into all purine molecules. Thesynthesis of creatine phosphate and of phosphatidylcholine do not require folate. Folate deficiencies during pregnancy can lead to neural tube defects (e.g., spina bifida) in the fetus.

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25. Excessive degradation of AMP and GMP would result in increase during excretion of: A. creatine B. urea C. uric Acid.√ D. thiamine E. thymine Answer: C. The purine bases, adenine (A) and guaninie (G), are oxidized to uric acid, which is excreted in the urine. Excessive production of uric acid can result in the condition known as gout.

26. The component that would be elevated to the greatest extent in the blood of a person suffering from Gout is: A. bilirubin. B. uric acid.√ C. creatine kinase. D. blood urea nitrogen. E. creatinine. Answer: B.

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27. De novo pyrimidine synthesis requires A. B. C. D. E.

phosphoribosyl pyrophosphate (PRPP) for the initial step. tetrahydrofolate for the incorporation of carbons 2 and 8. both carbon and nitrogen of aspartate to form the ring. NH4+ as a substrate for carbamoyl phosphatesynthetase II. glycine as the source of two nitrogens in the ring.

Ans: C. Options A and B are true for purine but not pyrimidine biosynthesis. During pyrimidine synthesis, the entire aspartate molecule is incorporated into the ring. Glutamine is the substrate for carbamoyl phosphate synthetase II, the enzyme involved in pyrimidine biosynthesis. (NH4+ is the substrate for synthetase I used in urea synthesis.) Glycine supplies one nitrogen for purine synthesis.

Sources of the Individual Atoms in Pyrimidine Ring

De novo pyrimidine synthesis requires: Both carbon & nitrogen of aspartate to form the ring.

28. The component that would be elevated to the greatest extent in the blood of a person suffering from Gout is_______. A. bilirubin. 23

B. uric acid.√ C. creatine kinase. D. blood urea nitrogen. E. ammonia. Answer: B. 29. A 56-year-old diabetic with end-stage renal disease receives a kidney transplant from his son. His nephrologist is concerned for the possibility of transplant rejection and puts the patient on mycophenolic acid, which inhibits which important enzyme in the synthesis of nucleotides? A. B. C. D. E.

PRPP. IMP dehydrogenase√ Adenylosuccinate synthase. Ribonucleotide reductase. Adenylosuccinase.

Answer:B. Mycophenolic acid, a potent immunosuppressant, is an inhibitor of IMP dehydrogenase, which normally converts IMP to xanthosine monophosphate. PRPP synthase catalyses the initial step in nucleotide metabolism, forming PRPP from ATP and ribose. Adenylosuccinate synthase and adenylosuccinase are sequential enzymes in the synthesis of AMP.

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Conversion of IMP to AMP & GMP Showing Feedback Inhibition

Mycophenolic acid is an Immuno suppressor used during transplantation / grafting.

30. A 58-year-old man is awoken by a throbbing ache in his great toe. He has suffered these symptoms before, usually after indulging in a rich meal. On exam, he is noted to have an angry inflamed great toe; also of note are several small nodules on the anti-helix of his ear. Inhibition of which of the following enzymes might prevent further occurrences of this man's ailments? A. Carbamoyl phosphate synthetase-II. B. Hypoxanthine-guanine phosphoribosyl transferase (HGPRT). C. PRPP synthetase. D. Xanthine oxidase. √ E. Orotate phosphoribosyl transferase. Answer: D. Gout is caused by either the increased production or reduced excretion of uric acid, leading to the deposition of urate crystals. Allopurinol, a xanthine oxidase inhibitor, decreases the production of urate from hypoxanthine and xanthine. Carbamoyl phosphate synthase-II is an enzyme in pyrimidine biosynthesis. HGPRT is an enzyme in the pathway for purine salvage. Orotate phosphoribosyi transferase is important in the synthesis of pyrimidines. PRPP synthetase is an important enzyme in the biosynthesis of purines; overactivity can cause gout.

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Allopurinol and Hypoxanthine

Allopurinol

Allopurinol

31. The genetic defect in the hypoxanthine guanine phosphoribosyl transferse (HGPRT) will leads to, A. B. C. D. E.

Parkinson’s disease. Cystinuria. Pellagra. Lesch-Nyhan syndrome. Hartnup’s disease.

Answer: D. The defect in the purine salvage enzyme HGPRT causes Lesch-Nyhan syndrome.

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Salvage Pathways of Purine Nucleotide Synthesis LeschLesch-Nyhan Syndrome is characterized by the deficiency of “hypoxanthine-guanine phosphoribosyl transferase (HGPRT)” (HGPRT)”, leads to accumulation of PRPP and uric acid, acid, the condition is know as “Hyperuricemia” Hyperuricemia”. Hyperuricemia is the presence of high levels of uric acid / sodium urate crystals in the blood. The upper end of the normal range of uric acid is 360 µmol/L (6 mg/dL mg/dL)) for women and 400 µmol/L (6.8 mg/dL) for men. PRPP=Phosphoribosyl PRPP=Phosphoribosyl pyrophospate. pyrophospate.

32. 5-Fluorouracil (5-FU) is an effective chemotherapeutic agent because interferes with DNA synthesis by directly inhibiting the reaction in which: A. FH2 → FH4. B. Deoxyuridine monophosphate (dUMP) → thymidinemonophosphate (dTMP).√ C. glutamine+ PRPP → Phosphoribosylamine. D. methyl B12 → B12. E. methionine → S-Adenosyl methionine. Answer: B. Methotrexate inhibits the reaction of dihydrofolate (FH2) → tetrahydrofolate (FH4), which is mediated by dihydrofolate reductase. 5-FU inhibits the reaction of deoxyuridine monophosphate (dUMP)→ thymidine monophosphate (dTMP). The remaining reactions are not directly affected by 5-FU. Glutamine to create phosphoribosylamine is the first step in purine biosynthesis. Methyl B12→ vitamin B12 provides the methyl group for the biosynthesis of methionine from homocysteine; and the reverse reaction does not occur in the cell because there is no demethylation for the reaction.

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Synthesis of dTMP from dUMP, Ilustrating Sites of Action of Antineoplastic Drugs

5-Fluorouracil: An antineoplastic agent, used especially in the treatment of cancers of the skin, breast, and digestive system. Methotrexate: An anticancer drug that acts as a folic acid antagonist to interfere with cellular reproduction and is used in the treatment of psoriasis, certain cancers, and certain inflammatory diseases, such as rheumatoid arthritis. Pregnant women frequently suffer from folate deficiencies. A deficiency of folate would decrease the production of: the thymine nucleotide required for DNA synthesis

33. Methotrexate is a potent anticancer agent that serves dividing cells of deoxyribonucleotides through direct inhibition of the following enzymes? A. B. C. D. E.

Ribnucleotide reductase. Xanthine oxidase. Carbamoyl phosphate synthase II. Anemia. Dihydrofolate Reductase.√

Answer: E.

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34. This enzyme recognizes allopurinol as a substrate and is then inhibited by the product of enzyme catalysis. A. PRPP synthetase. B. Hypoxanthine-guanine phosphoribosyl transferase. C. Adenosine deaminase. D. PRPP amidotransferase. E. Xanthine oxidase.√ Answer: E. 35. A 12-week-old infant with a history of persisent diarrhea and candidiasis is seen for a respiratory tract infection with Pneumocystis carinii. A chest x-ray confirms pneumonia and Reveals absence of a thymic shadow. Trace IgG is present in his serum, but IgA and IgM are absent. His red blood cells completely lack an essential enzyme in purine degradation. The product normally formed by this enzyme is: A. A. B. C. D.

Guanine monophosphate. hypoxanthine. inosine.√ xanthine. xanthine monophosphate.

Answer: C. The child most likely has severe combined immunodefciency (SCID) caused by adenosine deaminase (ADA) deficiency. This enzyme deaminates adenosine (a nucleotide) to form inosine (another nucleoside). Hypoxanthine and xanthine are both purine bases, and the monophosphates are nucleotides.

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Adenine Deaminase (ADA) Deficiency

Adenine Deaminase The importance of adenosine deaminase (ADA) ADA) for the catabolism of dA, dA, but not A, A, and the resultant accumulation of dATP when ADA is defective. defective.

36. A 57-year-old sales representative for a biotechnology firm has a history of alcohol abuse and hyperuricemia, He attends an out-of-town conference and gorges himself on appetizers, including liver pate, caviar, aid sweetbreads before dinner. Early the following morning he develops a painful swelling in his big toe. In addition to the alcohol consumed with his meals, which of the component may have contributed to this episode? A). Carbohydrate B). Cholesterol C). Nucleic acid D). Protein E). Triglyceride. Answer is C. Purines from ingested nucleic acids are converted to uric acid by intestinal epithelial cells and released into the blood for potential excretion by the kidney. Foods rich in DNA (caviar) or RNA (liver pate and sweetbreads [pancreas], derived from organs with active protein synthesis) are particularly rich sources of purines, and their ingestion coupled with excess alcohol consumption presumably increased his plasma urate levels to the point at which uric acid crystallized in a susceptible joint. The resultant painful swelling is gout. The pain is caused by precipitation of uric acid crystals in the joints with attendant inflammation. 37. A 2-year-old boys mother is concerned about his tendency to bite himself to the point of bleeding. The boy’s fingers show scarring and several scabs, and his lips are swollen and bruised. He exhibits poor coordination, poor muscle tone, and frequent jerking movements of his arms and 30

legs. He is significantly delayed in speech. His urine is orange in color and “gritty”. Which of the following is the most likely diagnosis? A. B. C. D. E.

Tay-Sachs disease. Gout. Lesch-Nyhan syndrome.√ Severe combined immunodeficiency. Cerebral palsy.

Answer: C.

38. A 47-year-old man complains of pain in the joints of his big toe, which are obviously swollen and tender. The pain has been chronic but became intolerable the day after thanksgiving when he had a large meal and several glasses of red wine. He is obese, and his past medical history is significant for removal of kidney stones. Which of the following involved in the pathophysiology of this patient’s condition? A. B. C. D. E.

Elevated orotic acid. Elevated uric acid. √ Deficiency of folic acid. Anemia. Hypoglycemia.

Answer: B. Wether his gout arises from impaired excretion of uric acid or is due to a mutation of PRPP synthase cannot be determined from the data. Analysis of his blood may confirm the gout if high concentrations of uric acid (hyperurecemia) are present.

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39. Severe combined immunodeficiency (SCID) arises from inhibition of lymphocyte proliferation because B and T cells are particularly sensitive to allosteric inhibition of which of the following enzymes of purine nucleotide metabolism? A. B. C. D. E.

Xanthine oxidase. Dihydrofolate reductase. Adenosine deaminase.√ Ribonucleotide reductase. Hypoxanthine-guanine phosphoribosyltransferase.

Ans: C.

Adenine Deaminase (ADA) Deficiency SCID

Adenine Deaminase The importance of adenosine deaminase (ADA) for the catabolism of dA, dA, but not A, A, and the resultant accumulation of dATP when ADA is defective. defective.

40. Methotrexate is a potent anticancer agent that starves dividing cells of deoxyribonucleotides through direct inhibition of which of the following enzymes? A. Ribonucleotide reductase B. Xanthine oxidase C. Carbamoyl phosphate synthetase II D. Thymidylate synthetase 32

E. Dihydrofolate reductase.√ The answer is E. Methotrexate is an analog of folic acid that binds with very high affinity to the substrate-binding site of dihydrofolate reductase, the enzyme that catalyzes conversion of DHF to THF, which is used in various forms by enzymes of both the purine and pyrimidine de novo synthetic pathways. Thus, synthesis of dTMP from dUMP catalyzed by thymldylate synthetase and several steps in purine synthesis catalyzed by formytransferase are indirectly blocked by the action of methotrexate because both those enzymes require THF coenzymes.

Synthesis of dTMP from dUMP, dUMP, Ilustrating Sites of Action of Antineoplastic Drugs

5-Fluorouracil: An antineoplastic agent, used especially in the treatment of cancers of the skin, breast, and digestive system. Methotrexate: An anticancer drug that acts as a folic acid antagonist to interfere with cellular reproduction and is used in the treatment of psoriasis, certain cancers, and certain inflammatory diseases, such as rheumatoid arthritis. Pregnant women frequently suffer from folate deficiencies. A deficiency of folate would decrease the production of: the thymine nucleotide required for DNA synthesis

41. Methotrexate is a potent anticancer agent that starves dividing cells of deoxyribonucleotides through direct inhibition of which of the following enzymes? A. Ribonucleotide reductase B. Xanthine oxidase C. Carbamoyl phosphate synthetase II D. Dihydrofolatre Reductase.√ E. Adenosine deaminase. Ans: D.

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42. The genetic defect in the hypoxanthine guanine phosphoribosyl transferse (HGPRT) will leads to, A. B. C. D. E.

Parkinson’s disease. Cystinuria. Pellagra. Lesch-Nyhan syndrome.√ Hartnup’s disease.

Answer: D. The defect in the purine salvage enzyme HGPRT causes Lesch-Nyhan syndrome. 43. Excessive degradation of AMP and GMP would result in increased urinary excretion of, ___________. A. creatine. B. urea. C. uric Acid.√ D. thiamine. E. thymine. Answer: C. The purine bases, adenine (A) and guaninie (G), are oxidized to uric acid, which is excreted in the urine. Excessive Production of uric acid can result in the condition known as gout.

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Degradation of Purine Nucleotides to Uric Acid

44. Which of the following is a compound formed from both a hydroxylation with an enzyme requiring vitamin C and subsequent methylation? A. B. C. D. E.

Histamine. Epinephrine.√ GABA. Carnitine. Creainine.

Answer: B. Epinephrine forms when dopamine is hydroxylated by an enzyme that requires copper and vitamin C, to form norepinephrine, which subsequently methylated. Histamine is derived by decarboxylation of histidine. GABA is decarboxylated glutamate. Creatinine is formed by a SAM methylation, and carnitine is an acyl group acceptor.

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Biosynthesis of Catecholamines PLP BH4

Catecholamins: Catecholamins: include adrenaline, noradrenaline & dopamine (DOPA), with roles as hormones & neurotransmitters. Catecholamines (epinephrine & norepinephrine) norepinephrine) are produced naturally in the body & functions as key neuroneuro-chemicals.

45. The biosynthetic pathway involves both BH4 and a pyridoxal phosphate (PLP)-dependent decarboxylation reaction to form which of the following? A. B. C. D. E.

Histamine. L-aminobutyric acid (GABA). Creatine. Epinephrine.√ Carnitine.

Answer: D. Epinephrine synthesis requires both BH4 and a pyridoxal phosphate (PLP)-dependent decaroxylation from Phenylalanine→ Tyrosine→DOPA→Dopamine→Epinephrine. Both Histidine→Histamine and Glutamate→L-Aminobutyric acid (GABA) require pyridoxal phosphate (PLP) for decarboxylation (but not BH4). Creatine is produced from glycine, arginine and S-adenosyl methionine (SAM).

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46. Patients with Lesch-Nyhan syndrome have hyperuricemia, indicating an increased biosynthesis of purine nucleotides, and markedly decreased levels of hypoxanthine phosphoribosyl transferase (HPRT). The hyperuricemia can be explained on the basis of a decrease in which regulator of purine biosynthesis? A. B. C. D. E.

ATP. GDP. Glutamine. IMP.√ PRPP.

Answer: D. IMP is a feedback inhibitor of PRPP amidophospho--ribosyl transferase, the first reaction in the biosynthesis of purines. IMP is formed by the HPRT reaction in the salvage of hypoxanthine.

47. Severe combined immunodeficiency arises from inhibition of lymphocyte proliferation because B and T cells are particularly sensitive to allosteric inhibition of which of the following enzymes of purine nucleotide metabolism? A. B. C. D. E.

Xanthine oxidase. Dihydrofolate reductase. Adenosine deaminase. √ Ribonucleotide reductase. Hypoxanthine-guanine phosphoribosyltransferase.

Ans: C.

48. Pregnant women frequently suffer from folate deficiencies. A deficiency of folate would decrease the production of, A. Creatine phosphate from creatine. B. All of the pyrimidines required for RNA synthesis. C. The thymine nucleotide required for DNA synthesis. √ D. Phosphatidylcholimne from diacylglycerol and CDP-choline. Answer: C. The only pyrimidine that requires folate for its synthesis is thymine (dUMP → dTMP). Folate is required for incorporation of carbons 2 and 8 into all purine molecules. The Synthesis of creatine phosphate and of phosphatidylcholine do not require folate. Folate deficiencies during pregnancy can lead to Neural tube defects (e.g., spina bifida) in the fetus.

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Synthesis of dTMP from dUMP, dUMP, Ilustrating Sites of Action of Antineoplastic Drugs

5-Fluorouracil: An antineoplastic agent, used especially in the treatment of cancers of the skin, breast, and digestive system. Methotrexate: An anticancer drug that acts as a folic acid antagonist to interfere with cellular reproduction and is used in the treatment of psoriasis, certain cancers, and certain inflammatory diseases, such as rheumatoid arthritis. Pregnant women frequently suffer from folate deficiencies. A deficiency of folate would decrease the production of: the thymine nucleotide required for DNA synthesis

49. A 63-year-old woman reports a long history of joint pain. Her fingers are severely deformed secondary to rheumatoid arthritis. Upon visiting a rheumatologist, she is started on methotrexate. This drug inhibits which of the following conversions? A. B. C. D. E.

Dopamine to norepinephrine. Tyrosine to dopa. Folate to dihydrofolate.√ Histamine to form-iminoglutamate (FIGLU). Norepinephrine to vanillylmandelic acid.

Ans: C. Folate to dihydrofolate. 50. A couple of African-American descent gives birth to a boy after an otherwise uneventful pregnancy. The child is exceptionally fair-skinned and has almost white hair. Further exam reveals red pupils. A postnatal screen is likely to confirm the deficiency of which of the following? A. B. C. D. E.

Peroxidase. Inducible nitricoxide synthase (iNOS). Glutathione reductase (GR). Tyrosinase.√ S-Adenosylmethionine (SAM).

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Ans: D. Tyrosinase.

51. A 59-year-old woman develops a shuffling gait and a pill-rolling tremor. She is referred to a neurologist for evaluation. After a thorough workup, a diagnosis of Parkinson disease is made., and the patient is placed on a manoamine oxidase inhibitor (MAOI). The drug, in this case, is given to decrease the degradation of which of the following? A. B. C. D. E.

Serotonin. Nicotinamide. 5-Hydroxy indole acetic acid (5-HIAA). Endothelium derived relaxing factor (EDRF). Dopamine.

53. A newborn develops jaundice (yellow skin and yellow scleras) that requires laboratory evaluation. Which of the following porphyrin derivatives is conjugated, reacts directly, and is a major component of bile? A. Bilirubin diglucuronide. √ B. Stercobilin. C. Biliverdin. D. Heme. E. Bilirubin. F. Urobilinogen. G. Urobilin. Ans: A. Bilirubin diglucuronide. (Bilirubin conjugated with glucuronic acid)

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54. A 12-year-old boy develops convulsions. After running an electroencephalogram (EEG), a neurologist determines that the child has epilepsy. He is started on a benzodiazepine, which promotes the activity of GABA. GABA is derived from the amino acid glutamate via which of the following biochemical reactions? A. B. C. D. E.

Deamination. Decarboxylation.√ Hydroxylation. Iodination. Methylation.

55. Which of the following compounds is an analogue of hypoxanthine? A. B. C. D. E.

Ara C. Allopurinol. √ Ribose phosphae. 5-phosphpribosylpyrophosphate (PRRPP). 5-FU.

Ans: B. Allopurinol. 56. Which one of the following contributes nitrogen atoms to both purine and pyrimidine ring? A. B. C. D. E.

Aspartate.√ Carbamoyl phosphate. Carbon dioxide. Glutamine. Terahydrofolate.

Ans: A. Aspartate.

57. Which of the following would rule out hyperuricemia in a patient? A. B. C. D. E.

Lesch-Myhan syndrome. Gout. Xanthine oxidase hyperactivity. Carbamoyl phosphate synthase deficiency. Purine overproduction secondary to Von Gierke’s disease.

Ans: D. Carbamoyl phosphate synthase deficiency .

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Biomedical Genetics 58. Huntington’s disease is due to a CAG triplet repeat expansion. It is useful to know the number of repeats in any person. The number of CAG repeats is indicative of the severity of the disease progression. Which of the following statements is TRUE regarding the number of CAG repeats? A. B. C. D. E.

Higher the number of repeats, lesser is the severity of the disorder Higher the number of repeats, higher is the severity of the disorder.√ Higher the number of repeats, the age of onset of the disease is delayed There is no relation between the number of repeats and the age of onset There is no relation between age of onset and severity of the disorder.

Ans: B. 59. A 20-year-old man has had no retinoblastomas but has produced two offsprings with multiple retinoblastomas. In addition, his father had two retinoblastomas as a young child, and one of his siblings had three retinoblastomas. What is the most likely explanation for the absence of retinoblastomas in this individual? A. A new mutation in the unaffected individual, which has corrected the disease-causing mutation. B. Highly variable expression of the disorder. C. Incomplete penetrance.√ D. Multiple mutations in other family members. E. Pleiotropy. Ans: C. Incomplete Penetrance. Because multiple family members are affected & because mutations at the retinoblastoma genes are known to be sometimes non-penetrant, the man in an obligate carrier of the mutation who did not experience a second mutation in this gene during his fetal development.

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Normal I

Affected Mother

Obligate Carrier II No Phenotypic Expression III

v Autosomal Dominant.

Incomplete Penetrance v The concept that even

though several members of a family or population may have the same DNA mutation, but not everyone with that mutation will have or develop the disease.

v No Skipped generations (atleast (atleast 1 or more will affect in each generations). v Male to male transmission (II (II--3 to IIIIII-5). v But IIIIII-1 and IIIIII-3 do not have an affected parent. v IIII-1 must have inherited the mutant allele from I-2 and has given the mutant allele to IIIIII-1 and IIIIII-3. v IIII-1 has the disease genotype (Carrier) but does not display the disease phenotype: Example of Nonpenetrance. Non-penetrance.

60. In a family with myotonic dystrophy, a child is born with infantile myotonic dystrophy. The mother on examination is observed to have ptosis and mild facial weakness. The grandmother has bilateral cataracts and no demonstrable muscle weakness. The phenomenon in this family is: A. B. C. D. E.

Pleiotropy. Variable expression. Anticipation.√ Incomplete penetrance. Delayed age of onset.

Ans: C.

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Fragile X Syndrome vTriplet repeat expansion (CGG repeat). vCharacteristic features: Mental retardation, prominent ears, elongated face, macro orchidism (enlarged testis).

vFemales are less severely affected than males. vShows anticipation in successive generations. vHigher the number of repeats, greater is the severity of the disease.

Anticipation 57 yrs (40 repeats) 36 yrs (50 repeats)

v Individuals in the recent generations of a pedigree develop disease at an earlier age and with greater severity. v Huntington’ Huntington’s disease. v Triplet repeat expansion: CAG repeat. repeat. v Greater the number of repeats, earlier is the age of onset of symptoms. symptoms. 61. A 30-year-old man is phenotypically normal, but two of his siblings died from infantile TaySachs disease, an autosomal recessive condition that is lethal by the age of five. What is the risk that this man is a heterozygous carrier of the disease-causing mutation? A. 1 / 4.

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B. C. D. E.

1 / 2. 2 / 3.√ 3 / 4. Not elevated above that of the general population.

Ans: C. Because two of the man’s siblings had Tay-sachs disease, his parents must both be carriers. 62. A man who is affected with hemophilia A (X-linked recessive) mates with a woman who is a heterozygous carrier of this disorder. What proportion of this couple’s daughters will be affected, and what proportion of the daughters will be heterozygous carriers? A. B. C. D. E.

0 % ; 50%. 100% ; 0%. 0 % ; 100%. 50 % ; 50 %.√ 2 / 3 ; 1/3.

Ans: D. 63. A man and woman are both affected by an autosomal dominant disorder that has 80% penetrance. They are both heterozygotes, for the disease-causing mutation. What is the probability that they will produce phenotypically normal offspring? A. B. C. D. E.

20 %. 25%. 40 %.√ 60 %. 80%.

50 X 80 100

Ans: C. 64. In assessing a patient with osteogenesis inperfecta, a history of bone fractures, as well as blue sclerae, are noted. These findings are an example of: A. B. C. D. E.

allelic heterogeneity. gain-of-function mutation. locus heterogeneity. multiple mutations. pleiotropy.√

Ans: E.

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Pleiotropy v A disease causing mutation affects multiple organ systems. v Marfan syndrome (autosomal dominant). v Mutation in the fibrillin gene. v Skeletal abnormalities (long limbs, pectus excavatum). excavatum). v HyperHyper-mobile joints. v Ocular abnormalities (myopia, lens dislocation). v Cardiovascular disease (Mitral valve prolapse, prolapse, Aortic aneurysm). v Osteogenesis imperfecta (Bone, Sclera).

65. Waardenberg syndrome is an autosomal dominant disorder in which patients may exhibit a variety of clinical feature, including patches of prematurely grey hair, white eyelashes, a broad nasal root, and moderate to severe hearing impairments. Occasionally, affected individuals display two eyes of different colors and and cleft lip and/or palate. Patient who proposes a mutation in the PAX3 gene on chromosome 2 can present with all of these disparate signs and symptoms. Which of the following characteristics of genetic traits is illustrated by this example? A. B. C. D. E.

Anticipation. Imprinting. Incomplete penetrance. Locus heterogeneity. Pleiotropy.√

Ans: E. Pleiotropy: refers to the appearance of apparently unrelated characteristics resulting from a single genetic defect. It is often the result of the presence of a single altered molecule in multiple locations in the body, so that the single mutation has effect in multiple organ system. In Marfan syndrome, for e.g. A defect in the fibrillin gene causes manifestations of the disease in the eye, aorta and joints.

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Marfan Syndrome Pleotropy

Ocular abnormalities (myopia, lens dislocation

Skeletal abnormalities (long limbs, pectus excavatum). excavatum).

HyperHyper-mobile joints

66. The severe form of alpha-1 antitrypsin deficiency is the result of a single nucleotide substitution thus produces a single amino acid substitution. This is best described as a: A. B. C. D. E.

frameshift mutation. In-frame mutation. missense mutation.√ nonsense mutation. splice-site mutation.

Ans: C. A missence mutation results in the change of only a single amino acid.

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Mutations Mutation: is an alteration in DNA sequence, thus mutations produce new alleles. Missence Mutations: result in the substitution of single AA in the polypeptide chain. Substitution of valine Missence Mutation for glutamine in Sickle Cell Disease β-Globin polypeptide. Nonsense Mutations: produce a stop codon → premature termination of translation & truncated protein. InIn-frame Mutation: If the number of inserted or deleted bases is multiples of 3, the mutation is said to be inin-frame. frame. If not multiples of 3 the mutation is a frameframe-shift , which alters all codons down stream → truncated / severely altered proteins. proteins.

67. An 18 year old female has presented to the physician with primary amenorrhea. On examination she has a characteristic webbing of the neck. Karyotype analysis of the subject would most likely reveal, A. B. C. D. E.

47, XX,+21. 46, XX. 47, XXY. 47, XYY. 45, X0. √

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68. A newborn infant with a single palmar crease and depressed nasal bridge is suspected to have a genetic disorder. The karyotype analysis shows the presence of 47, XX,+21. The abnormality that results in this karyotype is, A. B. C. D. E.

Nondisjunction during spermatogenesis. Nondisjunction during oogenesis.√ Presence of an extra X chromosome. Faulty X–inactivation during embryogenesis. Presence of an extra Y chromosome.

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Nondisjunction Leads to Trisomy

Responsible for trisomy after fertilization (oogenesis) Nondisjunction during meiosis I: I: Homologous chromosomes do NOT separate. separate.

Nondisjunction during meiosis II: II: Sister chromatids of a chromosome do NOT separate.

69. Jack and Jane are first cousins. The proportion of genes that may be common in them is, A. B. C. D. E.

¼. ½. 1/3. 1/8.√ 1/32.

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Consanguinity

IIII-1, IIII-3 share ½ of their genes (Siblings).

IIIIII-1 and IIIIII-5 share ⅛ of their genes (first cousins). IVIV-1 and IVIV-2 share 1/32 of their genes (second cousins).

Michael and Jane are second cousins. The proportion of genes that may be common in them is, F. G. H. I. J.

¼. ½. 1/3. 1/8. 1/32.√

70. A given population has 2 alleles ‘a’ and ‘b’ for a specific gene. The frequency of allele ‘a’ is 0.7. What is the frequency of allele ‘b’?. A. B. C. D. E.

0.1 0.2 0.3.√ 0.4 0.5

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The sum of frequency of the alleles = 1 Allele frequency ‘a’+ Allele frequency ‘b’ = 1 p+q=1 0 .7 +0 .3 = 1

71. A gene exists in two allelic forms in the population ‘a’ and ‘b’. The genotypes that can be present in the population are, A. B. C. D. E.

aa, bb. aa, ab, bb.√ ab, ba, aa. aa, aab, bba. Aaa, bbb.

Ans: B.

PRACTICE QUESTIONS: GENETIC DIAGNOSIS 1. A 50 year old male has a loss of motor control and tremors of arms. The physician suspects a diagnosis of Huntington’s disease. The family history revealed that his mother had similar symptoms which began at the age of 60 years. His maternal grandfather also had similar complaints at an age of 75 years. PCR amplification and Southern blot of chromosome 4 shows the following results.

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Patient

Normal Control

190 bp

104 bp

101 bp 90 bp

The results indicate that, A. B. C. D. E. 2.

The patient has triplet repeat expansions on both copies of chromosome 4 The patient has a triplet repeat expansion on one copy of the chromosome 4.√ The patient’s DNA is of similar length when compared to normal control The blot does not give an indication of the size of the repeat expansion The results are inconclusive Jim and Jane are second cousins and they have decided to get married. They know that one of their uncles had died of cystic fibrosis at an early age. They want to know their risk of being carriers of the mutant cystic fibrosis gene. The genetic counselor performs an ASO test for the mutant cystic fibrosis gene (D508). The results of the test are depicted below.

Jim

Jane

The results indicate that, A. Both are homozygous normal B. Both are carriers of the mutant gene.√ C. Both are homozygous for the mutant gene 52

D. Jim is homozygous for the normal gene E. Jane is homozygous for the mutant gene 3. Use the ASO test in the previous question to answer this question. What is the risk that the children of Jim and Jane would be affected with cystic fibrosis?. A. B. C. D. E.

100% 50% 25%.√ 5% Nil

4. Hemochromatosis is an autosomal recessive disorder. The most common mutation causing hemochromatosis is the C282Y mutation. The normal HFE allele has the sequence 5’TATACGTGCCAGGTG3’. The C282Y allele has the sequence (Mutated) 5’TATACGTAGCAGGTG5’ The allele specific oligonucleotide (ASO) probe that would bind to the mutant allele is, A. B. C. D. E.

5’ATATGCACGGTCCAC3’ 5’ATATGCATCGTCCAC3’ 5’CACCTGGCACGTATA3’ 5’CACCTGCTACGTATA3’.√ 5’TATACGTGCCAGGTG3’

5. Huntington’s disease is due to a CAG triplet repeat expansion. It is useful to know the number of repeats in any person. The number of CAG repeats is indicative of the severity of the disease progression. Which of the following statements is TRUE regarding the number of CAG repeats?. F. G. H. I. J.

Higher the number of repeats, lesser is the severity of the disorder Higher the number of repeats, higher is the severity of the disorder.√ Higher the number of repeats, the age of onset of the disease is delayed There is no relation between the number of repeats and the age of onset There is no relation between age of onset and severity of the disorder.

6. A couple seeks genetic counseling, as they have a son, aged 10 years affected with sickle cell anemia. They also have a daughter, aged 8 years, who is phenotypically normal. The parents are both carriers of the sickle cell gene and are not affected. The lady is pregnant, and is worried that the unborn child may have the disease. An RFLP analysis was done to check whether the fetus was going to be affected. Which of the following conclusion may be drawn from the RFLP pattern of the family?

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A. B. C. D. E.

The father is not a carrier of the mutation The daughter is heterozygous for the mutant gene The son is heterozygous for the mutant gene The unborn fetus is a carrier of the mutant gene.√ The daughter is a carrier of the mutant gene

CYTOGENETICS 1. Observe the karyotype below and write the karyotype of the person.

A. 46, XY. B. 45, XO.

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C. 47, XXY. D. 46, XX.√ E. 47, XX, +21. 2. Observe the karyotype below and write the karyotype of the person.

F. G. H. I. J.

46, XY. 45, XO. 47, XXY. 46, XY.√ 47, XX, +21.

3. An infant has been brought to the outpatient clinic by its mother who is 45 year old. The child has developmental delay. The child also has a depressed nasal bridge, and a prominent epicanthal fold. Karyotyping of the child reveals a 47, XX, +21. Which is the Karyotype you would observe in the child?.

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A

B

C

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D

A. B. C. D.

A B C. D

4. A physician observes gynecomastia in a 22 - year old man. On examination, he also notices that the man has reduced body and facial hair and testicular atrophy. The physician suspects a diagnosis of Klinefelter syndrome and wants a karyotyping to confirm the diagnosis. What is the karyotype that may be expected in the individual? A. B. C. D. E.

45, XO. 46, XY. 47, XYY.√ 47, XXX. 47, XXY.

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Klinefelter Syndrome (47,XXY)

An Extra X Chromosome

47,XXY

Presence of a Barr body in the buccal mucosal cells

Barr body: A microscopic feature of female cells due to the presence of two X chromosomes in the female. One of these X chromosomes is inactive and is crumpled up to form the Barr body.

Klinefelter Syndrome (47,XXY) v Testicular atrophy. v Gynecomastia. Gynecomastia. v Female distribution of hair. v Infertility.

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5. Karyotype analysis of a patient with chronic myelogenous leukemia, reveals the presence of an abnormal chromosome 22, which is shorter than the actual chromosome 22. The abnormal Philadelphia chromosome is formed by A. B. C. D. E.

Reciprocal translocation.√ Robertsonian translocation. Inversion. Chromosomal deletion. Chromosomal Imprinting.

6. A 3 month old infant is brought to the clinic by its parents. The child has a peculiar cat-like cry. The pediatrician notices developmental delay and microcephaly. What chromosomal abnormality would you suspect in the child? A. B. C. D. E.

Trisomy 21. Edward syndrome. Deletion of paternal 15q. Deletion of chromosome 5p.√ Angelman syndrome.

Cri-Du-Chat Syndrome 46, XX, del(5p) or 46, XY, del(5p).

Chromosome 5

Cri-du-chat syndrome

v Chromosome loses some of its genetic material. v Deletion of chromosome 5p → “CriCri-dudu-chat Syndrome” Syndrome” v 46, XX, del(5p) or 46, XY, del(5p). v Features: Ø High pitched, Cat like cry. cry. Ø Mental retardation, Microcephaly. Microcephaly.

7. A 4-year old obese child with a low IQ has been diagnosed to have the Prader–Willi syndrome. This syndrome is due to

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A. B. C. D. E.

Deletion of maternal chromosome 15q Deletion of paternal chromosome 15q.√ Deletion of chromosome 5p. Robertsonian translocation of chromosome 14 and 21. Reciprocal translocation of chromosome 9 and 22.

Prader Willi Syndrome Chromosome 15 p arm

q arm Deletion of 15q11-13 Paternal

Maternal

“All the Best” “Knowledge is Power” T ha nk U

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