UW 2024 Biochemistry Past Paper (OCR) - Medical Study Zone PDF

= - Item 1 of 40 \" Mar1< Question Id: 1756 ~~ @ i' ~ ~ I" ~ © Previous Next Full Screen Tutonal Lab values Notes Galrolator Reverse Color Text Zoom Settings Foreign antigen recognition at the cell surface by cytotoxic T-lymphocytes stimulates a response that results in rapid cell death. Which of th...

= - Item 1 of 40 \" Mar1< Question Id: 1756 ~~ @ i' ~ ~ I" ~ © Previous Next Full Screen Tutonal Lab values Notes Galrolator Reverse Color Text Zoom Settings Foreign antigen recognition at the cell surface by cytotoxic T-lymphocytes stimulates a response that results in rapid cell death. Which of the following act as effectors of this response? 0 0 0 O O A. Matrix metalloproteinases B. Acid hydrolases C. Caspases D. Phospholipase A 2 E. Protein kinase A Submit Block Time Elapsed: 00:00:43 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block = - Item 1 of 40 \" Mar1< Question Id: 1756 ~~ @ i' ~ ~ I" ~ © Previous Next Full Screen Tutonal Lab values Notes Galrolator Reverse Color Text Zoom Settings Foreign antigen recognition at the cell surface by cytotoxic T-lymphocytes stimulates a response that results in rapid cell death. Which of the following act as effectors of th is response? 0 A. Matrix metalloproteinases (6%,) X @ B. Acid hydrolases (9%) ✓ 0 O O C. Caspases (78%) D. Phospholipase A 2 (3%,) E. Protein kinase A (2%) Incorrect I 1, 1 78% ~ Answered correctly Correct answer C ('j\ 55 secs \.::) Tfflle Spent Explanation The two primary modes of cell death are necrosis (which is induced by inj ury) and apoptosis (which is initiated by the host organism in response to cell damage, age-related atrophy, or specific stages of embryogenesis). Apoptosis is helpful in the elimination of cells that are no longer necessary and in the maintenance of a constant number of cells in rapidly growing tissues. Tumor cells, for example, die by apoptosis. The death of Band T lymphocytes and virally-infected cells also occurs through apoptosis. The remains of cells that die by this method are phagocytized by macrophages. The process of apoptosis involves the following steps: 1. Initiation: Apoptosis is triggered by different stimuli and can occur through either the intrinsic, mitochondriamediated pathway or the extrinsic, receptor-initiated pathway. Cells damaged by ultraviolet light, heat, hypoxia, toxins, or radiation display intrinsic apoptotic signals (eg, phosphatidylserine or thrombospondin) on Block Time Elapsed: 00:00:55 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block = - Item 1 of 40 \" Mar1< Question Id: 1756 ~~ @ i' ~ ~ I" ~ © Previous Next Full Screen Tutonal Lab values Notes Galrolator Reverse Color Text Zoom Settings Explanation The two primary modes of cell death are necrosis (which is induced by inj ury) and apoptosis (which is initiated by the host organism in response to cell damage, age-related atrophy, or specific stages of embryogenesis). Apoptosis is helpful in the elimination of cells that are no longer necessary and in the maintenance of a constant number of cells in rapidly growing tissues. Tumor cells, for example, die by apoptosis. The death of Band T lymphocytes and virally-infected cells also occurs through apoptosis. The remains of cells that die by this method are phagocytized by macrophages. The process of apoptosis involves the following steps: 1. Initiation: Apoptosis is triggered by different stimuli and can occur through either the intrinsic, mitochondriamediated pathway or the extrinsic, receptor-initiated pathway. Cells damaged by ultraviolet light, heat, hypoxia, toxins, or radiation display intrinsic apoptotic signals (eg, phosphatidylserine or thrombospondin) on their plasma membranes. Extrinsic apoptosis, in contrast, is induced by the tumor necrosis factor (TNF) when bound to tumor necrosis factor receptor 1 (TNFR1) or the Fas ligand when bound to cell surface receptor Fas. 2. Control: Intrinsic apoptosis is mediated by a group of bcl-2 proteins. Some of the components of this system are pro-apoptotic (eg, Bak, Bax, and Bim proteins), while others are anti-apoptotic (eg, Bcl-x and Bcl-2 proteins). Apoptotic signals tip the balance between these two forces, resulting in changes of the inner mitochondrial membrane. These changes are responsible for the formation of the mitochondrial permeability transition (MPT) and the release of cytochrome c and other pro-apoptotic proteins into the cytoplasm, which then activate caspases. In extrinsic apoptosis, the binding of the death ligand and the death receptor allows for pro-caspase molecules to be brought into close proximity. 3. Destruction: Both the intrinsic and extrinsic pathways converge at this step, resulting in caspase activation. Caspases are proteolytic enzymes that destroy cell components. They contain cysteine and are able to cleave aspartic acid residues (cysteine-aspartic-acid-proteases). The eleven caspases that have been identified are classified as either initiator or effector caspases. Initiator caspases activate the effector caspases, which then cleave the cellular proteins. Block Time Elapsed: 00:00:55 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block = - Item 1 of 40 \" Mar1< Question Id: 1756 ~~ @ i' ~ ~ I" ~ © Previous Next Full Screen Tutonal Lab values Notes Galrolator Reverse Color Text Zoom Settings proteins). Apoptotic signals tip the balance between these two forces, resulting in changes of the inner mitochondrial membrane. These changes are responsible for the formation of the mitochondrial permeability transition (MPT) and the release of cytochrome c and other pro-apoptotic proteins into the cytoplasm, which then activate caspases. In extrinsic apoptosis, the binding of the death ligand and the death receptor allows for pro-caspase molecules to be brought into close proximity. 3. Destruction: Both the intrinsic and extrinsic pathways converge at this step, resulting in caspase activation. Caspases are proteolytic enzymes that destroy cell components. They contain cysteine and are able to cleave aspartic acid residues (cysteine-aspartic-acid-proteases). The eleven caspases that have been identified are classified as either initiator or effector caspases. Initiator caspases activate the effector caspases, which then cleave the cellular proteins. (Choice A) Metalloproteinases are zinc-containing enzymes that degrade the components of the extracellular matrix (eg, collagen, laminin, fibronectin). Metalloproteinases are essential for proper tissue remodeling during wound healing. (Choice B) Acid hydrolases do not participate in apoptosis. (Choice D) Phospholipase A2 does not participate in apoptosis. (Choice E) Protein kinase A is a component of the cAMP-associated signaling system. The binding of a ligand to a G-protein-l inked receptor resu lts in adenylyl cyclase activation and the release of cAMP. Elevated levels of cAMP activate protein kinase A. Educational Objective: Apoptosis can occur through either the intrinsic (mitochondria-mediated) pathway or the extrinsic (receptorinitiated) pathway. Both pathways converge in the activation of caspases. Caspases are proteolytic enzymes that cleave cellular proteins. Biochemistry Biochemistry (General Principles) Cell mediated immunity Subject System Topic Block Time Elapsed: 00:00:55 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block = - ltem2of40 D Mar1< Question Id: 1501 \ ~~ @ i' ~ ~ I" ~ © Previous Next Full Screen Tutonal Lab values Notes Galrolator Reverse Color Text Zoom Settings A 5-day-old boy is diagnosed with hyperphenylalaninemia by newborn screening. He is placed on a special phenylalanine-restricted diet with tyrosine supplementation. The parents are extensively counseled on the patient's condition and informed of the necessary dietary restrictions. They are also instructed to return to the clinic for regular follow-up visits. Several months later, laboratory results indicate that the infant has a normal serum phenylalanine level. Carefu l examination, however, shows some neurologic abnormalities, including axial hypotonia. Further work-up is notable for elevated prolactin. Impaired synthesis of a cofactor essential for metabolism of aromatic amino acids is suspected. Which of the following enzymes is most likely deficient in this patient? 0 0 0 O 0 A. Dihydropteridine reductase B. Dopamine beta-hydroxylase C. PhenylaJanine hydroxylase D. Phenylethanolamine N-methyltransferase E. Tyrosinase Submit Block Time Elapsed: 00:00:55 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block = - ltem2of40 D Mar1< Question Id: 1501 \ ~~ @ i' ~ ~ I" ~ © Previous Next Full Screen Tutonal Lab values Notes Galrolator Reverse Color Text Zoom Settings A 5-day-old boy is diagnosed with hyperphenylalaninemia by newborn screening. He is placed on a special phenylalanine-restricted diet with tyrosine supplementation. The parents are extensively counseled on the patient's condition and informed of the necessary dietary restrictions. They are also instructed to return to the clinic for regular follow-up visits. Several months later, laboratory results indicate that the infant has a normal serum phenylalanine level. Carefu l examination, however, shows some neurologic abnormalities, including axial hypotonia. Further work-up is notable for elevated prolactin. Impaired synthesis of a cofactor essential for metabolism of aromatic amino acids is suspected. Which of the following enzymes is most likely deficient in this patient? ✓ @ A. Dihydropteridine reductase (33%) 0 0 O O B. Dopamine beta-hydroxylase (15%) C. PhenylaJanine hydroxylase (23%) D. Phenylethanolamine N-methyltransferase (13%) E. Tyrosinase (13%) I 1.1 33% l!!!!.. Answered correctly Correct IT\ 04 secs \..:.) Tnne Spent Explanation Disorders of phenylalanine metabolism Phenylketonuria En zyme ~- Block Time Elapsed: 00:00:59.............. Phenylalanine hydroxylase ~ =... BH4 deficiency Dihydropteridine reductase Normally re generates the cofactor BH4 https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block = - ltem2of40 D Mar1< Question Id: 1501 \ ~~ @ i' ~ ~ I" ~ © Previous Next Full Screen Tutonal Lab values Notes Galrolator Reverse Color Text Zoom Settings Explanation Disorders of phenylalanine metabolism Phenylketonuria Enzyme deficiency Phenylalanine hydroxylase Normally converts phenylalanine to tyrosine j Phenylalan ine o Pathophysiology of disease Neurologic abnormalities (eg, developmental delay, microcephaly, seizures) o Hypopigmentation o Musty odor BH4 deficiency Dihydropteridine reductase Normally regenerates the cofactor BH4 (essentia l for reactions involving phenylalanine, tyrosine & tryptophan) r Phenylalanine (same as in PKU) ! Neurotransmitters (eg, dopamine, epinephrine, serotonin) o dystonia), feeding difficu lties, autonomic dysfunction o Treatment Extrapyramidal signs (eg, axial hypotonia, Dysregulation of appetite/mood/sleep Phenylalanine-restricted diet Phenylalanine-restricted diet ± Tyrosine supplementation BH4 supplementation BH4 = tetrahydrobiopterin. This patient had elevated phenylalanine levels at birth, wh ich is most commonly caused by deficiency of phenylalanine hydroxylase (phenylketonuria [PKU]). Management of presumed PKU includes a phenylalanine- restricted diet, which prevents manifestations of disease (eg, neurologic abnormalities). However, this patient developed progressive neurologic findings (eg, axial hypotonia) despite normalization of serum phenylalanine levels, which is suggestive of a less common cause of impaired phenylalanine metabolism such as impaired tetrahydrobiopterin (BH,) synthesis. Block Time Elapsed: 00:00:59 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block = D Mar1< ltem 2 of40 - Question Id: 1501 - - I - \ , , I ,- e e I., -e 1 e 1e. I e ~~ @ i' ~ ~ I" ~ © Previous Next Full Screen Tutonal Lab values Notes Galrolator Reverse Color Text Zoom Settings. ,,..... 1. developed progressive neurologic findings (eg, axial hypotonia) despite normalization of serum phenylalanine levels, which is suggestive of a less common cause of impaired phenylalanine metabolism such as impaired tetrahydrobiopterin (BH 4) synthesis. BH4 is an essential cofactor in several hydroxylation reactions and is converted from its unoxidized form by the enzyme dihydropteridine reductase. Deficient dihydropteridine reductase and therefore impaired BH4 synthesis cause the following processes to be defective: Conversion of phenylalanine to tyrosine: Impaired conversion leads to hyperphenylalaninemia, as seen in th is patient's orig inal presentation. A phenylalanine-restricted diet normalizes phenylalanine levels. Conversion of tyrosine to L-dopa: L-dopa is the precursor for catecholamine (dopamine, norepinephrine, epinephrine) production, and impairment of neurotransmitter production leads to progressive neurologic injury (eg, axial hypotonia, dystonia, autonomic dysfunction) despite a phenylalanine-restricted diet and tyrosine supplementation. Moreover, dopamine normally inhibits prolactin synthesis and secretion, and without this negative feedback, lactotrophs in the anterior pituitary secrete excess prolactin, as seen in this patient Conversion of tryptophan to serotonin: Impaired serotonin production can lead to dysregulation of mood, appetite, sleep, and muscle contraction. (Choices Band 0) Dopamine beta-hydroxylase converts dopamine to norepinephrine, and phenylethanolamine N-methyltransferase converts norepinephrine to epinephrine. Deficiency of either enzyme would not result in elevated phenylalanine levels, as seen in this patient's initial presentation. (Choice C) Phenylalanine hydroxylase deficiency resu lts in PKU. A normalized phenylalanine level would be expected with a phenylalanine-restricted diet, as seen in this patient, but neurologic manifestations of disease wou ld not progress. Moreover, tyrosine supplementation in PKU allows for dopamine production, and this patient's hyperprolactinemia suggests persistent dopamine deficiency. (Choice E) Melanin is synthesized from tyrosine derivatives by tyrosinase in melanocytes. Tyrosinase deficiency (absent melanin production) causes albinism, characterized by a lack of pigment in skin and hair. A history of Block Time Elapsed: 00:00:59 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block = - D Mar1< ltem2of40 - Question Id: 1501. - \.... - ~... -........... -......... Previous Next... -...... -. ~~ @ i' ~ ~ I" ~ © Full Screen Tutonal Lab values Notes Galrolator Reverse Color Text Zoom Settings supplementation. Moreover, dopamine normally inhibits prolactin synthesis and secretion, and without this negative feedback, lactotrophs in the anterior pituitary secrete excess prolactin, as seen in this patient. Conversion of tryptophan to serotonin: Impaired serotonin production can lead to dysregulation of mood, appetite, sleep, and muscle contraction. (Choices Band 0) Dopamine beta-hydroxylase converts dopamine to norepinephrine, and phenylethanolamine N-methyltransferase converts norepinephrine to epinephrine. Deficiency of either enzyme would not resu lt in elevated phenylalanine levels, as seen in this patient's initial presentation. (Choice C) Phenylalanine hydroxylase deficiency resu lts in PKU. A normalized phenylalanine level would be expected with a phenylalanine-restricted diet, as seen in this patient, but neurologic manifestations of disease wou ld not progress. Moreover, tyrosine supplementation in PKU allows for dopamine production, and this patient's hyperprolactinemia suggests persistent dopamine deficiency. (Choice E) Melanin is synthesized from tyrosine derivatives by tyrosinase in melanocytes. Tyrosinase deficiency (absent melanin production ) causes albinism, characterized by a lack of pigment in skin and hair. A history of hyperphenylalaninemia and progressive neurologic findings would not be seen. Educational objective: Although hyperphenylalaninemia is most commonly due to phenylketonuria, progressive neurologic injury despite appropriate treatment (phenylalanine-restricted diet, tyrosine supplementation) should raise concern for impaired tetrahydrobiopterin synthesis. This distinct disorder of phenyla lanine metabolism is often due to dihydropteridine reductase deficiency, resulting in the inability to synthesize critical neurotransmitters (eg, dopamine, serotonin). References Tyrosine, phenylalanine, and catecholamine synthesis and function in the brain. Biochemistry Biochemistry (General Principles) Phenylketonuria Subject System Topic Block Time Elapsed: 00:00:59 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block = - ltem 2 of40 D Mar1< Question Id: 1501 \ ' 'f o ~~ @ i' ~ ~ Previous Next Full Screen Tutonal Lab values Notes Galrolator , Reverse Color ~ © Text Zoom Settings Oysregulat1on of appetite/mood/sleep Impaired tetrahydrobiopterin (BH 4 ) synthesis This p pheny Dihydropteridine reductase Dihydropteridine reductase 8 v2 BH. restri devel levels tetrah T Phenylalanine--------,► Tyrosine--~--------;► Phenylalanine hydroxylase BH4 is Melanin BH2 Tyrosine h}'droxylase l-DOPA 7t Catecholamines enzy Dihydropteridine reductase Tryptophan - - - - - -T,yptophan hydroxylase e ( 5-hydroxytryptophan - - Serotonin (S-HT) ~Phenylalanine hydroxylase deficiency results in phenylketonuria, another disorder of phenylalanine metabolism. BH2 = Dihydrobiopterin; BH, : Tetrahydrobtopterin, L-DOPA : Levodopa s n (OUWor1d ~ zoom In 0.. Zoom Out C Reset C::, New I~ Existing ~ My Notebook N-methyltransferase converts norepinephrine to epinephrine. Deficiency of either enzyme would not result in Block Time Elapsed: 00:00:59 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook fro Flashcards CJ Feedback @ Suspend 0 End Block = - ltem 2 of40 D Mar1< Question Id: 1501 \ ~~ @ i' ~ ~ Previous Next Full Screen Tutonal Lab values Notes Galrolator , Reverse Color ~ © Text Zoom Settings BH4 = tetrahydrobiopterin. ~ X Exhibit Display This pheny Prolactin restri devel TRH levels Dopamine VIP L_ Antipsychotics 1 Antidepressants Hypothalamus - - - - tetrah BH4 is enzy Hypophyseal portal system e ( s Anterior __ __ pituitary n a Systemic circulation (Choi N-met TRH = lhyrottop,n-relea5inO hormone VIP = vasoactm ,nle5bnal pel)lide ~ zoom In elevat 0.. Zoom Out C Reset C::, New O UWorld I~ Existing ~ My Notebook (Choi expected with a phenylalanine-restricted diet, as seen in this patient, but neurologic manifestations of disease Block Time Elapsed: 00:00:59 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook fro Flashcards CJ Feedback @ Suspend 0 End Block = - ltem3of40 D Mar1< Question Id: 12278 \ ~~ @ i' ~ ~ I" ~ © Previous Next Full Screen Tutonal Lab values Notes Galrolator Reverse Color Text Zoom Settings A 54-year-old man comes to the emergency department with a 3-month history of fatigue and exertional dyspnea. He has early satiety and frequent upper abdominal discomfort. On physical examination, the patient has palpable splenomegaly but no lymphadenopathy. Laboratory tests are as follows: Complete blood count Hemoglobin 9.2 mg/dl Platelets 80,000/mm3 Leukocytes 56,000/mm3 Reverse transcription polymerase chain reaction is used to diagnose chronic myelogenous leukemia in this patient. Which of the following is most likely to be detected by this test? 0 O O O 0 A. Chromosomal position of the BCR and ABL genes B. DNA rearrangement in the BCR promoter region C. Fusion protein contain ing BCR and ABL domains D. Messenger RNA transcript containing BCR and ABL exons E. Point mutation in the ABL enhancer region Submit Block Time Elapsed: 00:00:59 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block = - ltem3of40 D Mar1< Question Id: 12278 \ ~~ @ i' ~ ~ I" ~ © Previous Next Full Screen Tutonal Lab values Notes Galrolator Reverse Color Text Zoom Settings A 54-year-old man comes to the emergency department with a 3-month history of fatigue and exertional dyspnea. He has early satiety and frequent upper abdominal discomfort. On physical examination, the patient has palpable splenomegaly but no lymphadenopathy. Laboratory tests are as follows: Complete blood count Hemoglobin 9.2 mg/dl Platelets 80,000/mm3 Leukocytes 56,000/mm3 Reverse transcription polymerase chain reaction is used to diagnose chronic myelogenous leukemia in this patient. Which of the following is most likely to be detected by this test? 0 A. Chromosomal position of the BCR and ABL genes (26%,) x @ B. DNA rearrangement in the BCR promoter region (10%,) O ✓O 0 C. Fusion protein contain ing BCR and ABL domains (30%) D. Messenger RNA transcript containing BCR and ABL exons (28%) E. Point mutation in the ABL enhancer region (3%) Incorrect Correct answer D I 1,1 28% IT\ 06 secs l.!!!.!. Answered correctly \.::,)Time Spent @ 2023 Version Explanation Reverse transcription polymerase chain reaction (RT-PCR) is used to detect and quantify levels of messenger Block Time Elapsed: 00:01 :05 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block = - ltem3of40 D Mar1< Question Id: 12278 \ ~~ @ i' ~ ~ I" ~ © Previous Next Full Screen Tutonal Lab values Notes Galrolator Reverse Color Text Zoom Settings Explanation Reverse transcription polymerase chain reaction (RT-PCR) is used to detect and quantify levels of messenger RNA (mRNA) in a sample. It is similar to regular PCR in that it uses sequence-specific primers, thermostable DNA polymerase, and a pool of deoxyribonucleoside triphosphates to amplify a DNA template. In RT-PCR, this template is generated by the action of reverse transcriptase on the mRNA sample, producing a complementary DNA (cD NA) strand that can then be amplified by PCR. Because cDNA is complementary to the mRNA sequence, it contains the exons of a gene along with the 5' and 3' untranslated regions. Chronic myeloge-nous leukemia (CML) is characterized by uncontrolled proliferation of the myeloid stem cell line due to a chromosomal translocation. This translocation causes the BCR gene on chromosome 22 to fuse with the ABL gene on chromosome 9, forming the BCR-ABL fusion gene. The BCR-ABL fusion protein product is a constitutively active tyrosine kinase that accelerates cell division and increases genetic instability. RT-PCR can be used to identify mRNA transcribed from the BCR-ABL fusion gene and therefore diagnose CML. (Choice A) Fluorescence in situ hybridization (FISH) techniques al low direct localization of genes to their respective chromosomes by using a labeled DNA probe complementary to the sequence of interest. (Choices Band E) RT-PCR amplification uses an mRNA template, so it cannot detect changes in the parts of the gene that are not transcribed (eg, promoter and enhancer reg ions). Other PCR techniques that use chromosomal DNA can detect changes in these nontranscribed regions. (Choice C) RT-PCR is used to detect levels of mRNA expression; it does not identify proteins. A Western blot study can detect the BCR-ABL protein by using monoclonal antibodies directed against BCR or ABL. Educational objective: Reverse transcription polymerase chain reaction (RT-PCR) is used to detect and quantify levels of mRNA in a sample. It uses reverse transcription to create a complementary DNA template that is then amplified using the standard PCR procedure. RT-PCR can be used to diagnose chronic myelogenous leukemia by identifying an mRNA transcript containing both BCR and ABL exons in affected cells. Block Time Elapsed: 00:01 :05 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block = - ltem3of40 D Mar1<......... , ·........ , Question Id: 12278 · ,.......... -.. Previous Next -. ~~ @ i' ~ ~ I" ~ © Full Screen Tutonal Lab values Notes Galrolator Reverse Color Text Zoom Settings _,.:, (cD NA) strand that can then be amplified by PCR. Because cDNA is complementary to the mRNA sequence, it contains the exons of a gene along with the 5' and 3' untranslated regions. Chronic myelogenous leukemia (CML) is characterized by uncontrolled proliferation of the myeloid stem cell line due to a chromosomal translocation. This translocation causes the BCR gene on chromosome 22 to fuse with the ABL gene on chromosome 9, forming the BCR-ABL fusion gene. The BCR-ABL fusion protein product is a constitutively active tyrosine kinase that accelerates cell division and increases genetic instability. RT-PCR can be used to identify mRNA transcribed from the BCR-ABL fusion gene and therefore diagnose CML. (Choice A) Fluorescence in situ hybridization (FISH) techniques al low direct localization of genes to thei r respective chromosomes by using a labeled DNA probe complementary to the sequence of interest. (Choices Band E) RT-PCR amplification uses an mRNA template, so it cannot detect changes in the parts of the gene that are not transcribed (eg, promoter and enhancer reg ions). Other PCR techniques that use chromosomal DNA can detect changes in these nontranscribed regions. (Choice C) RT-PCR is used to detect levels of mRNA expression; it does not identify proteins. A Western blot study can detect the BCR-ABL protein by using monoclonal antibodies directed against BCR or ABL. Educational objective: Reverse transcription polymerase chain reaction (RT-PCR) is used to detect and quantify levels of mRNA in a sample. It uses reverse transcription to create a complementary DNA template that is then amplified using the standard PGR procedure. RT-PCR can be used to diagnose chronic myelogenous leukemia by identifying an mRNA transcript containing both BCR and ABL exons in affected cells. References Real time polymerase chain reaction in diagnosis of chronic myeloid leukemia. Biochemistry Biochemistry (General Principles) Chronic myeloid leukemia Subject System Topic Block Time Elapsed: 00:01 :05 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block = - ltem4of40 D Mar1< Question Id: 1070 \ ~~ @ i' ~ ~ I" ~ © Previous Next Full Screen Tutonal Lab values Notes Galrolator Reverse Color Text Zoom Settings Nutrition researchers investigating the relationsh ip between fructose consumption and cardiovascular disease conduct a prospective cohort study on a population of randomly selected young adults. Study participants undergo semiannual measurement of waist circumference, blood pressure, and serum cholesterol and triglyceride concentrations. Dietary fructose consumption is assessed through the use of questionnaires and by measuring urinary fructose excretion. A 23-year-old man enrolled in the study is found to excrete large amounts of fructose in his urine compared to other study participants despite maintaining a moderate fructose intake. Further evaluation shows a hereditary defect in fructose metabolism, but he is asymptomatic and has no other medical problems. This patient most likely remains able to metabolize fructose due to the compensatory activity of wh ich of the following enzymes? Q A. Aldolase B 0 0 0 O B. Aldose reductase C. Fructokinase D. Hexokinase E. UDP-galactose-4-epimerase Submit Block Time Elapsed: 00:01 :07 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block = - ltem4of40 Question Id : D Mar1< 1 0 70 Previous \ Next F ull Screen @ T u to nal i' Lab values ~ Notes ~ Galrola t I" or Reverse Color ~ Text Zoom © Settings Nutrition researchers investigating the relationsh ip between fructose consumption and cardiovascular disease conduct a prospective cohort study on a population of randomly selected young adults. Study participants undergo semiannual measurement of waist circumference, blood pressure, and serum cholesterol and triglyceride concentrations. Dietary fructose consumption is assessed through the use of questionnaires and by measuring urinary fructose excretion. A 23-year-old man enrolled in the study is found to excrete large amounts of fructose in his urine compared to other study participants despite maintaining a moderate fructose intake. Further evaluation shows a hereditary defect in fructose metabolism, but he is asymptomatic and has no other medical problems. This patient most likely remains able to metabolize fructose due to the compensatory activity of wh ich of the following enzymes? x @ A. Aldolase B (28%) 0 O ✓0 O B. Aldose reductase (15%) C. Fructokinase (21%) D. Hexokinase (30%>) E. UDP-galactose-4-epimerase (4%) Incorrect I 1.1 30% L!!!!. Answered correctly Correct answer D IT\ 04 secs VTmeSpent Explanation Disorders of fructose metabolism l Glucose Hexokinase Block Time Elapsed: 00:01 :09 ~ = -. https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block = - ltem4of40 D Mar1< Question Id: 1070 \ ~~ @ i' ~ ~ Previous Next Full Screen Tutonal Lab values Notes Galrolator , Reverse Color ~ © Text Zoom Settings Explanation Disorders of fructose metabolism l Glucose Hexokinase l Glucose 6-phosphate Sucrose l ::::~:~UCO· t Hexoklnase Fructose - - - - - - - - - - - - - - - - - - - ► Fructose 6-phosphate 5 I f~ c~:~~1~ ~ (benign condition) J Fructose 1-phosphate Aldolase B +--0---- DHAP 6- ]j Fructoklnase Fructose-1, bisphosphatase Hereditary fructose intolerance Hypoglycemia, vomiting PFK-1 Fructose-1 6-bisphosphatase A/dolase A & B after fructose ingestion Failure to lhrive, liver & renal failure Glyceraldehyde Glyceraldehyde DHAP ~phosphate I Tnose 1 ~hosphate 1 1 ,somerase y Pyruvate ©uwortd Fructose is obtained in the diet primarily from fru its and food sweeteners such as table sugar (sucrose) and highfructose corn syrup. Fructose is absorbed in the proximal intestine through the GLUTS fructose transporter. It is normally phosphorylated by fructokinase in the liver, yielding fructose-1-phosphate, which is converted by aldolase B to dihydroxyacetone phosphate (DHAP) and glyceraldehyde (Choice C). Glyceraldehyde and DHAP can be converted to glyceraldehyde-3-phosphate, which can then be metabolized in the glycolytic pathway. Block Time Elapsed: 00:01 :09 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block = - ltem4of40 D Mar1< Question Id: 1070 \ ~~ @ i' ~ ~ Previous Next Full Screen Tutonal Lab values Notes Galrolator , Reverse Color ~ © Text Zoom Settings Exp Bl X Exhibit Display Disorders of fructose metabolism l Glucose Hexokinase I Glucose 6-phosphate Sucrose l Phosphog/uco1somerase ~t Hexol..... conce :~ (Choi - nonco E 0 ro Q) >N C enzy Noncompetitive inhibitor w and c (Choi enzy Substrate concentration enzy ~ zoom In 0.. Zoom Out C Reset C::, New I~ Existing ~ My Notebook inhibitors do not affect enzyme function; therefore, maximal velocity (Vmu.) is unchanged in their presence. Block Time Elapsed: 00:01 :27 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook fro Flashcards CJ Feedback @ Suspend 0 End Block = - ltem 9 of40 D Mar1< Question Id: 12066 \ ~~ @ i' ~ ~ Previous Next Full Screen Tutonal Lab values Notes Galrolator , Reverse Color ~ © Text Zoom Settings impede substrate binding. Because these inhibitors compete with the substrate for binding to the active site, additi Bl X Exhibit Display effect In this malat Noncompetitive inhibition (Choi molec conce (Choi tVmax Normal nonco enzy ?;- ·5 ~ ro Q) E >, N C enzy Noncompetitive inhibitor w enzy Substrate concentration Comp to ach inhibit ~ zoom In 0.. Zoom Out C Reset C::, New I~ Existing ~ My Notebook Subject Block Time Elapsed: 00:01 :27 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook fro Flashcards CJ Feedback @ Suspend 0 End Block = - ltem10 of40 D Question Id: 1378 Mar1< \ ~~ @ i' ~ ~ I" ~ Previous Next Full Screen Tutonal Lab values Notes Galrolator Reverse Color Text Zoom © Settings Two graphs illustrating the transport rate of solutes across the plasma membrane are shown on the slide below. (/) 1 (/) 2 a, (.) C t:.c a, a, 8. E (/) a, CE ~ t- Solute concentration Which of the following best explains the difference in the shape of the curves? O A. Different amounts of membrane surface area for diffusion 0 O O O B. Different degrees of membrane thickness C. The 2 solutes have different molecular weights D. The 2 solutes have different oil/water partition coefficients E. The presence of a protein transporter Submit Block Time Elapsed: 00:01 :30 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block = - ltem10of40 Question Id: 1378 D ~~ @ i' ~ ~ I" ~ Previous Next Full Screen Tutonal Lab values Notes Galrolator Reverse Color Text Zoom Mar1< \ © Settings Two graphs illustrating the transport rate of solutes across the plasma membrane are shown on the slide below. 1 Solute concentration Which of the following best explains the difference in the shape of the curves? O A. Different amounts of membrane surface area for diffusion (3%) 0 O B. Different degrees of membrane thickness (1%) x @ ✓ O C. The 2 solutes have different molecular weights (3% ) D. The 2 solutes have different oil/water partition coefficients (10%) E. The presence of a protein transporter (80%) Incorrect I 1.1 80% L!!!!. Answered correctly Correct answer E (i'\ 06 secs ~ TnneSpem 2023 Version Explanation Characteristics of simple & facilitated diffusion ~ 0 Block Time Elapsed: 00:01 :33 ~ = Facilitated diffusion (saturable) ' https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block = - ltem10 of40 Question Id: 1378 D Mar1< \ I -... t II I ; ,,... I I I ~~ @ i' ~ ~ Previous Next Full Screen Tutonal Lab values Notes Galrolator , Reverse Color ~ © Text Zoom Settings I I Bl X Exhibit Display Characteristics of simple & facilitated diffusion The i simpl Facilitated diffusion (saturable) There F Carri Solute concentration Bindin ©UWorld releas trans ~ zoom In 0.. Zoom Out C Reset C::, New I~ Existing ~ My Notebook Lower molecular weight Block Time Elapsed: 00:01 :33 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook fro Flashcards CJ Feedback @ Suspend 0 End Block = - ltem10 of40 D Question Id: 1378 - I.. I t I ~~ @ i' ~ ~ I" ~ Previous Next Full Screen Tutonal Lab values Notes Galrolator Reverse Color Text Zoom Mar1< \. I I ''.. I I., I. © Settings simple diffusion (line 2) and carrier-mediated transport (line 1). There are two types of diffusion: Simple diffusion - molecules move through a membrane without the help of carrier proteins. Faci litated diffusion - requires carrier proteins. Carrier proteins are typical ly transmembrane proteins that possess binding sites for the substrate they transport. Binding is followed by movement of the substrate across the cell membrane to the intracellular space, where it is released into the cytoplasm. Because there is a fin ite number of carrier proteins in the cell membrane, transporter saturation occurs with facilitated diffusion, and can be seen as a flattening of the curve (maximum diffusion speed), even as solute concentration continues to increase. This maximum rate of transport is referred to as the transport maximum (Tm) and is similar in principle to the Vmax in standard enzyme kinetics. (Choices A, B, C, D) These other factors are important for determining the rate of diffusion (ie, slope of the line), but would not explain the flattening of the curve in line 1 (which is best accounted for by saturation of a protein carrier). In general, the rate of diffusion increases with: Higher concentration gradients across the membrane Lower molecular weight Larger diffusion surface area Thinner membrane thickness For molecules that move through the membrane via simple diffusion, the degree of lipophilicity is also important; molecules with a high oil-water partition coefficient are more easily able to cross the membrane and will diffuse faster. Educational objective: Carrier-mediated transport includes facilitated diffusion and active transport. Movement of substrate across the cell membrane by these mechanisms depends on the presence of carrier proteins that can become saturated at high substrate concentrations. Block Time Elapsed: 00:01 :33 ~ = https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block - - Item 11 of40 Question Id: 1244 f Mark ~~ @ i' ~ ~ Next Full Screen Tutonal Lab values Notes Galrolator , Reverse Color ~ © Text Zoom Settings...... -....... -..... "' ) D. Homocysteine (8%) E. Lactate (8%) F. Palmitate (27%) Incorrect Correct answer A Block Time Elapsed: 00:02:18 ~ = I 1,1 44% (j'\04secs I.!!!!. Answered correctly \.::,) Time Spent https://t.me/USMLEWorldStep1 11;] My Notebook q,o Flashcards CJ Feedback @ Suspend 0 End Block = - ltem18 of40 Questio n Id: 1886 \" Mar1< ~~ @ i' ~ ~ Previous Next Full Screen Tutonal Lab values Notes Galrolator , Reverse Color ~ © Text Zoom Settings Bl X Exhibit Display Fatty acid oxidation Fatty acid Acyl-CoA synthetase l Primary camitlne deficiency Muscle weakness Card,omropathy Hypoketctic hypoglycemia Elevated muscle triglycendes Acyl-CoA Carrubne -- +- ~ Acyt-carm11ne Cytoplasm Carmbne ( -=II. Acyl-cam,t,ne Mitochondrial matrix Acyl-CoA ~ Acyl-CoA FAOH2- i dehydrogenase ~ans-Enovl C