SNAPP Week 1: Concepts and Questions Flashcards
Which of the following is an expected result of a drug that enhances the action of normal p53 in response to ultraviolet exposure?
Increased apoptosis in damaged cells.
-p53 is a tumor suppressor gene so it arrests the cell cycle, promotes DNA repair and induces apoptosis in damaged cells.
Which of the following is a true statement regarding the regulation of cholesterol synthesis?
Insig binds SCAP only when cholesterol is high.
- SREBP needs to move into the Golgi to be cleaved and released as a transcription factor.
- Insig binds SCAP when cholesterol levels are high, this complex blocks SCAP signaling.
- SCAP signal domain is recognized by COPII for vesicles to move from ER to Golgi.
- As cholesterol concentration drops, Insig no longer binds SCAP and SCAP/SREBP complex gets packaged into vesicles to go to the golgi.
Which of the following is true regarding the body’s normal pH levels?
Venous blood is more acidic than arterial blood.
- Veins carry more CO2 and are slightly more acidic than arteries
- The inside of cells is MORE acidic than the outside.
- Blood pH = 7.4, [H+] = 40nmol/L
- The major regulatory organs are the kidneys and the lungs.
- Venous blood pH = 7.28-7.42
- Arterial blood pH = 7.34-7.44
You calculate the pKa for a compound and find it to have an extremely high value. Without knowing anything about the compound, which of the following conclusions is most correct?
It is a strong base.
-High pKa = strong base
-Low pka = strong acid
-pKa indicates the tendency of a species to be a proton donor (acid) or acceptor (base)
pKa = -log ([H+][A-] / [HA])
-think pKa = -log (dissociated concentration/undissociated concentration)
If you wanted to calculate the pH of a given solution and had the pKa, the concentration of the base, and the concentration of the acid, which equation would you use?
pH= pKa+log ([base]/[acid])
This is the Henderson Hasselbalch equation. In the question stem, you have ever variable except for the pH. This would be the appropriate equation to use.
Your Aunt Betty came over last night asking for help from her ‘soon to be doctor.’ She says she had a stomach ache last night and took some antacids, but is worried that she may have taken too many. Assuming she is correct, which of the following physiologic functions has been impaired?
facilitation of passive diffusion of plasma across the mucosal barrier.
- Low pH of gastric juice facilitates passive diffusion of plasma across mucosal barrier
- H. Pylori is a bacteria that can thrive in the gut because it uses urease to neutralize its environment. –For ionizing drugs, uptake can be determined by pH of local environment
If [A-]/[HA] = 2, what percentage of the solution is deprotonated?
66% (2/3) If [A-]/[HA] = 2 then you know that there are 2 A- molecules for every HA The deprotonated:protonated ratio is 2:1. Therefore, 2/3 are deprotonated. (Note: the number 3 comes from adding [A-] and [HA-] to account for the whole solution.
As a result of ketoacidosis, blood HCO3- concentration is reduced from 24 to 8mM. Hyperventilation causes pCO2 to drop to 15mmHg. What is the patient’s blood pH?
- 3
7. 3 pH = 6.1 + log (8mM)/(.03x15) = 7.34
Which of the following is a characteristic of malignant tumors but NOT a characteristic of benign lesions?
metastatic.
benign tumors are not invasive or metastatic but they are:
- undifferentiated
- unresponsive to growth control signals
- immortalized
p53, BRCA1 and BRCA2 mutations are considered risk factors for cancer because they:
encode DNA repair factors
p53, BRCA1 and BRCA2 encode DNA repair factors
Which of the following is true regarding the APC (adenomatous polyposis coli) gene?
loss of heterozygosity is associated with Familial Adenomatous Polyposis.
loss of heterozygosity is associated with FAP It has a dominant inheritance pattern APC (adenomatous polyposis coli) is a tumor suppressor gene individuals who are born as heterozygotes are at increased risk but normal individuals will need 2 mutations in the same cell to develop cancer (odds are much lower)
In the presence of elevated CDK (Cyclin-Dependent protein kinase) levels:
more cells will proceed into S phase.
increasing CDK will decrease RB (Retinoblastoma) inhibition (double negative by phosphoralating) and more cells will proceed from G1 into S phase (proliferation)
An 8 year old patient is diagnosed with a unilateral retinoblastoma in her right eye. A biopsy and genetic sequencing of normal tissue surrounding the tumor is homozygous for the RB (Retinoblastoma) gene, with normal protein expression. Which of the following findings is MOST likely, given this patient’s history and exam findings?
both copies of RB (Retinoblastoma) are inactivated in tumor cells.
Unilateral retinoblastomas indicate a spontaneous RB mutation. Since the surrounding normal tissue is expressing normal RB protein levels, the patient does not have familial RB heterozygosity. Retinoblastoma cells will be homozygous for mutated RB , but all other cells will be homozygous for normal RB. Bilateral retinoblastomas are rarely the result of spontaneous mutations because the likelihood of a single cell getting 2 mutations to RB is rare and will probably not occur in more than one place. Bilateral retinoblastoma is a trademark of familial RB.
Which of the following correctly describes the function of the protein normally transcribed by APC (adenomatous polyposis coli) gene?
degradation of free Beta-catenin in the cytoplasm.
APC codes for a protein that degrades free beta-catenin If APC is lost, free beta catenin will go to the nucleus and produce c-myc oncogenes Beta-catenin is normally held outside the nucleus by E-cadherin
BRCA1 and BRCA2 code for proteins involved in which of the following functions?
DNA repair.
BRCA1 and BRCA2 are involved in DNA repair Inherited BRCA mutations exhibit LOH, but acquired cases do not. Other genes likely influence BRCA function indirectly
Mutations to p53 are present in what percentage of all cancers?
50% .
50% of all cancers have a p53 mutation. However, 100% of all cancers have mutations that either directly alter p53 or interfere with one of its pathways.
95% of p53 mutations alter p53 function by:
inhibiting the DNA binding domain
95% of all p53 mutations inhibit its binding domain (so p53 can’t bind DNA)
Human Papilloma virus is an oncogenic virus in humans meaning that it:
inhibits p53 and RB (Retinoblastoma).
oncogenic viruses inhibit RB (Retinoblastoma) and p53
Which of the following correctly characterizes the genetic basis of Retinoblastoma?
autosomal recessive disorder with a dominant inheritance pattern
recessive disorder (requires 2 hits) but has a dominant inheritance pattern A single cell that has lost heterozygosity will turn into a tumor. Familial cases = 36% of the time bilateral Spontaneous = 6% of the time is bilateral Surgically repaired
A cell that is homozygous for mutated RB (Retinoblastoma gene) is in S phase of the cell cycle. If a signal from outside the cell arrives to the plasma membrane in order to trigger inhibition of cell growth, which of the following will occur?
the cell to continue to divide.
if RB (Retinoblastoma) is mutated, it is constitutively turned off, and the cell is free to proliferate without regulation. The outside of the cell can no longer communicate with the inside and RB (Retinoblastoma) cannot be activated. In normal cells, if you want to divide you must turn off RB (Retinoblastoma)
p53 mutations are most often:
missense.
p53 mutations are mostly missense (75%)
Internal virion proteins are encoded by which type of viral gene?
gag.
-gag: codes for internal virion proteins
-env: encodes for membrane glycoproteins
-pol: encodes for reverse polymerase
v-myc: mimics the c-myc proto-oncogene (cell growth/division)
If the RB (Retinoblastoma) protein is hypophosphorylated in a cell, what will happen?
the cell will be arrested in G1 phase.
-hypophosphorylation means RB (Retinoblastoma) is active and cells cannot go from G1 to S phase of the cell cycle
A cell containing only water and 300mM of a non-permeating protein is placed in a solution of 600 non-permeating sucrose. The membrane is permeable to water. What will be the volume of the cell (relative to its initial volume) after all net fluxes have stopped?
the cell will be 1/2 its initial volume
the cell will shrink to 1/2 its initial volume, at which point the osmolarity inside will equal the osmolarity outside, which is another way of saying that the concentration of water on the two sides is the same
A cell containing only water and 300mM of a non-permeating protein is placed in a solution of 300 mM non-permeating NaCl, and 600 permeating glycerol. The membrane is permeable to water. What will be the volume of the cell (relative to its initial volume) after all net fluxes have stopped?
the cell will be 1/2 its initial volume
the cell will shrink to 1/2 its volume. 300 mM NaCl - 600 mosM solution of Na and Cl ions [See above for explanation] You can ignore glycerol completely.
Which of the following DKA patients has the lowest risk for cerebral edema?
a 10 year old diabetic diagnosed 5 years ago
established diabetics have less risk of cerebral edema than new onset.
- Risk factors for cerebral edema:
- Administration of bicaRB (Retinoblastoma)onate
- insulin during first hour
- smaller than normal increase in Na
- increased volume of fluid treatment increases risk.