Block 7 Pace Quiz 2 Part 1

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Block 7 Pace Quiz 2 Part 1 - Quiz

Pace quiz 6 sem 2 december 2011 final week of semester


Questions and Answers
  • 1. 

    In the accompanying drawing, pick the cell (A, B, C, D, or E) that is part of your immune system:

    • A.

      A

    • B.

      B

    • C.

      C

    • D.

      D

    • E.

      E

    Correct Answer
    C. C
    Explanation
    Based on the given information, it is not clear what the drawing represents or how it relates to the immune system. Therefore, an explanation cannot be provided.

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  • 2. 

    Dr. Moore Which of the cell types in the accompanying drawing is directly responsible for the formation of a keloid?

    • A.

      A

    • B.

      B

    • C.

      C

    • D.

      D

    • E.

      NONE OF THE ABOVE

    Correct Answer
    E. NONE OF THE ABOVE
    Explanation
    The correct answer is NONE OF THE ABOVE because a keloid is formed due to an overgrowth of scar tissue. None of the cell types shown in the drawing are directly responsible for the formation of keloids.

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  • 3. 

    Dr. Kalliecharan Milk protein is produced by which of the following mechanisms?

    • A.

      Paracrine secretion

    • B.

      Autocrine secretion

    • C.

      Holocrine secretion

    • D.

      Apocrine secretion

    • E.

      Merocrine secretion

    Correct Answer
    E. Merocrine secretion
    Explanation
    Merocrine secretion is the correct answer because milk protein is produced and secreted by the mammary glands through the process of merocrine secretion. In this mechanism, the protein is synthesized and packaged into vesicles within the Golgi apparatus. These vesicles then fuse with the cell membrane, releasing the protein into the extracellular space without any damage to the cell. This allows for continuous production and secretion of milk protein without harming the mammary gland cells.

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  • 4. 

    Which of the following statements is characteristic of hair follicles?

    • A.

      They contain epidermal keratinocytes

    • B.

      They are always associated with an eccrine sweat gland

    • C.

      Their associated arrector pili muscles are composed of skeletal muscle fibers

    • D.

      They are inserted into the papillary layer of the epidermis

    • E.

      They do not extend into the dermis

    Correct Answer
    A. They contain epidermal keratinocytes
    Explanation
    Hair follicles are composed of multiple layers, including the outermost layer called the epidermal keratinocytes. These keratinocytes are responsible for producing the hair shaft. Therefore, the statement "They contain epidermal keratinocytes" is characteristic of hair follicles.

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  • 5. 

    Dr. Martin A 45 year old woman is diagnosed with a malignant tumor in the superolateral quadrant of the left breast, including the axillary tail of Spence. If  metastasis occurs, the cancer will MOST LIKELY spread first to which of the following locations?

    • A.

      Deep cervical lymph nodes

    • B.

      Parasternal lymph nodes

    • C.

      Contra lateral breast lymph nodes

    • D.

      Lateral axillary lymph nodes

    • E.

      Anterior lymph nodes

    Correct Answer
    E. Anterior lymph nodes
    Explanation
    The axillary tail of Spence is an extension of breast tissue that extends into the axilla (armpit). When a tumor is located in this area, the cancer cells are most likely to spread to the lymph nodes in the axilla, which are the lateral axillary lymph nodes. The other options, such as the deep cervical lymph nodes, parasternal lymph nodes, contra lateral breast lymph nodes, and anterior lymph nodes, are not the most likely locations for metastasis in this case.

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  • 6. 

    Dr. Wright MM, a 38 year old female, is in her 19th week of pregnancy.  For this time period, what protein produced by her placenta is the most important in replacing the function of LH in her nonpregnant state to maintain her progesterone secretion at appropriate levels?

    • A.

      Estradiol

    • B.

      Human chorionic gonadotropin

    • C.

      Prolactin

    • D.

      Human chorionic somatomammotropin

    • E.

      Dopamine

    Correct Answer
    B. Human chorionic gonadotropin
    Explanation
    Human chorionic gonadotropin interacts with the LHCG receptor and promotes the maintenance of the corpus luteum during the beginning of pregnancy, causing it to secrete the hormone progesterone. Progesterone enriches the uterus with a thick lining of blood vessels and capillaries so that it can sustain the growing fetus. Due to its highly-negative charge, hCG may repel the immune cells of the mother, protecting the fetus during the first trimester. It has also been hypothesized that hCG may be a placental link for the development of local maternal immunotolerance. For example, hCG-treated endometrial cells induce an increase in T cell apoptosis (dissolution of T cells). These results suggest that hCG may be a link in the development of peritrophoblastic immune tolerance, and may facilitate the trophoblast invasion, which is known to expedite fetal development in the endometrium. It has also been suggested that hCG levels are linked to the severity of morning sickness in pregnant women.
    Because of its similarity to LH, hCG can also be used clinically to induce ovulation in the ovaries as well as testosterone production in the testes. As the most abundant biological source is women who are presently pregnant, some organizations collect urine from pregnant women to extract hCG for use in fertility treatment.[10]
    Human chorionic gonadotropin also plays a role in cellular differentiation/proliferation and may activate apoptosis
    Human chorionic gonadotropin (hCG) is a hormone produced during pregnancy that is made by the developing embryo after conception, and later by the placental component syncytiotrophoblast. Some cancerous tumors produce this hormone; therefore, elevated levels measured when the patient is not pregnant can lead to a cancer diagnosis. However, it is not known whether this production is a contributing cause or an effect of tumorigenesis. The pituitary analog of hCG, known as luteinizing hormone (LH), is produced in the pituitary gland of males and females of all ages.

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  • 7. 

    MM a 38 year old female, is in her 19th week of pregnancy.  She has a family history of type 2 diabetes, so prior to pregnancy she had maintained a very healthy lifestyle with regular exercise and a low-fat, high fiber diet.  To date during the pregnancy she has maintained an exercise regimen similar to that of her pre-pregnant state, and only adjusted her diet in accordance with her physician’s instructions.  When her blood pressure (BP) is taken, what would be the expected result for this female, relative to that of her (normal) pre-pregnant BP?

    • A.

      Her BP is slightly elevated (high normal range, pre-hypertensive)

    • B.

      Her BP is greatly elevated (hypertensive)

    • C.

      Her BP is greatly decreased (hypotensive)

    • D.

      Her BP remains unchanged

    • E.

      Her BP is slightly decreased (low normal range)

    Correct Answer
    E. Her BP is slightly decreased (low normal range)
    Explanation
    During pregnancy, it is common for blood pressure to decrease due to hormonal changes and increased blood volume. This is known as physiological hypotension. Since MM has maintained a healthy lifestyle and exercise regimen, it is expected that her blood pressure would be slightly decreased (low normal range) compared to her pre-pregnant state.

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  • 8. 

    Every morning, a woman runs 12 kilometers.  Today, conditions are mild (no wind, 3% humidity, overcast skies but no rain forecast).  When she reaches the 6 kilometer turnaround point and stops for one minute, her body is producing much more heat than it does at rest.  By which of the following heat loss mechanisms will this person dissipate the most heat during this one minute interval?

    • A.

      Radiation

    • B.

      Conduction

    • C.

      Convection

    • D.

      Evaporation

    • E.

      Countercurrent

    Correct Answer
    D. Evaporation
    Explanation
    During the one-minute interval when the woman stops at the 6 kilometer turnaround point, her body will dissipate the most heat through evaporation. Evaporation is the process by which sweat on the skin's surface turns into water vapor, taking away heat from the body in the process. Since the conditions are mild with low humidity, the sweat will evaporate quickly, allowing for efficient heat loss. Radiation, conduction, and convection are less effective in dissipating heat in this scenario. Countercurrent is not a heat loss mechanism.

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  • 9. 

    Dr. Adebiyi Which of the following immune cell types in the skin is most likely to be the dominant antigen processing and presenting cell?

    • A.

      Langerhans cells

    • B.

      Plasmacytoid dendritic cells

    • C.

      Dermal dendritic cells

    • D.

      Macrophages

    • E.

      Epidermal cells

    Correct Answer
    C. Dermal dendritic cells
    Explanation
    Dermal dendritic cells are the most likely immune cell type in the skin to be the dominant antigen processing and presenting cell. These cells are specialized in capturing antigens, processing them, and presenting them to other immune cells, such as T cells. They are strategically located in the dermis, where they can efficiently encounter and interact with antigens. Langerhans cells and plasmacytoid dendritic cells also play important roles in immune responses in the skin, but dermal dendritic cells are considered to be the main antigen-presenting cells in this context. Macrophages and epidermal cells are not as specialized in antigen processing and presentation as dendritic cells.

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  • 10. 

    Dr. Buxbaum An 18-month old with a history of recurrent bacterial and viral infections, failure to thrive, developmental delay, and tremors is submitted to your care. On exam you notice a lack of peripheral lymphoid tissue. Blood analysis reveals lymphopenia with normal B-cell count and normal immunoglobulin levels. Patient is most likely suffering from

    • A.

      Hypoxanthine guanine phosphoribosyl transferase (HGPRT)

    • B.

      Purine nucleoside phosphorylase

    • C.

      Adenine phosphoribosyl transferase (APRT)

    • D.

      Adenosine desaminase (ADA)

    • E.

      Adenosine kinase

    Correct Answer
    B. Purine nucleoside phosphorylase
    Explanation
    The correct answer is Purine nucleoside phosphorylase. This is because the patient's lack of peripheral lymphoid tissue, along with the symptoms of recurrent infections, failure to thrive, developmental delay, and tremors, are indicative of a severe combined immunodeficiency (SCID) disorder. Purine nucleoside phosphorylase deficiency is a type of SCID that is characterized by lymphopenia (low lymphocyte count) and normal B-cell count and immunoglobulin levels. This deficiency leads to impaired T-cell function and can result in a variety of clinical manifestations, including recurrent infections and developmental delays.

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  • 11. 

    A child with penicillin allergy is given a sulphonamide against otitis media. Human cells are not affected by sulphonamides because sulphonamides are specific for bacterial

    • A.

      Folate synthesis

    • B.

      DNA polymerases

    • C.

      RNA polymerases

    • D.

      Ribonucleotide reductase

    • E.

      Mismatch repair

    Correct Answer
    A. Folate synthesis
    Explanation
    Sulphonamides are specific for bacterial folate synthesis. Folate is a vitamin that is essential for the synthesis of DNA, RNA, and proteins. Bacteria have a different pathway for folate synthesis compared to human cells. Sulphonamides inhibit the enzyme dihydropteroate synthase, which is involved in the bacterial folate synthesis pathway. This inhibition disrupts the production of folate in bacteria, leading to their death. Since human cells do not have the same folate synthesis pathway, they are not affected by sulphonamides. Therefore, a child with a penicillin allergy can safely be given a sulphonamide for the treatment of otitis media.

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  • 12. 

    You are a neonatologist working in a tertiary care hospital in a state capital. A 4-week old male is transferred to you with severe jaundice that appeared at birth and has been worsening ever since. The boy is the first child of a healthy Jewish couple, pregnancy and vaginal delivery were unremarkable. The boy is of average height and weight for his age, in no acute distress, but shows marked jaundice and slight hepatomegaly. Lab: CBC normal, indirect bilirubin high, low fecal urobilinogen. What is the most likely diagnosis?

    • A.

      Neonatal jaundice

    • B.

      Dubin-Johnson syndrome

    • C.

      Rotor (-Manahan-Florentin) syndrome

    • D.

      Uridyl glucuronyl transferase deficiency

    • E.

      Glucose-6-phosphate dehydrogenase deficiency

    Correct Answer
    D. Uridyl glucuronyl transferase deficiency
    Explanation
    The most likely diagnosis for the 4-week old male with severe jaundice, high indirect bilirubin, low fecal urobilinogen, and slight hepatomegaly is Uridyl glucuronyl transferase deficiency. This condition, also known as Gilbert syndrome, is a genetic disorder that affects the liver's ability to process bilirubin, leading to an accumulation of bilirubin in the blood and causing jaundice. It is more common in individuals of Jewish descent and typically presents in infancy or early childhood. The other options, such as neonatal jaundice, Dubin-Johnson syndrome, Rotor syndrome, and Glucose-6-phosphate dehydrogenase deficiency, do not fit the given clinical presentation and laboratory findings.

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  • 13. 

    Dr. Sands What mutational event is often associated with Burkitt’s lymphoma?

    • A.

      The ras protooncogene is mutated at one nucleotide so that it is less able to hydrolyze bound GTP

    • B.

      The 3’ end of the c-src gene is deleted so that the SRC protein is always activated.

    • C.

      There is a translocation of the c-myc gene to one of the immunoglobulin loci.

    • D.

      There is a deletion of the p16 gene which allows unregulated passage through the G1 restriction point.

    • E.

      EBV causes the Her-2 protein to dimerize without a ligand binding to the receptor

    Correct Answer
    C. There is a translocation of the c-myc gene to one of the immunoglobulin loci.
    Explanation
    To assure the rapid loss of c-Fos and c-Myc after their induction in normal cells, both proteins and their corresponding mRNAs are intrinsically unstable. Some of the changes that turn c-fos from a normal gene to an oncogene involve loss of sequences in the gene that make the Fos mRNA and protein short-lived. Conversion of the c-myc proto-oncogene into an oncogene can occur by several different mechanisms. In cells of the human tumor known as Burkitt’s lymphoma, the c-myc gene is translocated to a site near the heavy-chain antibody genes. An analogous transloca- tion in the mouse genome is also involved in mouse myelomas. In both cases, the tumor cells arise from antibody-producing cells, which carry out DNA rearrangements during their maturation. The c-myc translocation is a rare aberration of the normal rearrangement events, bringing it from its normal, distant chromosomal location into juxtaposition with the enhancer of the antibody genes. The translocated myc gene, now regulated by the antibody enhancer, is continually expressed, causing the cell to become cancerous. Localized reduplication of a segment of DNA containing the myc gene, which occurs in several human tumors, also causes inappropriately high expression of the otherwise normal Myc protein.

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  • 14. 

    When DNA damage occurs in a cell that has a functional p53 protein, the activation of p53 enhances oxidative pentose phosphate flux.   This causes an increase in the amount of NADPH and R5P, both of which are needed for nucleotide synthesis and DNA repair.   How does the cell get enough NADPH for these processes when the cell does not have a functional p53 protein?

    • A.

      Transketolase-like 1 (TKTL1) is up-regulated and it catalyzes the nonoxidative pentose phosphate flux.

    • B.

      Fructose 1,6 bisphosphate (F1,6biP) increases concentration in cancer cells and inhibits glucose-6 phosphate dehydrogenase (G6PD).

    • C.

      Pyruvate kinase-M2 (PK-M2) is up-regulated and allows glycolytic intermediates to accumulate.

    • D.

      Malic enzyme (ME) is up-regulated and works with lactate dehydrogenase (LDH-A) to convert malate to lactate.

    • E.

      Hexose kinase II (HKII) is up-regulated and binds to the voltage dependent anion channel (VDAC).

    Correct Answer
    D. Malic enzyme (ME) is up-regulated and works with lactate dehydrogenase (LDH-A) to convert malate to lactate.
    Explanation
    Glucose provides cells with a source of Ac-CoA for fatty acid synthesis (blue arrows).    Fatty acid synthesis is enhanced in tumors due to oncogene-driven expression of the lipogenic enzymes ATP-citrate lyase (ACL), acetyl-CoA carboxylase-1 (ACC) and fatty acid synthase (FAS).
    However, continuous citrate export introduces a deficit to the TCA cycle, and this must be replaced by an anaplerotic flux in order for fatty acid synthesis and cell growth to continue.
    Metabolism of glutamine (red arrows) provides a mitochondrial oxaloacetate pool for continued citrate synthesis. After citrate cleavage by ACL in the cytoplasm, the resulting oxaloacetate can be converted to malate and ultimately lactate by the low NAD+/NADH ratio created by rapid glycolysis.
    Glutamine may also be converted to lactate if mitochondrial malate is exported to the cytoplasm and decarboxylated by malic enzyme (ME). This pathway appears to be a major source of NADPH for fatty acid synthesis and other activities in tumor cells.
    (slide 36 Cancer-Sands)

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  • 15. 

    Dr. Blanchetot The pedigree below segregate for Duchenne muscular dystrophy (DMD) and X-linked recessive condition lethal in males.  What is the risk for III3 of being a carrier for DMD? (use Bayesian calculation)

    • A.

      1/20

    • B.

      1/10

    • C.

      1/5

    • D.

      1/3

    • E.

      1/2

    Correct Answer
    B. 1/10
    Explanation
    Prior risk to II-2 is 1/2. BUT two healthy boys decreases her risk. the chance of having 2 healthy is.
    1/2 x 1/2 or 1/4. then multiply 1/4 by prior risk (1/2) this gives 1/8.
    on the other colum you get 4/8 therefore II-2 risk is (1/1+4) or 1/5 and child has 1/2 chance =1/10

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  • 16. 

    How could the following pedigrees be explained by genetic imprinting?

    • A.

      Maternal uniparental disomy (UPD)

    • B.

      Paternal uniparental disomy (UPD)

    • C.

      Maternal expressed gene

    • D.

      New mutation

    • E.

      Paternal expressed gene

    Correct Answer
    E. Paternal expressed gene
    Explanation
    notice when mom has disease the children dont but when dad has it the kids do too.

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