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1. a) Why do couples who can taste PTC have children who cannot?
PTC- is phenylthiocarbamide. The ability to taste PTC is a dominant character. Humans are
diploid, so they have 2 copies of same gene (in this case to taste PTC), also referred as
alleles. PT- dominant allele for PTC and pt- recessive allele. So the parents who have ability
to taste PTC have PT pt (mother) and PT pt (father). Such parents can have children with
different combination of alleles i.e. PT PT, PT pt, pt PT and pt pt. The children with PT PT,
PT pt and pt PT combination of alleles will have ability to taste PTC. While children with pt
pt alleles can’t taste PTC. Therefore, couples who can taste PTC have children who cannot
taste PTC.
b) What is the significance of the fact that couples who cannot taste PTC never have children
who can?
As mentioned above tasting PTC is dominant character so the parents that can’t taste PTC
will have pt pt (mother) and pt pt (father) alleles. They will have children that have pt pt
alleles, hence they can’t taste PTC.
Question 2:
One strand of a section of DNA isolated from E. coli reads:
5’-GTAGCCTACCCATAGG-3’
a)Suppose that an mRNA was transcribed using the complement of this DNA strand as the
template. What would the sequence of the mRNA in this region be?
The given DNA sequence is 5’-GTAGCCTACCCATAGG-3’. The complementary DNA
sequence from which mRNA is made is 3’ CATCGGATGGGTATCC 5’. Therefore, mRNA
made from above complementary sequence will be 5’-GUAGCCUACCCAUAGG-3’.
b) How many different peptides could potentially be made from this sequence of RNA? Would
the same peptides be made if the other strand of the DNA served as the template for
transcription?
Theoretically speaking, 3 peptides can be made from the above sequence. mRNA can be read
from G or U or A so that in each case the frame of codons is different
Case 1: mRNA can be read from G- the codons are GUA GCC UAC CCA UAG G (VAYP)
Case 2- mRNA can be read from U- the codons are G UAG CCU ACC CAU AGG(PTHR)
Case 3- mRNA can be read from A- the codons are GU AGC CUA CCC AUA GG (SLPI)
If the other strand is used for mRNA synthesis, the peptides formed will be different.
c)What peptide would be made if translation started exactly at the 5’ end of the mRNA in part
A? When rRNAAla
leaves the ribosome, what tRNA will be bound next? When the amino
group of alanine forms a peptide bond, what bonds, if any, are broken, and what happens to
tRNAAla
?
VAYP is formed if translation started exactly at the 5’ end of the mRNA in part A. When
rRNAAla
leaves the ribosome, tRNAtyr
will be bound next. When the amino group of alanine
forms a peptide bond, what bonds, tRNAAla
is released from Acceptor site. The dissociation
of tRNAAla
results in breaking of hydrogen bonds between tRNA Ala
and mRNA Ala
.

2. Question 3:
You have a mixture of 4 proteins with the following characteristics:
Protein a: pI = 6.2, Molecular Weight 43,000
Protein b: pI = 7.1, Molecular Weight 61,000
Protein c, pI = 5.5, Molecular Weight 60,000
Protein d, pI = 5.1, Molecular Weight 79,000
At pH = 8.0 you chromatograph them using an anion exchange column, with a gradient of
NaCl to elute the proteins: what will be the order of elution?
a, c, b then d
b, a, c, then d
d, b, c then a
d, c, a, then b
The order of elution of proteins using increasing gradient of Na Cl from anion exchange will
be b, a, c, then d. Since, it is an anion exchange column hence the elution of the proteins will
depend on the net negative charge on the respective proteins and independent of molecular
weight of the proteins. At a pH below pI (isoelectric point), the proteins are positively
charged; at pH=pI the net charge on the protein is zero and at pH > pI the proteins are
negatively charged. So, the proteins will carry more negative charge in the order d>c>a > b.
Therefore, d will be binding most strongly with the anion exchange column followed by c,a,
and b. Hence, we need more amount of NaCl to remove d from the column in comparison to
other 3 proteins. Therefore, d elutes last.
Conclusion: The order is elution of proteins is b, a, c, then d.