Unit 2 Practice Exam Questions Flashcards Preview

MSI Unit II > Unit 2 Practice Exam Questions > Flashcards

Flashcards in Unit 2 Practice Exam Questions Deck (45)
Loading flashcards...
1

Small signaling molecules such as NO bind to surface receptors on target cells (T/F)

False, these small molecules interact directly with enzymes in the cytoplasm

2

If chorionic villus testing returns a result of 47,XY,+21, then it can be concluded that the fetus has down syndrome. TF?

False, results from CVS testing could reflect an anomoly in the placenta only (confined placental mosacism). All abnormal results from CVS must be confirmed by amniocentesis

3

23
.
For the enzymatic reaction shown below, the following rate constants were determined:
k1 = 10 mM-1 sec
k2 = 20sec
k3 = 40sec-1

What is the dissociation constant of the
enzyme-substrate complex?

dissociation constant = k2/k1

20/10 = 2nM

4

It has been shown that chromosomal breakage and rearrangement are key
characteristics of
cancer.
A.
True
B.
False

true

5

Hexokinase and glucokinase are isozymes that carry out the initial step of glucose metabolism, conversion of glucose to glucose-6-phosphate. Hexokinase has a Km for glucose of 0.1 mM and glucokinase has a Km
for glucose of 5 mM. Cytosolic glucose
(substrate) concentration is 2 mM in a fasting subject and 6 mM in a subject who has just eaten a large meal. Which of the following conclusions can be drawn from this in formation?

A. The rate of the hexokinase reaction will triple when the glucose concentration goes
from 2 mM to 6 mM.
B. The rate of the glucokinase reaction will triple when the glucose concentration goes
from 2 mM to 6 mM.
C. At 2 mM glucose, 2/7 of the active sites on glucokinase are bound to substrate.
D. All of the above are true.
E. None of the above are true

C, use equation Fe=[s]/(km+[s])

6

total enzyme concentration = 1nm
inhibitor = 2nm

Uninhibited enzyme = slope 4, y int of 1
+2mM inhibitor A = slope 8, y int of 1
+2mM inhibitor B = slope 8, y int of .5

what can be concluded from this data?

A.Both inhibitor A and inhibitor B are competitive inhibitors.
B. Both inhibitor A and inhibitor B have the same Ki
C.The plot for inhibitor A will have the same x intercept as the plot for the uninhibited
enzyme.
D.The Ki for inhibitor B will double if the total enzyme concentration is doubled

both have same Ki

both have same slope and same concentration

plug into alpha = 1+ I/Ki

7

Activated T cells express _______________, which binds to B7 with 20 times higher affinity than CD28 and results
in _____________ of T cell activity and proliferation.

CTLA, Suppression

8

Superantigens are...

bacterial toxins that activate T-cells at a 100x greater rate than other antigens

9

Describe the drug used for immune rejection

CTLA fused to Fc region of an Ig molecule

10

Patient had a helminth infection resulting in high levels of IFN gamma and activated macrophages. Patients serum had high levels of IgG. Pronogis?

Poor, due to prescence of a TH1 response.

Need TH2 for helminths (IgE)

11

A cyclin molecule can best be described as...

the regulatory subunit of a protein kinease

12

After prolonged exposure to a high concentration of a hormone, you find that a specific tissue shows a greatly
decreased response to the hormone (adaptation). What is most likely to be responsible?

a change in the number of receptors on each cell

13

patient presents with tumor, DNA reveals novel mutations in kknown/suspected receptors associated with tumor tissue. Which type of receptor is frequently the target of oncogenic mutations and therefor most likely to be linked to the tumor growth in this patient?

Receptor Tyrosine Kineases

14

EBV vs CMV?

EBV causes severe sore throat more often

15

Which of the following statements about herpesvirus latency is CORRECT?
A: Latency is a hindrance to vaccine development because the virus genome is maintained in the cell for the life of the infected
person.
B: Latency is established before symptoms of the primary infection are apparent.
C: Reactivation from latency may not be apparent to the infected person.
D: The genome is present in a cell but infectious virions are absent.
E: All of the above
F: None of the above

all of the above

16

After polio binds GI receptors. When does infection begin and what happens next?

infection begins when genome enters cytoplasm

Polio's genome is +ssRNA which is recognized by ribosomes, so the next step involves translation of genomic mRNA into viral polyprotein by host ribosomes

17

child has recurrent infections with labs showing absence of large granular lymphocytes. Flow cytometry shoes normal numbers of CD3+, CD19+, and lack of CD56+ cells. What ist he probable mechanism?

mutation of an essential transcription factor

NK cells are "large granular lymphocytes" with a cd56+ receptor

18

Which statement about rotovirus is correct?
1. incidence peaks during summer
2. Gastroenteritis is most common in infants and toddlers
3. icosahedral, enveloped, -ssRNA virus
4. heat killed injectable rotovirus vaccine is used in the US; the live, attenuated, rotavirus vaccine is used in other countries

b - Gastroenteritis caused by the virus is most common in infants/toddlers

19

icosahedral capsid with spikes at verticies with no envelope, double stranded DNA

most likely is?

adenovirus

20

the bronchus of a dead smoker has respiratory epithelium likely replaced with?

stratified squamous epithelium (like skin)

21

protection from tumor cells of the cervix has been demonstrated in women who have
received the Gardasil. The major components of the vaccine are capsid proteins from a DNA virus. The
pre-neoplastic lesions do not express capsid proteins, but instead express viral oncoproteins
analogous to the

t antigen of merkel cell polyomavirus

22

Both chromosome breakage syndromes and HNPCC (hereditary non-polyposis colon cancer) are associated with

mutations in dna repair genes

23

To kill bacteria that are adapted to survive inside of phagocytes, the most important immune response
is____________________.

TH1 response and activation of macrophages

24

Which of the following junction types is MOST IMPORTANT for determining epithelial tissue permeability

tight

25

Expression of the ribozyme called telomerase in cancer cells may prevent cellular senescence by

stimulating RNA templated extension of the parental DNA strand

26

which is not a component of the BCR? cd2, cd19, cd21, cd81

cd2 - this is an adhesion molecule found on T cells and NK cells

27

14 year old gets a thymectomy, likely outcome?

will likely survive with functional immune system

28

A newborn in the neonatal intensive care unit has features consistent with Patau syndrome. The perinatologist requests that the
diagnosis be confirmed by karyotype analysis. It is important to determine the karyotype of the baby because .
􁅞􀀃A: it will reveal which parent is the source of the abnormality, which is important for future family planning
􁅚􀀃B: it is possible that this child has a normal chromosome constitution and is actually affected by an autosomal recessive phenocopy. In addition, if the child has a chromosome rearrangement involving chromosome 13, the chromosome imbalance could place future pregnancies at risk
􁅞􀀃C: if the child is a mosaic, the mutation cannot be due to an error in meiosis, and therefore the recurrence risk in future pregnancies will be negligible
􁅞􀀃D: if the disease causing duplication can be identified, it will reveal which of the parents is a carrier of that mutation which is important for future
family planning
􁅞􀀃E: it will aid in his/her care

it is possible that this child has a normal chromosome constitution and is actually affected by an autosomal recessive phenocopy. In addition, if the child has a chromosome rearrangement involving chromosome 13, the chromosome imbalance could place future pregnancies at risk

Answer B: The diagnosis of Patau syndrome at birth would be done by clinical assessment. It is always a good idea to have the
chromosome complement confirmed by karyotype analysis, but, more important, is the confirmation that the trisomy is the result of a meiotic
nondisjunction event giving rise to 3 individual copies of chromosome 13. The recurrence risk for this is quite low. However, the trisomy could also be
the result of a nondisjunction error if one of the parents is a carrier of a Robertsonian translocation involving chromosome 13. Clinically there is no way
to tell the difference between a translocation trisomy and a three chromosome trisomy, but if a parent is a carrier of a rearrangement, then he/she is at
risk of having another child with a trisomy. Knowing that there is a recurrence risk is an important piece of information for future family planning.
Learning Outcomes: Genetics

29

A 24-year-old woman who is 18 weeks pregnant is meeting with her obstetrician to discuss prenatal screening for chromosomal
abnormalities. This is her first pregnancy and she has no family history of chromosomal disorders. Which of the following should
be offered and discussed with the patient?

amniocentesis


This patient is in her 2nd trimester and amniocentesis is the only listed procedure that is carried out in the 2nd trimester. All others are 1st
trimester procedures/screens.
Learning Outcomes: Genetics

30

A 45-year-old patient presents with an acute leg cramp that has become worse over several days, along with evidence of restricted
circulation in his leg. After further investigation, it is determined that he has a deep vein thrombosis (a clot has formed in a leg
vein). He is prescribed the drug coumadin. Coumadin has a structure similar to Vitamin K and acts as an anticoagulant. This was
not the first time that this patient had a deep vein thrombosis. After he recovered, his physician suggested that he be tested for a
deficiency in one of the factors involved in regulation of coagulation. Genetic deficiency in which of the following
is MOST LIKELY to cause the tendency toward thrombosis (excessive clotting) in this patient?

anti-thrombin ATIII deficiency

Decks in MSI Unit II Class (46):