Biology Assignment 3: Exploring DNA, Mutations, and Genetic Processes

This biology assignment delves into fundamental concepts of genetics, covering DNA structure, replication, and mutation. It addresses key questions on DNA interpretation, factors influencing mutation rates, and the principles of eugenics. The assignment compares the contributions of Levene and Chargaff to understanding DNA structure and explores the differences between cellular DNA replication and PCR. Furthermore, it analyzes the genetic code, including translation and the use of tRNA molecules. The assignment also explores the effects of different mutations on a gene's protein and differentiates between genetic heterogeneity and allelic disorders. Overall, the assignment covers a wide array of topics providing a detailed understanding of DNA and its related processes.

Running head: QUESTIONS 0
GENETIC
SEPTEMBER 28, 2019
STUDENT DETAILS

QUESTIONS 1
Assignment 3:
1. DNA-
a) Is always interpreted using the same code with minor exception
(This is correct option because it is always interpreted with the help of similar genetic code)
(Taylo, Bright & Buckleton, 2013).
b) Incorporates any type of the five nitrogenous bases
(It is incorrect because they are not five but have only four nitrogenous bases)
c) Bases occur in random order
(it is incorrect because bases always occur in systematic result. If it will occur in random
order, it will result into mutation)
d) Joins bases to each other covalently
(It is incorrect because only nucleotides are joined by the covalent bond)
e) Is not the genetic material in bacteria
(It is same that genetic material is found in bacteria)
2. Which of these factor do not influence rate of mutation-
a. The presence of repetitive sequence
(It is incorrect because presence of repetitive sequence repeats the influence rate of mutation)
b. The protein encoded by gene
(incorrect because it also influence the mutation rate)
c. The presence of palindrome
(it is correct because it is the region of DNA where nucleotide sequence is identical with the
reversed sequence of the stand, therefore it cannot influence the mutation rate)
d. A gene’s length =
(it is incorrect because as large the gene sequence is, higher will be the mutation probability)
(Smith, 2008).
e. The ability to repair DNA
(this greatly influence mutation by the ability or inability of the DNA repair mechanism)

QUESTIONS 2
3. Eugenics-
a. Depends on the voluntary participation
(incorrect because most of the times it is carried out to the unawareness of individual.
therefore, government can also decide for removing the bad genes in the society).
Government is therefore said to be the significant person in eliminating the bad genes.
b. Is successful at eliminating recessive alleles from the population
(it is correct because it can control the inheritance of the harmful genes that is passed by the
parents to their descendants)
c. Is a form of natural selection
(incorrect because it cannot clearly stated that it is a form of the natural selection)
d. Is intend to alleviate individual suffering
(incorrect because almost all the genes presented in a person is not harmful)
e. May involve sterilization
(It can also be correct because it may involve the sterilization rendering the people unable to
pass their genes to the next generation)
4. Analysis of which source of data does not contribute to tracing human and
ancestral human migration pattern?
a. Mitochondrial DNA:
(it can be used to search or trace the mitochondrial DNA from the mothers. In the recent time,
it can be used to regulate the ideal mother of the child. Moreover, it is also seen that it can be
passed from mother to the offspring.
b. Admixture info:
(It is not correct because once people intermarry with the foreigners tends to introduce genes
in the genetic pool. In such a situation, it also becomes quite difficult to trace the encestry.
c. Chromosome banding:
(it can also be used because same ancestry will have same chromosomal banding if it is not
same exactly)
d. Haplogroups :

QUESTIONS 3
(human from the similar ancestry will be into the group of similar halogroup by the
phylogenecists.
Question 5: Compare the contributions of Levene and Chargaff in the determination of the
structure of DNA. Why wasn’t their information enough to determine the correct structure?
Answer: In the initial career, neither Leven nor any scientist were known to the component of
DNA. Afterwards, several scientist came up with the suggestions (Thierry, El Messaoudi,
Gahan, Anker & Stroun, 2016). From the work of biochemist Levene as well as others,
scientist has known hat DNA is composed of some subunits that are known as nucleotide.
Nucleotide is made up of the phosphate group, sugar (deoxyribose), one of the four
nitrogenous bases named thymine, adenine, cytosine and guanine. In addition to this, Erwin
Chargaff, an Austrain biochemist stated that the several species in determining their
composition not found in the similar quantities. Their bases amount differ among the species
but not between the individual of those species. However, amount of C has always equal the
amount of G, and amount of A always equal to the amount of T. these findings therefore
turned to be significant to the Watson and Crick model of the DNA (Marini, Limongi,
Moretti,Tirinato & Di Fabrizio, 2017).
In this, Chargaff has proved to be one such scientist who expanded on the work of Levene by
coming up with some of the extra benefits of the DNA structure (Copes & Kimbel, 2016).
However, as compare to Levene, Chargaff rule was that DNA is double stranded helix having
double strands connected by the hydrogen bond.
DNA structure that is presented in the model of Watson and Crick is antiparallel, double-
stranded and right handed helix. “Super phosphatre” backbone of the DNA tends to make up
the helix outside. As compare to this, nitrogenous bases is found on the inside which in turn
form the hydrogen pairs. This helps in holding the DNA together (Osborn-Gustavson,
McMahon, Josserand & Spamer, 2018). Therefore, the two chemists is not able to clearly
showcase the exact structure.
Question 6

QUESTIONS 4
1. How many double helices will contain only higher-density
nitrogen?
Answer: only one double helices contain the higher density nitrogen. Double helices
describes the presence of double stranded DNA that is composed of the two linear strands. It
runs opposite to each other and the twist together. Each of the DNA strand is long that is
made up of the smaller units.
2. How many double helices will have intermediate density?
In double helices, two will have the intermediate density.
3. How many double helices will contain only lighter-density nitrogen
Only one double helices will contain the lighter-density nitrogen.
Question 7: Compare cellular DNA replication to PCR replication.
Answer- The major difference between DNA replication and PCR replication are :
 Machinery involved: DNA replication is brought at the body temperature that is 37C
in the humans. It is carried out by taking the assistance of complex machinery. As compare to
this, PCR utilizes the temperature cycle with the 70 to 90 C in cause the DNA strands and
denaturation.
 Polymerase type: In eukaryotes there are several DNA polymerases. As compare to
this, PCR DNA polymerases derived from the archaea and bacteria.
 Feature of the polymerase used: speed, high fidelity, repair, and proofreading are
some of the desirable feature that is required for the DNA replication. PCR reaction tends to
use the simple polymerases that do not seen as the feature rich.
 DNA synthesis depends on the well defined but complex set of co-factors and
enzymes. As compare to this, PCR facilitates in the vitro DNA synthesis by using the set of
defined reaction and ingredients conditions involving the higher temperature.
Question 8: The following sequence is part of the DNA coding strand of a gene. 5’…
ATGCGTTCAGCTACTTTAGAGCGAATCC… 3’
Give the sequence that will be produced in replication.
Answer: 5’…AUGCGUUCAGCUACUUUAGAGCGAAUCC…3’

QUESTIONS 5
What sequence will be produced as a result of transcription?
Answer: 5’…AUGCUUCAGCUACUUUAG…3’
c. Translate the sequence in part b, using all three possible reading frames. Do not translate
the partial codons.
Answer: Metionine, Leucine, Glycine, Leucine, and Terminator
d. What feature of the genetic code other than the presence of control sequences is revealed
by these sequences? Identify a specific example.
Answer: Non overlapping genetic code. Specific example is AUG, CUU, CAG.
Question 9: There are 49 nuclear genes that code for tRNA molecules. There are 61 codons for
amino acids. The 49 different tRNA molecules can translate all 61 amino acid codons. Answer the
following theoretical questions and explain your answers. Include the amino acids encoded by
each of the codons noted.
a. Could a single type of tRNA molecule be used in the translation of CAC and CAG?
Answer: No, a single TRNA cannot be used to translate the single code for several amino
acids.
b. Could a single type of tRNA molecule be used to translate AUU and AUA?
Answer: Both are the code for Isoleucone. Therefore, it can be easily translated by the similar
tRNA.
c.
d. Could a single type of tRNA molecule be used to translate UCU and AGC?
Answer: Both these are the code for the similar amino acid. Due to this, a tRNA single can be
easily used to translate them (Yoo, Kim, Aksimentiev & Ha, 2016).
10. Do these descriptions support or refute the statement that a region of DNA can code for more
than one polypeptide? Explain.
Answer: Exons can be joined in different combinations.

QUESTIONS 6
It is correct that they can be combined at the time of rearrangement of the pre-mRNA before
ensuring the transcription (DeLouize, Coolidge & Wynn, 2017). Exons tend to form the
codion area of the DNA.
a. An intron can become an exon.
Yes, introns are the non-protein coding genes that helps in coding helpful molecules that
are significant in transcription. Due to this, it can become an exon by mutation.
b. Both strands of the DNA can be used as the coding strand for different gene
Yes, it is absolutely correct because they are upon rearrangement and complementary
into codon varied genes result.
Question 11: compare and contrast the genetic heterogeneity and allelic disorders.
Answer: genetic heterogeneity is said to be situation where single genetic disorder as well as
phenotype can be caused by any one of the varied number of alleles and non-allele mutation.
Therefore, genetic heterogeneity is said to be the disease that comes from the multiple
abnormalities of genes. As compare to this, allelic disorder is the situation where varied
mutation at the similar locus tends to cause the diverse phenotypic variations (Fragkos,
Ganier, Coulombe & Méchali, 2015). These kind of allelic variation bring due to the natural
selection process as a result of the genetic drift, exogenous mutagens and genetic migration.
A number of genetic disorders are being caused whenever an individual tends to inherits the
two recessive alleles.
Question 12: Describe the possible effects of each of these mutations on a gene’s protein. Which
of these mutations in a gene would likely have the most severe effect on its protein?
a. missense: mutation is the change in one base pair of DNA that further results in substitute
of one amino acid to the other in protein that is made gene. Missense mutation is said to be
the point mutation where single nucleotide change as a result codon codes for the different
amino acid. Not every missense mutation lead to the appreciable changes in the protein.
b. point mutation in an intron
If point mutation tends to occur in the intron of the coding gene, it do not bring any change in
the coded message. However, there might be the possibility of the RNA splicing. As a result,

QUESTIONS 7
it can affect the resulting protein. Point mutation can also affect the other regions of the non-
coding such as promoter sequence. It further results in the failure of protein transcription.
Besides this, it can also affect the regulatory sequence that results in the decreased gene
product and translation. Therefore, these kinds of mutation are very common that are known
as single nucleotide polymorphism (Mazur, 2016).
c. a deletion of three bases
It is known as the deletion of the entire gene as three bases core for one particular gene.
c. a translocation whose break point is within the gene.
It occurs when the segment called chromosomal is shifted from one position to the other
within the similar chromosome or other chromosome (Alazami et al, 2016).
d. a nonsense mutation at the end of the gene
nonsense mutation is also said to be the alteration in one of the DNA base pair. Inaltered
DNA is failed to signal the cell to discontinue building the protein. This kind of mutation
results into the big protein that might function inappropriate.
Question 13: A population of 5545 people is assumed to be in Hardy-Weinberg equilibrium. Of this
population, 3607 have the dominant trait of lactase persistence. The recessive form of this trait is
lactose intolerance. Showing all calculations, determine
a. the frequency of the homozygous recessive genotype in the population.
Answer: 1938/5545=0.35-p
b. the frequency of the recessive allele.
Answer: 3607/5545=0.65=q
c. the frequency of the dominant allele.
Answer: P2= 0.1225
d. the number of homozygous dominant people.
answer: P2=0.1225
e. the number of heterozygous people.
answer: 0.455
Question 14: A rural population in Quebec has a significantly higher frequency of the
phenylketonuria allele compared to the urban population of Montreal. There are multiple
explanations for this phenomenon. Propose four hypotheses that relate to Hardy-Weinberg
equilibrium, and specifically describe how they could explain the difference between the rural and

QUESTIONS 8
urban populations. What type of information or research would be used to determine which
explanation/s is/are most likely?
Answer: 1. Mental retardation in the PKU occur due to the inability of the OKY fetus as well
as the relative inability of the heterozygous method in synthesizing the tyrosine from the
phenylalanine. Further, it can also be counteracted by the extra tyrosine as compare to the
less phenylalanine.
2. Reserve of the hydroxylation of tryptophan and tyrosine in bring by high phenylalanine
can be undone by the high doses of the tryptophan and tyrosine.
3. Properly treated PKY patient have to face high number of school problem that do not
attribute to the abnormal IQ (Anton, 2018).
4. LNAA that is tyrosine, phenylalanine, methionine, tryptophan, isoleucine, histidine and
valine hasve some common system of carrier across the brain. This tends to overloading the
system including one of the LNAA.
Question 15: Choose any two species from the list of Key Terms in unit 13 lesson 1. Create a table
or other format to compare their times of existence, geographic ranges, physical features,
habitats, and socio-cultural features.
Homo habilis Homo erectus
Geographical areas These found only in the east
Africa
These found across Africa,
Europe, and near and far
from east.
Time for which these
species exist
About 2.3 to the 1.44 million
years ago
These species founded
around 1.9 to 143,000
million years ago
Physical feature  These are robust,
short and their height
is not more than the 4
feet and 3 inch.
 Shorter legs and
longer arms is their
characteristics
 They are taller having
average height of 5
feet and 10 inch.
 They have flatter face
with less cheekbones.
Besides this, they
have large brow-

QUESTIONS 9
 They have prominent
cheekbones with
protruding face
ridges.
 They have slender
legs and arms
Habitats These species caves during
the day. They trees during
the night.
They lived in the caves at the
day time. in night, they caves
on trees.
Socio cultural feature Used primitive oldowan tool
for the use of scavenging as
compare to hunting.
Used the advanced tools for
the defense as well as
hunting purpose.
Man who discovered them Marry and Louis Leakey in
the year 1960
Eugene Dubois in the year
1891
Poster at the time of
walking
Stopped over Upright posture
Staple diet They are large sized
predators
Meat including fruits, nuts
and berries
Teeth Bigger as compare to those
of the modern humans
They are smaller than homo
habilis

QUESTIONS 10
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QUESTIONS 11
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