Answer determining factor which helps in bringing hair


Answer 1

Four 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.

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To 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.

Not 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:

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

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.

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

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

Step 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.

Gibbs free energy If ?G < 0, this reaction will proceed in the forward direction, as written. If ?G > 0, however, the reaction will proceed
in the reverse direction and B will be converted to A.
This aids to bring the equilibrium point of the reaction,
thus for every positive change there must be a preceding equal
negative change. For example 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.


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.




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.

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

Main 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:

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,

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.

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

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:

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.

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′.



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 KJ.

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 depleted.

Triose phosphate Isomerase; ?G= +7.6 KJ.

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


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.

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,

Synonymous codons that
signal the initiation of synthesis: AUG.



Bonus Question:

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.








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