BRS Biochemistry Clinical Correlates Questions Semester 1 Flashcards
Interference in generalized cytokine signaling can lead to which one of the following disorders?
(A) Adrenoleukodystrophy
(B) SCID
(C) Influenza
(D) Myasthenia gravis
(E) Type 1 diabetes
The answer is B.
X-linked SCID is due to the lack of a common cytokine receptor subunit (the γ subunit), which affects the ability of a variety of cytokines to transmit signals to hematopoietic cells. Adrenoleukodystrophy is due to the buildup of very long-chain fatty acids, for a variety of reasons. Influenza is due to a virus. Myasthenia gravis is due to the production of autoantibodies directed against the acetylcholine receptor. Type 1 diabetes results from an inability to produce insulin
Questions 118 to 121 are matching questions. A dietary deficiency of a vitamin can cause each of the conditions below. Match each condition with the appropriate vitamin. An answer may be used once, more than once, or not at all.
- Pellagra
(A) Vitamin C
(B) Niacin
(C) Vitamin D
(D) Biotin
(E) Thiamine
118 The answer is B. Pellagra is due to a dietary deficiency of niacin, beriberi is due to a lack of thiamine (vitamin B1), scurvy due to a lack of vitamin C, and rickets from a lack of vitamin D
A young woman (5’ 3” tall, 1.6 m) who has a sedentary job and does not exercise consulted a physician about her weight, which was 110 lb (50 kg). A dietary history indicates that she eats approximately 100 g of carbohydrate, 20 g of protein, and 40 g of fat daily.
What is this woman’s BMI?
(A) 16.5
(B) 17.5
(C) 18.5
(D) 19.5
(E) 20.5
The answer is D.
The BMI is calculated by dividing the weight of the individual (in kilograms) by the square of the height of the individual (in meters). For this woman, BMI 5 50/1.62 5 19.5
A 3-year-old girl has been a fussy eater since being weaned, particularly when fruit is part of her diet. She would get cranky, sweat, and display dizziness, and lethargy, after eating a meal with fruit. Her mother noticed this correlation, and as long as fruit was withdrawn from her diet the child did not display such symptoms. The problems the girl exhibits when eating fruit is most likely due to which one of the following?
(A) Decreased levels of fructose in the blood
(B) Elevated levels of glyceraldehyde in liver cells
(C) High levels of sucrose in the stool
(D) Elevated levels of fructose-1-phosphate in liver cells
(E) Decreased levels of fructose in the urine
The answer is D.
The patient has HFI, which is due to a mutation in aldolase B. Sucrose would still be cleaved by sucrase, thus it would not increase in the stool. Fructose would not be metabolized normally, therefore it would be elevated in the blood and urine. Aldolase B would not cleave fructose 1-phosphate, thus its levels would be elevated and the product, glyceraldehyde, would not be produced
A young woman (5’ 3” tall, 1.6 m) who has a sedentary job and does not exercise consulted a physician about her weight, which was 110 lb (50 kg). A dietary history indicates that she eats approximately 100 g of carbohydrate, 20 g of protein, and 40 g of fat daily.
What is the woman’s approximate DEE in calories (kilocalories) per day at this weight?
(A) 1,200
(B) 1,560
(C) 1,800
(D) 2,640
(E) 3,432
The answer is B.
This woman’s DEE is 1,560 calories (kcal). DEE equals BMR plus physical activity. Her weight is 110 lb/2.2 = 50 kg. Her BMR (about 24 kcal/kg) is 50 kg X 24 = 1,200 kcal/day. She is sedentary and needs only 360 additional kcal (30% of her BMR) to support her physical activity. Therefore, she needs 1,200 + 360 = 1,560 kcal each day.
Consider the section of the TCA cycle in which isocitrate is converted to fumarate. This segment of the TCA cycle can be best described by which one of the following?
(A) These reactions yield 5 moles of highenergy phosphate bonds per mole of isocitrate.
(B) These reactions require a coenzyme synthesized in the human from niacin (nicotinamide).
(C) These reactions are catalyzed by enzymes located solely in the mitochondrial membrane.
(D) These reactions produce 1 mole of CO2 for every mole of isocitrate oxidized.
(E) These reactions require GTP to drive one of the reactions.
The answer is B.
In the conversion of isocitrate to fumarate, 2 CO2, 2 NADH (which contains niacin), 1 GTP, and 1 FADH2 are produced. A total of approximately 7.5 ATP are generated. The enzymes for these reactions are all located in the mitochondrial matrix except succinate dehydrogenase, which is an inner mitochondrial membrane protein. GTP is not required in any of the reactions but is produced in the conversion of succinyl-CoA to succinate
Pyridoxine deficiency
(A) Megaloblastic anemia
(B) Hypochromic, microcytic anemia
(C) Hemolytic anemia
(D) Sickle cell anemia
The answer is B.
Pyridoxine is required for the formation of pyridoxal phosphate, the cofactor for the first reaction in heme formation. The lack of heme means that the red cells cannot carry as much oxygen (which gives them the pale color). The cells are small in order to maximize the concentration of hemoglobin present in the cells. A megaloblastic anemia is due to deficiencies in either vitamin B12 or folic acid (a lack of intrinsic factor will lead to a B12 deficiency named pernicious anemia). These cells are large because the vitamin deficiency interferes with DNA synthesis, and the cells double in size without being able to replicate their DNA. Once the anemia begins, the large blast cells are released by the marrow in an attempt to control the anemia. Hemolytic anemia occurs when the red cell membrane fragments, which can occur with pyruvate kinase deficiencies or a lack of glucose-6-phosphate dehydrogenase activity (which results in reduced NADPH levels). Sickle cell anemia is caused by a point mutation in the β-globin gene, substituting a valine for a glutamic acid
Match 115.
- Ehlers-Danlos syndrome
- Parkinson’s disease
- Tay-Sachs disease
- McArdle’s disease
- Maple syrup urine disease
(A) Glycogen
(B) Collagen
(C) Dopamine
(D) Valine
(E) A sphingolipid
The answer is E. Because a hexosaminidase is deficient in Tay-Sachs disease, partially degraded sphingolipids (gangliosides) accumulate in lysosomes. A defect in the synthesis or processing of collagen will lead to a variety of diseases, of which Ehlers-Danlos syndrome is one (osteogenesis imperfecta is another). Parkinson’s disease is due to low levels of dopamine in the nervous system. In the initial stages of Parkinson’s disease, giving DOPA can reduce the severity of the symptoms, as DOPA can be decarboxylated to form dopamine. DOPA can easily enter the brain, whereas dopamine cannot. McArdle’s disease is due to a defective muscle glycogen phosphorylase, such that the muscle cannot generate glucose from glycogen, leading to exercise intolerance. Maple syrup urine disease is due to the lack of branched-chain α-keto acid dehydrogenase activity, a necessary step in the metabolism of the branched-chain amino acids (leucine, isoleucine, and valine).
Which one of the following is a common metabolic feature of patients with anorexia nervosa, untreated type 1 DM, hyperthyroidism, and nontropical sprue?
(A) A high BMR
(B) Elevated insulin levels in the blood
(C) Loss of weight
(D) Malabsorption of nutrients
(E) Low levels of ketone bodies in the blood
The answer is C. All of these patients will lose weight—the anorexic patients because of insufficient calories in the diet, the patients with type 1 DM because of low insulin levels that result in the excretion of glucose and ketone bodies in the urine, those with hyperthyroidism because of an increased BMR, and those with nontropical sprue because of decreased absorption of food from the gut. The untreated diabetic patients will have high ketone levels because of low insulin. Ketone levels may be elevated in anorexia and also in sprue, due to a reduction in levels of gluconeogenic precursors. An increased BMR would be observed only in hyperthyroidism. Nutrient malabsorption would occur only in nontropical sprue and anorexia.
A 23-year-old male presents to the ER with a fracture of his humerus, sustained in what appeared to be a minor fall. He has a history of multiple fractures after a seemingly minor trauma. He also has “sky blue” sclera and an
aortic regurg murmur. His underlying problem is most likely due to a mutation in which one of the following proteins?
(A) Fibrillin
(B) Type 1 collagen
(c) Type IV collagen
(d) α1-Antitrypsin
(E) β-Myosin heavy chain
the answer is B.
The patient is exhibiting the signs of osteogenesis imperfecta, brittle bones, as exemplified by various mutations in type 1 collagen, the building blocks of the bones. The aortic regurgitation murmur is also due to a lack of type 1 collagen in the extracellular matrix of the aorta. Mutations in fibrillin give rise to Marfan syndrome, which does exhibit long bones, but not brittle or easily broken bones. Marfan syndrome would also be associated with lens dislocation, which is not occurring in this patient. Mutations in type IV collagen would lead to Alport syndrome, not brittle bones, and there is no mention of kidney/urine problems with the patient. A defect in α1-antitrypsin would lead to emphysema (not brittle bones), and a mutation in β-myosin heavy chain would lead to hypertrophic cardiomyopathy, not brittle bones.
A 1-year-old child, on a routine well child visit, was discovered to have cataract formation in both eyes. Blood work demonstrated elevated galactose and galactitol levels. In order to determine which enzyme might be defective in the child, which intracellular metabolite should be measured?
(A) Galactose
(B) Fructose
(C) Glucose
(D) Galactose-1-phosphate
(E) Fructose-1-phosphate
(F) Glucose-6-phosphate
The answer is D.
The child has a form of galactosemia. The elevated galactitol enters the lens of the eye, and is trapped. The difference in osmotic pressure across the lens of the eye leads to cataract formation. Galactose is phosphorylated by galactokinase to galactose 1-phosphate, which reacts with UDP-glucose in a reaction catalyzed by galactose-1-phosphate uridylyl transferase to form UDP-galactose and glucose 1-phosphate. An epimerase converts UDP-galactose to UDP-glucose. Deficiencies in either galactokinase (nonclassical) or galactose-1-phosphate uridylyl transferase (classical) result in galactosemia, with elevated levels of galactose and galactitol (reduced galactose) in the blood. An intracellular measurement of galactose-1- phosphate can allow a definitive diagnosis to be obtained (such levels would be nonexistent if the defect were in galactokinase, and the levels would be greatly elevated if the galactose1-phosphate uridylyl transferase enzyme were defective).
A patient is on a very low calorie liquid diet and must take supplements to ensure that he has the essential vitamins and minerals to maintain his health. Under the appropriate conditions, which one of the following compounds can be synthesized in humans, and would not need to be supplemented to the extent that the others are?
(A) Riboflavin
(B) Linoleic acid
(C) Leucine
(D) Thiamine
(E) Niacin
The answer is E.
Although niacin is a vitamin, it can be synthesized to a limited extent from tryptophan. None of the other vitamins indicated can be synthesized in humans to any extent.
A 6-month-old baby was doing well until he developed viral gastroenteritis and was unable to tolerate oral feeding for 2 days. He is admitted to the hospital with encephalopathy, cardiomegally and heart failure, poor muscle tone, and hypoketotic hypoglycemia. Blood work did not detect any medium-chain dicarboxylic acids.
Dietary supplementation of which one of the following would be beneficial to this patient?
(A) Pantothenic acid
(B) Niacin
(C) Riboflavin
(D) Carnitine
(E) Thiamine
The answer is D.
In many cases of primary carnitine deficiency, increasing the blood levels of carnitine is sufficient to allow some transport of carnitine into cells such that fatty acid oxidation can occur. While pantothenic acid (part of coenzyme A), niacin (the precursor for NAD1), and riboflavin (needed for FAD) are required for fatty acid oxidation, the rate-limiting step in these patients is the transport of the fatty acids from the cellular cytoplasm to the matrix of the mitochondria.
A patient presents with fatigue, and a blood count reveals a macrocytic, hyperchromic anemia. Which one of the following may account for this type of anemia?
(A) Lead poisoning
(B) Folate deficiency
(C) Hereditary spherocytosis
(D) Sideroblastic anemia
(e) Iron deficiency
The answer is B.
A folate deficiency leads to a megaloblastic anemia since DNA synthesis is inhibited in the absence of folate. Large cells are generated in the bone marrow because the cell enlarges in preparation for division, but the DNA does not replicate and the cell does not divide. All of the other answer choices suggested create a microcytic, hypochromic anemia. Lead poisoning interferes with heme synthesis, as does an iron deficiency. Under these conditions, the red cells released are small in size, as the final size is, in part, dependent on the intracellular concentration of heme (so if heme levels are low, the cell will be small). Sideroblastic anemia also results from a disruption in heme synthesis, for a variety of causes. Hereditary spherocytosis is due to mutations in red cell membrane proteins, which lead to early removal of these cells from the spleen.
A 33-year-old man had a screening colonoscopy, and was diagnosed with a right-sided, mucinous colon cancer, with no other lesions or polyps seen. The reason he had a colonoscopy at such an early age is that his father and paternal uncle had colon cancers diagnosed by age 40. His paternal grandmother had ovarian and uterine cancers. A likely defect in the patient is a reduction in the ability to carry out which one of the following processes?
(A) Removal of thymine dimers from the DNA
(B) Inability to remove the base U from DNA
(c) Loss of DNA ligase activity
(d) Inability to correct mismatched bases in newly synthesized DNA
(E) Inability to form a solenoid structure from individual nucleosomes
the answer is d.
The patient has HNPCC, which is due to specific mutations in proteins involved in mismatch repair (mutations in at least four different proteins have been identified that lead to HNPCC). Mismatch repair is not involved in thymine dimer removal, nor base excision repair (the removal of uracil from DNA). HNPCC does not involve a defective DNA ligase, nor does the disease result in defective DNA packaging (solenoid formation) in the nucleus.
A 42-year-old woman has slowly developed an inability to keep her eyes open at the end of the day. The eyelids droop, despite her best efforts to keep them open. This does not occur first thing in the morning. Further examination shows a generalized muscle weakness as the day progresses.
In order to diagnose the disease the patient is experiencing, a Western blot was run using the patient’s sera as the source of antibodies. The protein run on the gel would need to be which one of the following?
(A) Acetylcholinesterase
(B) Acetylcholine receptor
(C) Epinephrine receptor
(D) Catechol-O-methyltransferase
(E) Glucocorticoid receptor
(F) HMG-CoA reductase
The answer is B.
The woman has myasthenia gravis, which is due to the presence of autoimmune antibodies directed against the acetylcholine receptor in the neuromuscular junction. To confirm the diagnosis by Western blot, a sample of acetylcholine receptor would be run through a polyacrylamide gel, the protein transferred to filter paper, and the filter paper incubated with the patient’s sera. If the sera contain antibodies that bind to the acetylcholine receptor, the antibodies will be bound to the filter paper, and then visualized using a secondary antibody linked to a reporter enzyme. Controls would be done to indicate that sera from an individual who did not have myasthenia gravis did not allow for the formation of a band on the Western blot. Running acetylcholinesterase, the epinephrine receptor, catechol-O- methyltransferase, the glucocorticoid receptor, or HMG-CoA reductase on the gel would not allow detection of antibodies against the acetylcholine receptor in the patient’s blood sample.
An individual has been on a fad diet for 6 weeks, and has begun to develop a number of skin rashes, diarrhea, and forgetfulness. These symptoms could have been less severe if the diet contained a high content of which one of the following?
(A) Tyrosine
(B) Tryptophan
(C) Thiamine
(D) Thymine
(E) Riboflavin
The answer is B. The individual has developed pellagra due to a lack of dietary niacin. Although dietary niacin is the major source of the nicotinamide ring of NAD, it may also be produced from excess tryptophan. Tyrosine, thiamine, thymine, and riboflavin cannot contribute to the synthesis of the nicotinamide ring of NAD.
A 55-year-old Native American female had difficulty conceiving when she was young and was diagnosed with polycystic ovarian syndrome. When she did get pregnant, she was diagnosed with gestational diabetes, but successfully delivered a healthy 10-lb baby boy. She had difficulty controlling her weight over the next several years, but her blood glucose checks were always in the “normal” range and her only diagnosis prior to age 40 was sleep apnea. She has a strong family history of diabetes. At age 40, at a health fair, she had a finger-stick random blood glucose of 240. She made an appointment with her primary care doctor who ordered a fasting blood glucose (150) and a hemoglobin A1c (7.4). Other lab values were an HDL cholesterol of 35, total cholesterol of 210, triglycerides of 350, slightly elevated ALT, AST, and uric acid with normal BUN, creatinine, bilirubin, electrolytes, and alkaline phosphatase. Her blood pressure on repeated checks was consistently 150/90. For the next 15 years, she was tried on multiple different medications with only partial success. Currently, her height is 5′10″ (1.8 m), weight 220 lb (100 kg), BP 138/80, HbA1c 8.2, and creatinine 2.0. Her lipid values have not changed. Over the past 2 years, she has been hospitalized three times2for an MI, a community-acquired pneumonia (CAP), and gouty arthritis.
The woman’s BMI places her in which of the following categories?
(A) Underweight
(B) Normal weight
(C) Overweight
(D) Obese
(E) Morbidly obese
The answer is D.
A BMI <18.5 is considered underweight, 18.5 to 24.9 as normal weight, 25 to 29.9 as overweight, and >30 as obese. Some older classifications used >40 to signify morbidly obese. The calculated BMI and obesity classification can be important for insurance purposes especially if obesity surgery is considered.
A 4-year-old boy displays a failure to thrive, extreme sensitivity to the sun, hearing loss, severe tooth decay, pigmentary retinopathy, and premature aging. An analysis of fibroblasts from the boy demonstrated extensive DNA damage in cells trying to grow, but minimal damage in quiescent cells, which have a greatly reduced rate of transcription as compared to the growing cells. This child most likely has a defect in which one of the following processes?
(A) Repair of thymine dimers
(B) Base excision repair
(c) Nucleotide excision repair
(d) Mismatch repair
(E) Transcription-coupled DNA repair
the answer is E.
The child is exhibiting the symptoms of Cockayne syndrome, which is due to a defect in transcription-coupled DNA repair. During transcription of genes, if the RNA polymerase notices DNA damage, transcription will stop while the transcription-coupled DNA repair mechanism will correct the DNA damage. This syndrome can be due to mutations in either the ERCC6 or ERCC8 gene, and the protein products of both the genes are involved in repairing the DNA of actively transcribed genes. The key to answering the question is the amount of DNA damage in growing cells (which are transcriptionally active) versus the damage in quiescent cells (which express fewer genes). The symptoms described are also unique to individuals with this disorder. The repair of thymine dimers and the processes of base excision repair, nucleotide excision repair, and mismatch repair are all functional in individuals with this disorder.
Intrinsic factor deficiency
(A) Megaloblastic anemia
(B) Hypochromic, microcytic anemia
(C) Hemolytic anemia
(D) Sickle cell anemia
The answer is A.
Intrinsic factor is required for the absorption of dietary vitamin B12. Lack of B12 (or folate) results in a megaloblastic anemia. In a B12 deficiency, irreversible neurologic problems (due to demyelination) also occur. When decreased intrinsic factor causes a B12 deficiency, the condition is called pernicious anemia. An iron-deficiency anemia is characterized by small, pale red blood cells. The lack of iron reduces the synthesis of heme, so the red cells cannot carry as much oxygen (which gives them the pale color). The cells are small in order to maximize the concentration of hemoglobin present in the cells. Hemolytic anemia occurs when the red cell membrane fragments, which can occur with pyruvate kinase deficiencies or a lack of glucose-6-phosphate dehydrogenase activity (which results in reduced NADPH levels). Sickle cell anemia is caused by a point mutation in the β-globin gene, substituting a valine for a glutamic acid.
A 32-year-old female has developed breast cancer. Her mother and one maternal aunt had breast cancer and her maternal grandmother had ovarian cancer. Which of the following best describes the mechanism behind this inherited problem?
(A) A tumor suppressor leading to loss of apoptosis
(B) A tumor suppressor leading to an inability to repair DNA
(C) A tumor suppressor leading to a constitutively active MAP kinase pathway
(D) An oncogene leading to loss of apoptosis
(E) An oncogene leading to an inability to repair DNA
(F) An oncogene leading to a constitutively active MAP kinase pathway
The answer is B.
Hereditary breast cancer is due to inherited mutations in either of the tumor suppressor genes BRCA1 or BRCA2. These genes encode proteins that play important roles in DNA repair (primarily single- and double-strand breaks), and it is the loss of this function that predisposes the patient to breast and ovarian cancers. The inability to repair the breaks in the backbone leads to errors during replication, and mutations will develop that eventually lead to a loss of growth control. This is a loss-of-function disorder, which characterizes the genes involved as tumor suppressors. Inheriting one mutated copy of BRCA1 means that the other, normal copy of BRCA1 must be lost in a particular cell in order to initiate the disease (loss of heterozygosity). For breast cancer, this occurs 85% of the time (penetrance upon inheriting a BRCA1 or BRCA2 mutation). An oncogene is a dominant gene, so only one mutated copy can bring about the disease. BRCA1 or BRCA2 mutations do not directly lead to a loss of apoptosis, or to a constitutively active MAP kinase pathway.
A 55-year-old Native American female had difficulty conceiving when she was young and was diagnosed with polycystic ovarian syndrome. When she did get pregnant, she was diagnosed with gestational diabetes, but successfully delivered a healthy 10-lb baby boy. She had difficulty controlling her weight over the next several years, but her blood glucose checks were always in the “normal” range and her only diagnosis prior to age 40 was sleep apnea. She has a strong family history of diabetes. At age 40, at a health fair, she had a finger-stick random blood glucose of 240. She made an appointment with her primary care doctor who ordered a fasting blood glucose (150) and a hemoglobin A1c (7.4). Other lab values were an HDL cholesterol of 35, total cholesterol of 210, triglycerides of 350, slightly elevated ALT, AST, and uric acid with normal BUN, creatinine, bilirubin, electrolytes, and alkaline phosphatase. Her blood pressure on repeated checks was consistently 150/90. For the next 15 years, she was tried on multiple different medications with only partial success. Currently, her height is 5′10″ (1.8 m), weight 220 lb (100 kg), BP 138/80, HbA1c 8.2, and creatinine 2.0. Her lipid values have not changed. Over the past 2 years, she has been hospitalized three times2for an MI, a community-acquired pneumonia (CAP), and gouty arthritis.
Which one of the following is closest to her calculated BMI?
(A) 25
(B) 31
(C) 35
(D) 41
(E) 45
The answer is B.
BMI (body mass index) is calculated as weight in kilograms over height in meters squared. The calculation in this patient would be 100 kg/(1.8 m)2 or 100/3.24, which would equal 30.9
Folate deficiency
(A) Megaloblastic anemia
(B) Hypochromic, microcytic anemia
(C) Hemolytic anemia
(D) Sickle cell anemia
The answer is A. Folate deficiency results in a megaloblastic anemia because of decreased production of purines and the pyrimidine thymine. Thus, lack of folate causes decreased DNA synthesis. In contrast with a vitamin B12 deficiency, neurologic problems do not occur in a folate deficiency. A hypochromic, microcytic anemia can result from the lack of iron, or lack of pyridoxal phosphate. Both conditions lead to a reduction in the synthesis of heme, so the red cells cannot carry as much oxygen (which gives them the pale color). The cells are small in order to maximize the concentration of hemoglobin present in the cells. Hemolytic anemia occurs when the red cell membrane fragments, which can occur with pyruvate kinase deficiencies or a lack of glucose-6-phosphate dehydrogenase activity (which results in reduced NADPH levels). Sickle cell anemia is caused by a point mutation in the β-globin gene, substituting a valine for a glutamic acid.
A 40-year-old male is well controlled on warfarin for a factor V leiden deficiency and recurrent deep vein thrombosis. He presents today with a community-acquired pneumonia, and is placed on erythromycin. Three days later, he develops bleeding and his INR is 8.0 (indicating an increased time for blood clotting to occur, where INR is international normalized ratio). Which of the following best explains why this bleeding occurred?
(A) The erythromycin inhibited cytochrome P450
(B) The erythromycin stimulated cytochrome P450
(c) The causative agent of the pneumonia inhibited vitamin K utilization
(d) The causative agent of the pneumonia stimulated vitamin K utilization
(E) The erythromycin inhibited mitochondrial translation
(F) The erythromycin inhibited mitochondrial transcription
the answer is A.
Warfarin is metabolized by a specific subset of induced p450 isozymes. The p450 system is used by cells to modify the xenobiotic (in this case the warfarin) such that it can be more easily excreted. Erythromycin, along with other macrolide antibiotics, inhibits the p450 oxidizing system, which in this case would lead to a higher blood level of warfarin and, therefore, the balance of clotting and bleeding is shifted toward excessive bleeding. A stimulation of p450 production by erythromycin would lead to a lower level of warfarin (due to increased metabolism and loss of warfarin by p450) and the potential of excessive clotting. This effect of p450 is a common drug-drug interaction. The causative agents of communityacquired pneumonia do not affect vitamin K absorption in the small intestine, or distribution throughout the body. Erythromycin does not affect mitochondrial transcription, although it may affect mitochondrial translation. Inhibition of mitochondrial protein synthesis, however, will not alter the inhibition of cytochrome p450 activity, and the increased levels of warfarin present, which may lead to increased bleeding.
A homeless man was seen at a clinic because of bleeding gums and loosening teeth. His dietary history revealed that he has been consuming only chocolate milk and fast-food hamburgers for the past 6 months. 12. The patient is exhibiting these symptoms due to which one of the following?
(A) Reduced synthesis of fibrillin
(B) Reduced synthesis of collagen
(C) Reduced hydrogen-bond formation in collagen
(D) Increased hydrogen-bond formation in collagen
(E) Reduced disulfide-bond formation in collagen
(F) Increased disulfide-bond formation in collage
The answer is C. The homeless man has developed scurvy owing to a lack of vitamin C in his diet. Vitamin C is a required cofactor for the hydroxylation of proline and lysine within the collagen molecule. The lack of hydroxyproline reduces the stability of the collagen because of reduced hydrogen-bonding capabilities within the collagen triple helix. The lack of vitamin C does not affect disulfide-bond formation, which is required to initiate triple-helix formation within the cell. Fibrillin is not altered by the lack of vitamin C; it is the protein mutated in Marfan’s syndrome
A homeless man was seen at a clinic because of bleeding gums and loosening teeth. His dietary history revealed that he has been consuming only chocolate milk and fast-food hamburgers for the past 6 months.
The patient, due to his diet, had become deficient in which one of the following vitamins, which would lead to the symptoms observed?
(A) Vitamin A
(B) Vitamin C
(C) Vitamin B1
(D) Vitamin B2
(E) Vitamin B6
The answer is B. Scurvy is due to a lack of vitamin C, which is obtained from citrus fruits, which have been lacking in the diet. The patient may also become deficient in the other vitamins listed, but the lack of those vitamins will not lead to the symptoms characteristic of scurvy.
A 6-month-old baby was doing well until he developed viral gastroenteritis and was unable to tolerate oral feeding for 2 days. He is admitted to the hospital with encephalopathy, cardiomegally and heart failure, poor muscle tone, and hypoketotic hypoglycemia. Blood work did not detect any medium-chain dicarboxylic acids.
Once this baby is diagnosed and treated, his diet will need to be very restricted. Theoretically, which one of the following fatty acids will he be able to consume and metabolize?
(A) An 8-carbon fatty acid
(B) A 14-carbon fatty acid
(C) A 20-carbon fatty acid
(D) Only unsaturated fatty acids, regardless of chain length
(E) Only saturated fatty acids, regardless of chain length
The answer is A.
This baby has primary carnitine deficiency, an autosomal recessive disorder. The lack of medium-chain dicarboxylic acids in the blood rules out an MCAD deficiency. He is unable to transport blood-borne carnitine into the muscle and liver, thereby blocking fatty acid oxidation in those tissues. Carnitine is required to transfer most fatty acids from the cytoplasm to the matrix of the mitochondria. However, short- and medium-chain fatty acids (up to 10 or 12 carbons) are sufficiently water-soluble such that they can enter cells and be transferred into the mitochondria in the absence of carnitine. Once inside the mitochondria, an acyl-CoA synthetase will activate the fatty acid to an acyl-CoA such that β-oxidation can occur. The transfer is not affected whether the fatty acid is saturated or unsaturated; the chain length is the determining factor. Dietary restriction of long-chain fatty acids is essential to treat this disorder and alleviate the symptoms. The patient was doing well while feeding on a regular schedule because of the carbohydrate in the diet. Once the child had an extended fast, and needed to oxidize fatty acids for energy, the symptoms of carnitine deficiency became apparent. The hypoketotic hypoglycemia is a strong indication that the problem is in fatty acid oxidation
Patients with both Graves disease and Cushing syndrome are overproducing hormones that have which one of the following in common?
(A) Reacting with receptors in the cell membrane
(B) Utilizing second messengers
(C) Binding to intracellular receptors
(D) Binding to RNA to produce physiologically active proteins
(E) Inducing rRNA to ablate a particular genes expression
The answer is C.
Graves disease is due to the hypersecretion of thyroid hormone, whereas Cushing syndrome is an overproduction of cortisol. Steroid hormones and thyroid hormones cross the cell membrane and bind to intracellular receptors. The hormone–receptor complex binds to DNA, not RNA. Polypeptide hormones and epinephrine react with the receptors in the cell membrane triggering second messengers to transmit the signal that the receptor is occupied. Ribosomal RNA (rRNA) does not ablate gene expression; the induction of micro RNA has that ability
A patient complains of nervousness, palpitations, sweating, and weight loss without loss of appetite and has a goiter. Suspecting a defect in thyroid function, the physician orders a total serum T4. The test is performed by radioimmunoassay. The standard curve for the assay, which measures T4 in 0.1 mL of serum, is shown below. The normal levels of T4 are 4 to 10 μg/dL. In an assay of 0.1 mL of the patient’s
serum, 15% of the radioactive T4 was bound to the antibody.
The TSH levels in the patient’s blood were also measured by radioimmunoassay. If pituitary function is normal, the patient’s TSH levels will most likely be which one of the following?
(A) Higher than normal
(B) Normal
(C) Lower than normal
The answer is C.
Thyroid hormone suppresses TSH secretion by the anterior pituitary. If thyroid hormone levels are elevated, TSH levels will be lower than normal. This patient probably has Graves disease, in which thyroid-stimulating antibodies promote T3 and T4 production by the thyroid gland. These thyroid hormones inhibit the release of TSH from the anterior pituitary.
A 28-year-old female presents with fluctuating fatigue, drooping of her eyelids, difficulty swallowing, and slurred speech. The patient is given a drug that affects an enzyme’s activity, and kinetic analysis of the enzyme-catalyzed reaction, in the presence and absence of the drug, is shown below. The effect of this medication can best be described by which set of terms below?
the answer is A.
The patient has myasthenia gravis, and the treatment is pyridostigmine, a competitive, reversible inhibitor of acetylcholinesterase. Myasthenia gravis is caused by autoantibodies to the acetylcholine receptor, reducing the effectiveness of acetylcholine at the neuromuscular junction. By reversibly inhibiting acetylcholinesterase, the effective levels of acetylcholine are increased, thereby providing sufficient acetylcholine to bind to the few functional receptors that remain. The graph is classic for a competitive inhibitor. Competitive inhibitors display an increased apparent Km, and a constant Vmax.
