1.During prolonged fasting, rate of gluconeogenesis is determined by:
a) EFA in liver
b) Alanine in liver
c) Decreased c GMP
d) ADP in liver
2.After overnight fasting, levels of glucose transporters reduced in
a) Brain cells
In the fasting state, the glucose transporter of muscle and adipose tissue (GLUT-4) is in intracellular vesicles.
As insulin secretion falls in the fasting state, so the receptors are internalized again, reducing glucose uptake.
3.Crumpled tissue paper appearance of cells is seen in deficiency of
c) alpha Galactosidase
d) Beta Galactosidase
4.In well fed state, the activity of Carnitine Palmitoyl Transferase -1 in outer mitochondrial membrane is inhibited by
b) Acetyl CoA
c) Malonyl CoA
5.Regarding LDL receptors, all are true except
a) Found in Clathrin coated pits of cell membrane.
b) Found only in extrahepatic tissue.
c) Internalized by endocytosis.
d) High levels of celluar cholesterol down regulate LDL receptors.
Function uptake of LDL.
Present in the Liver and extrahepatic tissues(adipose tissue,Heart etc) .
High level of cholesterol upregulate LDL receptor ,causing an increase in the uptake LDL and hence the bad
Mechanism of uptake is receptor mediated uptake or absorptive pinocytosis.
Vesicles formed during absorptive pinocytosis are derived from invaginations (pits) are coated on the cytoplasmic
side with a filamentous material clathrin, hence the pits are named coated pits.
6.Eukaryotic plasma membrane is made up of all except
7.A patient was diagnosed with isolated increase in LDL .His father and brother had the same disease with increased
cholesterol. The likely diagnosis is
a) Familial type III hyperlipoproteinemia
c) Familial LPL deficiency(type1)
d) LDL receptor mutation
8.True about Ribozyme
a) Peptidyl Transferase activity
b) Cuts DNA at specific site
c) Participate in DNA Synthesis
d) GTPase activity
Ribozyme is RNA with catalytic activity.
Eg:-Sn RNA in Spliceosome-Takes part in splicing of exons and removal of introns
Ribonuclease P-Cuts the RNA
Peptidyl Tranferase-Peptide Bond formation
9.Test to differentiate in the chromosome of normal and cancer cell
b) Comparative Genomic Hybridisation
c) Western Blotting
d) Southern Blotting
10.Quantitative DNA analysis/estimation is done by
a) pH meter
Absorbance of uv light at 260 nm can be used to estimate DNA
Can be done using spectrophotometer or simply spectrometer.
11.There are 20 amino acids with three codons in spite of the no of amino acids could be formed is 64 leading to that an
amino acid is represented by more than one codon is called
d) Frame shift
12.All are added to PCR, except
c) Thermostat DNAP
d) Template DNA
Pre requisites of PCR
Sample DNA to be amplified
Thermostable Polymerase-Taq Polymerase obtained from Thermus Aquaticus found in hot springs.
13.The polypeptide from poly(A)is
Poly A codes for Lysine
Poly C codes for Proline
Poly G codes for Glycine
Poly U codes for Phenyl Alanine
14.If constitutive sequence of 4 nucleotide codes for 1 amino acid, how many amino acid can be theoretically formed?
15.Starting material for production of insulin from bacteria is:
a) Genomic DNA of lymphocytes
b) m RNA of lymphocytes
c) genomic DNA of beta cell of pancreas
d) mRNA of beta cells of pancreas
16.Not involved in translation in eukaryotes is:
a) Amino acyl t RNA
b) Peptidyl transferase
d) RNA polymerase
17.Apo B48 & Apo B100 is synthesized from the mRNA; the difference between them is due to : (May 2011)
a) RNA splicing
b) Allelic exclusion
c) Deamination of cytidine to uridine
d) Upstream repression
17.Histone acetylation cause
a) Increased Heterochromatin formation
b) Increased Euchromatin formation
c) Methylation of cystine
d) DNA replication
19.Triplet DNA is due to
a) Hoogsteen pairing
b) Palindromic sequences
c) Large no. of guanosine repeats
d) Polypyrimidine tracts
20.After digestion by restriction endonucleases DNA strands can be joined again by (May 2011)(Nov 2010)
a) DNA polymerase
b) DNA ligase
c) DNA topoisomerase
d) DNA gyrase
21.A four – year- old child is diagnosed with Duchene muscular dystrophy, an X-linked recessive disorder, Genetic analysis shows that the patient’s gene for the muscle protein dystrophin contains a mutation in its promoter region. What would be the most likely effect of this mutation?
a) Tailing of dystrophin mRNA will be defective
b) Capping of dystrophin mRNA will be defective
c) Termination of dystrophin transcription will be deficient.
d) Initiation of dystrophin transcription will be deficient
22.Methods of introducing gene in target cells are all except
c) Site directed recombination
23.Splicing activity is a function of