Answer 1

spiral chains of keratin twist to form keratin. Spiral chains cross linked by
hydrogen and disulfide bonds, changes of these linkage determine whether hair
is curly or straight.

We Will Write a Custom Essay Specifically
For You For Only $13.90/page!

order now

change the arrangement disulfide bonds of Keratin chains is to be disrupted,
reducing agents would oxidize keratin by disrupting the covalent bond between
cysteine molecules and favors formations of bonds between cysteine and hydrogen,
which is a weak covalent bond easily broken at this step hair can be rearranged.
In presence of oxidizing agent, the hydrogen ions will be removed so that
cysteine residues forms disulfide bonds.

all the broken disulfide bonds do not reform again and form weak spots in
keratin. So, the bonds formed were weak and likely to break and allows protein
to come back to original state.

Some type of hairs is curly as high number of disulfide bonds are
formed along different sites of nearby chains when compared to straight hair
and cause keratin to fold, and follicle shape also
a determining factor which helps in bringing hair to each other.

Answer 2:

Myo-inositol monophosphates catalyzes the hydrolysis
of inositol monophosphate is inhibited by Lithium is an Uncompetitive inhibition. Uncompetitive inhibition in which enzyme
substrate complex is binded with inhibitor (Lithium) to prevent formation of
final product as it takes longer for the substrate or
product to leave the active
site. At higher concentration of substrate, the inhibitor
works well.

Penicillin binds to DD-transpeptidase and prevents it from
binding and cross-linking
peptidoglycans in the bacterial cell wall, which will cause the bacterial cell
to burst. This is an example of Irreversible
inhibitor .An irreversible
inhibitor will bind to an enzyme so that no other enzyme-substrate complexes can form. It will bind to
the enzyme using a covalent bond at the active site which
therefore makes the enzyme

COX-2 catalyzes
formation of prostaglandins which is competitively inhibited by NSAIDs. Competitive inhibition in
which binding of an inhibitor prevents binding of substrate to enzyme. This is
done by blocking binding to active site of substrate. Addition of substrate displaces
inhibitor from the active site of enzyme and increase binding of substrate to
enzyme, this alters only the Km, leaving the Vmax unchanged.


Answer 3:

?G values which are negative values indicate forward
reaction and positive values indicate reverse reaction strongly favored. In
absence of inorganic phosphate which indicates that hexokinase (step 1) and
phosphofructokinase (Step 3) favor forward reactions. But Aldolase (step 4) and
phosphoglycerate mutase (step 8) never favor reaction even in presence of
inorganic phosphate as it has positive ?G but as a result of indirect coupling
these reactions are become favorable. Step 10 (PEP to pyruvate) is favorable that
low concentration of all intermediates prior to this pushes reaction in forward
direction to produce new substrate. Steps
6, 7, and 10 releases sufficient energy to drive the formation of NADH (step 6)
or the formation of ATP (steps 7 and 10).

Step 1 is formation of Glucose 6 phosphate (G6P) from
glucose with use of ATP molecule which is irreversible reaction. It is
catalyzed by hexokinase which is feedback inhibited by G6P, so phosphorylation
of glucose is controlled depending upon concentration of formed G6P. ?G is negative as the ATP used as phosphate
donor. The glucose is locked up in cell and It also enables the glucose to go on to step 2 of glycolysis.

3 is
conversion of fructose 6 phosphate to fructose 1,6 bisphosphate catalyzed by
phospho fructokinase by utilizing ATP. Phosphofructokinase is also activated by
fructose?2,6? bisphosphate, which is formed from fructose?1?phosphate by a
second, separate phosphofructokinase enzyme phosphofructokinase II. This step is activated under low energy when
cell has high AMP and ADP and fructose
1,6 bisphosphate.

1 and 3 consumes ATP energy to take but prepares substrates for the production
of energy in later steps.

7 Involves 1,3-bisphosphoglycerate
to 3- phosphoglycerate catalyzes by phosphoglycerate Kinase. The first ATP
forming step in glycolysis. ATP is formed by transfer of a phosphoryl group
from 1,3-bisphosphoglycerate to ATP is called as substrate level
phosphorylation. Coupling of step 6 with step 7 produces one NADH per glucose
are to be reoxidized to regenerate NAD that is needed for the oxidation of
glyceraldehyde 3 phosphate.

Step 4 is formation of trioses G-3 P (glyceraldehyde
3 phosphate) and DHAP (Dihydroxy acetone phosphate) from fructose 1,6
bisphosphate catalyzed by Aldolase which has high positive gibbs free energy. Aldolase
reaction is a near equilibrium reaction in cells, which shows concentration
of fructose 1,6 bisphosphate is highly relative to two triose groups G-3 P and
DHAP. The concentration of fructose 1,6 bisphosphate is very high in the
cells which is formed due to high -ve (negative) gibbs free energy. The
concentration of these trioses is low in cells when compared to fructose
1,6 bisphosphate. Flux due to high concentration of fructose 1,6 bisphosphate
which goes to next steps for pyruvate synthesis.

Answer 4:

In TCA cycle total of 20 ATP
produced from 2 Acetyl COA molecules. For one it produces 10 ATPs.

Conversion of Isocitrate to alfa Ketoglutarate produces one 1

Enzyme: isocitrate dehydrogenase.

Conversion of Alpha
ketoglutarate to succinyl CoA produce 1 NADH

Enzyme: alpha ketoglutarate dehydrogenase.

Oxidation of Malate forms oxaloacetate, 1 NAD+ is reduced to NADH.

 Enzyme: malate

 1 NADH = 2.5 ATP: Total 3 NADH for 1 Acetyl COA = 3*2.5 = 7.5 ATP

Oxidation of Succinate to fumarate. (FAD) is
reduced and forms FADH2 . Enzyme: succinate dehydrogenase.

1 FADH2 = 1.5 ATP


succinyl CoA converted to Succinate by
removing the COA using GTP and generate ATP in process

Enzyme: succinyl-CoA synthetase.



Total ATP for 1
Acetyl COA = 3 NADH + 1 FADH2 + 1 ATP = 7.5+ 1.5+ 1= 10 ATP.


For 2 Acetyl COA = 20 ATP.

Answer 5:

Glucose 1 phosphate conversion to lactate yields 3 ATP

1 ATP from Phosphofructokinase reaction

2 ATP form Phosphoglycerate Kinase reaction


Converting two
molecules of lactate to one molecule of glucose 1 phosphate needs 6 ATP.

2 ATP for Pyruvate
carboxylase reaction

2 ATP in the PEP
carbokinase reaction

2 ATP for
Phosphoglycerate Kinase reaction.



Answer 6:


rRNA are processed
and assembled into their ribosomal subunits within nucleus and exported so it
is resistant to nucleases.

tRNA are processed
from a primary transcript and heavy modification in nucleoside was seen and
have an extensive secondary structure which makes them
resistant to ribonuclease degradation.

of mRNA was conducted during the transcription by an enzyme complex.

reason for capping is protection of mRNA from 5′ exonuclease enzymes. Removal
of one nucleotide from mRNA
synthesizes an inefficient protein so that preservation of the mRNA translation
is very important.

Main functions of 5′ capping is
regulating transport of mRNA from nucleus and protection from exonucleases and
promotion of translation, 5′ proximal intron excision.


Answer 7:

mRNA: It has codons for peptide synthesis. It makes up to 3-5 % of
total RNA. Show base relationship to DNA.

tRNA: It has anticodons that can base pair or link the exact required
amino acid corresponding to mRNA codon. Also called as Adapter molecules. It
makes up to 15-20 % of total RNA.

rRNA: It is a molecule in cell that form a part of ribosome and then
exported to cytoplasm and help in translation process. Catalyzes the formation of the
peptide bond. It makes up 80% of total RNA. I have no base relationship to DNA.

Synthesis of mRNA, tRNA, rRNA:

mRNA: It is
formed by transcription process from DNA with the help of RNA polymerase which
make a copy of gene from DNA form to mRNA. It is transferred from Nucleus to Cytoplasm.

tRNA: It is
also formed by transcription process from DNA with help if RNA polymerase three
enzyme in the nucleus. It is formed by nuclear processing of precursor

rRNA: It is
formed by transcription from DNA using RNA polymerase one into large RNA
molecule. Later addition of sequences is added by many other polymerases
simultaneously to form giant RNA molecules.

Structure and function of mRNA, tRNA, rRNA:

mRNA: It is
linear shape molecule. IT carries genetic information from DNA to ribosome in Cytosol
which serves as a template for proteins synthesis and unpaired bases are binded
to mRNA and tRNA. 5’end terminal is capped by the 7 – methyl guanosine
triphosphate cap. It helps in recognizing the mRNA by the translation
machinery. Capping prevents cleavage by 5′ exonucleases. 3′ end have a polymer
of adenylate residues which protect from 3′ exonucleases.

tRNA: It
has a primary and secondary structure. Primary structure the nucleotide
sequence of all tRNA molecules allow intrastand complementary that forms a
secondary structure. Each tRNA have extensive internal base pairing and forms a
clover like structure. The hydrogen bonding stabilizes the structure. Clover
leaf structure have 5 arms

1. Acceptor arm – It is 3′ end, the
hydroxyl of Adenine binds with carboxyl groups of Amino acid.

2. Anticodon arm- Opposite ends of
Acceptor arms, it binds specifically with mRNA by hydrogen bonding.

3. DHU arm – Serves as site to
recognize enzymes that helps to add amino acid to acceptor arm.

4. T, C arm- Involves binding of
tRNA to ribosomes.

5. Extra arm – Only 75% of tRNA has
extra arm.

The tertiary structure is also
formed by internal bonding of hydrogen in clover leaf between T and D arms.

rRNA: large,
small rRNA combine along ribosomal proteins to form large, small subunit of ribosome.
 These complex structures,
which physically move along an mRNA molecule. Also help in binding tRNAs and
accessory molecules that are required for synthesis of proteins.

Answer 8:

After DNA strands are separated two
strands were formed one is Leading and other is called as lagging strands.
Leading strands always lead from 5′ to 3′ and lagging strand reads from 3′ to 5′. As DNA
strands are antiparallel only one continuous strand can synthesis at 3′ end of
the leading strand because of DNA polymerase property to start synthesis from
5′ to 3′. DNA polymerase is highly specific for 3′- OH terminal of new strand. DNA polymerase attacks by nucleophilic by the 3′-OH of the
nucleotide at the 3′ end of the strand on the 5′-?-phosphorus of the deoxy
nucleoside 5′-triphosphate. A primer (segment of new strand) is needed opposite to leading
strand to which nucleotide are attached. The primer should be in place before
DNA polymerase start to act. The polymerase can only add nucleotides to a
preexisting strand.

So, the lagging end is unavailable
for the DNA polymerase to interact. Lagging strand forms a short section of DNA
a result of discontinuation replication. Many RNA primers are made by primase
and bind to many sites of lagging strand and forms chunks of DNA called as
Okazaki fragments and then added to lagging strand in 5′ to 3′.



Answer 9:


Step 2: Isomerization of glucose-6-phosphate
to fructose 6- phosphate.

Enzyme: Phosphoglucomutase; ?G= +2.8 KJ.

Phosphoglucomutase belongs to Isomerases. Isomerase catalyzes the shifting of a functional group from
one carbon to other within a molecule.


Step 4: Fructose-1,6-bisphosphate is break
down to: dihydroxyacetone phosphate
(DHAP) and glyceraldehyde 3-phosphate.

Enzyme: Aldolase; ?G= +24.6

belongs to class Lyases. Aldolase catalyze an aldol cleavage reaction.


Step 5: DHAP and GAP are isomers
of are readily inter-converted.  GAP
is a substrate for the next step in glycolysis so all of the DHAP is eventually

Triose phosphate Isomerase; ?G=
+7.6 KJ.

phosphate Isomerase belongs to Isomerases
class. Interconverting of aldolases and ketoses are Involved.


Step 6: GAP is
dehydrogenated to form 1,3-bisphosphoglycerate.

Glyceraldehyde 3-phosphate dehydrogenase (GAPDH); ?G= +2.6 KJ.

3-phosphate dehydrogenase belongs to class


Step 8: Conversion of 3-phosphoglycerate to 2-phosphoglycerate. The
phosphate shifts from C3 to C2 to form 2-

Phosphoglycerate mutase; ?G=
+6.4 KJ.

mutase belongs to class Isomerases.


Answer 10:


Synonymous codons
that instruct ribosome complex to add arginine are:


Synonymous codons
for Methionine:  AUG

Synonymous codons
for termination of proteins synthesis: UAA,

codons that signal
the initiation of synthesis: AUG.



Tumor cells grow under limited oxygen supply initial
stages as they devoid of capillary supply. So, the cells depend on glycolysis
by converting glucose to pyruvate and lactate for ATP production but the ATP
produced is only 2 ATP per glucose, to compensate that tumor cells absorb more
glucose than normal cells. More lactic acid is formed there will be change in pH
to acidic. Increase in glycolysis is achieved by increased synthesis of
glycolytic enzymes and plasma membrane transporters. With high rate of
glycolysis, the tumor cells can survive anaerobic conditions.

High rate of glucose uptake used in pinpoint
location of tumors. In positron emission tomography isotope labelled glucose
analog is taken up and not metabolized by tissue. The decay of isotope yields
positrons that will be detected by a detector and help in pinpointing the
location of tumor precisely. The intensity of the positrons emitted is detected
in the pet scan is translated from green to red.