1. 20% and yz = 10%. The distance between

the only one that is recessive and double crossed.


map – the distance between A and C would be the equal.

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map – less than the sum of recombination frequencies due to multiple crossovers.


it would be 50% parental and 50% new combination.


spite of the distance between the genes, they will assort a minimum of 50% of
the time due to multiple crossovers. Even numbers bring the genes together,
while odd numbers bring them apart.


½ markers – the best loci placement is at every 10 million base pairs.


 xy = 20% 
and yz = 10%. The distance between x and z is recombination percentage.


genotypes with the lowest recombination % are the ones who went through double
crossovers. Reverse the double crossover and you end up with gene A in the
middle. The phenotype represents the parental.




a.     Proved that one gene =
one enzyme through the use of four different strains of arginine-dependent
Neurospora. All of the Neurospora had lost the ability to use a unique gene
that an enzyme needed to produce arginine.


b.     Experiment with Strep. pneumoniae  that used rough strains that
were non-pathogenic and smooth strains that were pathogenic. He injected a
mouse with living smooth strains in which the mouse died. Another mouse was
injected with rough strains and the mouse was still healthy. He also injected a
mouse with a heat killed smooth strain, in which the mouse lived. Finally, he
did a new combination with a mix of heat-killed smooth strain and rough strain in
which the mouse died. This proved that DNA was a genetic material and that the
rough strain had been “transformed” into a virulent strain by the smooth

Proved that DNA carried
genetic specificity and that it was the reason behind the transformation of
non-virulent strains into virulent ones by heat-killed strains. DNA was
isolated and purified from the smooth strains and were added to the rough cells
to transform them. When they added an enzyme, deoxyribonuclease, the
transformation was interrupted. They disproved that RNA carried genetic
specificity by adding ribonuclease or a protease, but the transformation kept

They wanted to prove if nucleic acid was a carrier of viral genes.
They studied a T2 bacteriophage that has a DNA core and a capsid. Phage
proteins were labeled 35S and the DNA was labeled 32P. These labels were used
to follow where the protein and the nucleic acid would end up in the progeny
phage. They found that the 32P labeled nucleic acid was still prominent in the
progeny and that a very small amount of the 35S labeled protein had actually
entered the bacteria. This led them to prove that all viruses contain nucleic

Found that there was an exact ratio of adenine, thymine, cytosine
and guanine that varied through species and that the bases had a complementary
relationship with each other. He calculated the nucleotide percentages in DNA
through paper chromatography and UV spectrophotometry.  ?

Used Atsbury’s X-ray diffraction of DNA to prove the helical, double
stranded structure of DNA and she noted that the strands had a major and minor
groove along the helical axis.

Watson anUsed Franklin’s X-ray diffraction and used Chargaff’s
rule to conclude that DNA chains are helical and complementary. This showed
that DNA is a template for its own replication and eventually became a part of
the Central Dogma.

Worked with extracts of bacteria to study DNA synthesis. 32P—cytosine
was added to unlabeled nucleotides and he used a 3′ phosphodiesterase to locate
the 32P-cytosine at a 3′ end. With this experiment he was able to isolate DNA
polymerase I and that it links 3′ hydroxy to a 5′ phosphate, therefore growing
in the 5′ à3′ direction. This
proved that DNA strands were anti-parallel.

A density gradient centrifugation was used to separate DNA into a heavy
15N15N DNA, a light 14N14N DNA, and a hybrid 14N15N DNA. The heavy DNA made a
band high in density near the bottom of the density gradient, the light DNA
stayed near the top of the gradient where the density equaled the density of the
salt solution, and the hybrid chain was intermediate. The 15N heavy was grown
in the 14N and so on. The results showed that after three generations, all of
the bands were different, and it was concluded that DNA replication is

Concluded that sickle cells had an amino acid change in B-globin
by doing a protein fingerprint on a wild type of hemoglobin and a sickle celled
hemoglobin. The hemoglobin was cleaved, and it ran in two directions. The electric
field caused it to move from the right to the left and the cells were chromatographically

Showed that DNA served as a template for its own replication. He
showed arrows from DNA to RNA and stated that this was called “transcription”
and after that RNA would “translate” the information into proteins.

Reticulocytes were used to measure protein synthesis. As the
protein synthesis started, a radiolabeled leucine was added, and the synthesis
was stopped before a new globin chain was synthesized. The proteins were then
separated through chromatography and then trypsin was used to make peptide
fragments. Then the globin chains that had completed synthesis were separated
from incomplete chains using gel electrophoresis. A plot that showed the length
of each peptide along with the label was made to show the position of the peptide
in the completed chain. The peptides on the B-globin regions had a small amount
of radiolabel and would be the N terminus and the end with the highest was near
the C terminus.

10.  It was believed that RNA did not
have the ability to differentiate between side groups of amino acids and
solubility so there had to be “something” that translated it.


11.  Since rRNA is about the same in
every species, it cannot explain the genetic variations between the species.


12.  5′ AUGCCAUUCGAA 3′ – from the 3′ – 5′
template strand.


13.  Each amino acid has multiple but
specific codons and each of those codons codes for only one amino acid.


14.  Inosine helps with t-RNA stability
at the wobble position and allows to read the codons as an A, U or C in this
position. This enhances variation and also follows the degenerate but not ambiguous


15.  Purification removes any
non-functional material from the polypeptide. If they’re not removed, the
results may be unclear.


16.  Weak bonds are further apart while
strong bonds are closer together. Strong bonds are hard to break but they release
a high amount of energy when they are broken whereas weak bonds are reversible
but release a smaller amount of energy when they’re broken.


17.  Peptide bonds are rigid and planar
because of resonance interactions.


18.  Strong bonds are harder to break
due to activation energy, and since all reactions must move towards higher
entropy, strong bonds need to be coupled with an energy releasing source that
allows them to reach their activation energy without violating the laws of


30. At Keq = 100, the reaction will
favor the products if G is negative, and it will favor the reactants if G is positive.


31. Free energy is noted as G and it is the energy available for a reaction to
occur. When G is negative, the reaction is considered spontaneous and
it will favor the products. When G is positive, it is considered less organized and not
spontaneous and it will favor the reactants unless it is positive at a high


32. At a high temperature, a positive G will be spontaneous but it will not be spontaneous
at a low temperature.


33. Although the given G value is negative, H2 and O2 will not react
spontaneously to form water because energy needs to be added to the system for
it to overcome activation energy, become less orderly, and for the molecules to
collide. This is due to the second law of thermodynamics, which states that reactions
move towards a more disordered system.


Weak bonds can be broken through a small amount of energy because kinetic
energy is dispersed.




             a) The smaller the van der Waals
bonds radius is, the stronger the van der Waal interaction is.


             b) Hydrogen bonds are strongest
when they are linearly aligned with other molecules. Any bond angle greater
than 30° will weaken the hydrogen bond.


             c) Ionic bonds occur between
partially charged molecules. If an ionically bound group is submerged in water,
the ionic forces would weaken.


             d) Hydrophobic bonds strengthen the
primary structure of proteins and they promote coiling and folding in alpha
helices and beta sheets.



             e) Nonpolar molecules will cluster if
submerged in water due to the van der Waals forces and hydrophobic groups.


             f) Polar molecules are soluble in
water and may participate in hydrogen bonds and interact with the lattice
structure of water.



In water, the nonpolar molecules are kept together by the hydrophobic effect. The
hydrogen atoms will break apart from the water molecules, releasing energy into
the system.  The water molecules then form
more hydrogen bonds and encapsulate the nonpolar molecule, therefore decreasing
the entropy in the system.


Hydrophobic proteins would be within the protein because the hydrophobic
interactions would help with the stability and further folding and coiling of
the protein.



When two molecules fit each other according to their shape, they are able to
have multiple interactions and therefore, stronger binding energies.


They mediate reactions between enzymes and substrates because of their ability
to break and form again under physiological conditions. Weak bonds also give
macromolecules their shape.


The formation of peptide bonds is achieved through the breakdown of ATP to AMP
and pyrophosphate, which yields a net of – G. Activation energy keeps proteins from easily coming


Weak bonds are thermodynamically unstable and can give off more energy than a
stronger bond.


G is the actual energy change in the cell, while G° is measured under standard-state conditions. G°’ is measured under standard conditions and also where
the pH is at 7. Biochemists prefer G°’ because it closely resembles cellular conditions.


The overall pathway has a -G because citrate synthase keeps oxaloacetate concentrations


Covalent bonds are strong and cannot be broken or rearranged because the
activation energy is too large to overcome without the help of enzymes.


Enzymes lower activation energies and speed up the rate of reactions. They do
not affect equilibrium.


The phosphoanhydride bonds are negatively charged and constantly fighting over
resonance. Hydrolysis is favored because ADP is higher in -G than ATP is.


Molecules exchange functional groups through group transfer and this allows
high energy bonds to be transferred between molecules.


ATP activates amino acids by using group transfer to give amino acids an AMP,
which gives the overall pathway of the reaction more entropy.


ATP uses group transfer to activate nucleotides and then pyrophosphate is
released and then split to make the overall reaction – G.


ATP hydrolysis is used to help proteins make the conformational changes needed
to carry out their specific functions.