Biology 341: Genetics Assignment on DNA Structure and Mutations

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This genetics assignment delves into various aspects of human genetics, starting with the fundamental structure of DNA, including its replication and the impact of mutations. It explores the contributions of scientists like Levene and Chargaff in understanding DNA, and compares cellular DNA replication with PCR. The assignment analyzes the genetic code, including translation and the role of tRNA molecules. It also addresses concepts like genetic heterogeneity, allelic disorders, and the effects of different mutations on proteins. Furthermore, the assignment incorporates Hardy-Weinberg equilibrium calculations and examines the reasons for the higher frequency of the phenylketonuria allele in a rural population. Finally, it compares the characteristics of Homo sapiens and Homo habilis, providing a comprehensive overview of key genetic concepts and evolutionary principles.
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Running head: ASSIGNMENT GENETICS
Assignment Genetics
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ASSIGNMENT GENETICS
Genetics assignment
1. DNA
a. Is always interpreted using the same code, with minor exceptions - DNA codes vary from one
species to another and hence this is a false assumption
b. Incorporates any type of the five nitrogenous bases- DNA is made up of four bases: guanine,
adenine, thymine and cytosine
c. Bases occur in random order – bases do not occur in random order and are usually arranged in
an order that is species specific.
d. Joins bases to each other covalently- this is a true statement as nucleotide bases are joined
together covalently.
e. Is not the genetic material in bacteria- this is a false statement as bacterial genetic material is
double-stranded DNA helix. (5 marks)
2. Which of these factors does not influence the rate of mutation?
a. The presence of a repetitive sequence – repetition on a DNA strand makes it more prone to
mutations
b. The protein encoded by the gene – a protein gene does not determine if the DNA sequence
will be more prone to mutations or not.
c. The presence of a palindrome – will increase the DNA's susceptibility to mutation
d. A gene's length – the longer the gene the more the chance of mutation
e. The ability to repair DNA – repair of genetic codes makes this region of DNA vulnerable to
replacement by other bases hence more likely to be mutated. (5 marks)
3. Eugenics
a. Depends on voluntary participation – it is ethical that each individual is given an equal chance
to determine what they would want for the future.
b. Is successful at eliminating harmful recessive alleles from a population – recessive alleles
have not been eliminated from the population even when each member practices eugenics.
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c. Is a form of natural selection – Natural selection is not deliberate and affects those intended
and those not intended, as eugenics is deliberate this statement is false.
d. Is intended to alleviate individual suffering - eugenics does not remove the possibility of
someone suffering as mutations may occur on healthy genetic codes.
e. May involve sterilization – this is unethical and would not be allowed to go through as it takes
away the right to choose if you want to be a parent or not. (5 marks)
4. Analysis of which source of data does not contribute to tracing human and/or ancestral human
migration patterns?
a. Fossils – give detailed information that can reveal migration patterns.
b. Mitochondrial DNA – since it is passed on unaffected to offspring it can be used to trace
ancestry
c. Admixture info – these mixtures could be used to study ancestral migration of individuals
d. Chromosome banding – chromosome banding cannot be used to study ancestral migration
patterns
e. Haplogroups – by looking at the different haplogroups of an individual it is possible to infer
ancestry on either side, paternal haploid or maternal haploid. (5 marks)
Biology 341: Human Genetics (Rev. C14) 2
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? (4 marks)
Levene's data gave clear information on the chemical makeup of the DNA and how the bonding
of the bases occurred. Chargaff was able to give information such as the variation of DNA
sequences from one species to another, the bonding of the different building blocks and how
variation occurred within the DNA sequence. This information was however not useful in
determining the structure of DNA as the bonding of the bases could affect the outcome of the
structure in different forms. They had not studied its physical characteristics hence did not study
DNA structure.
6. Repeat the exercise described in Unit 9 Lesson 3 Study Question 2 for a third replication (do
not submit this drawing). How many double helices will contain only higher-density nitrogen?
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How many double helices will have intermediate density? How many double helices will contain
only lighter-density nitrogen? (3 marks)
In reality, the experiment by Meselson-Stahl started with DNA which had only higher-density
nitrogen. Cells were provided with normal lighter-density nitrogen as they proceeded through the
cell cycle(s). The parental double helix has only higher-density nitrogen, thus it has a heavy
density weight. The newly synthesized strands contain normal (lighter-density) nitrogen, causing
the double helix to have an intermediate density. Meselson and Stahl allowed other cells to
continue growth and replication in the lighter-density nitrogen media for another (second) cell
cycle. Draw this next generation of daughter DNA molecules using one color to indicate the
parental strands of DNA and another color to indicate newly synthesized strands with lighter-
weight nitrogen. Your drawing should consist of four double helices. How many parental strands
are present in total? How many strands contain lighter-weight nitrogen? How many double
helices will have a light density weight?
7. Compare cellular DNA replication to PCR replication.
DNA replication involves three key steps initiation, elongation, and termination. DNA
replication on a cellular level happens even at low temperatures while in PCR the temperature is
usually raised to 70-90C for denaturation and annealing of strands. In cellular conditions,
different polymerases are used according to species while PCR is depended on Taq polymerase
obtained from bacteria. Under cellular conditions, whole genome replication is done while PCR
replicates portions of the DNA. High fidelity and proper proofreading happen on a cellular level
rather than in PCR. (6 marks)
8. The following sequence is a DNA coding strand representing a portion of a gene. (9 marks)
5'…ATGCGTTCAGCTACTTTAGAGCGAATCC… 3'
a. Give the sequence that will be produced in replication.
3'…TACGCAAGTCGATGAAATCTCGCTTAGG…5'
b. What sequence will be produced as a result of transcription?
3'…UACGCAAGUCGAUGAAAUCUCGCUUAGG…5'
c. Translate the sequence in part b, using three reading frames. Do not translate the partial
codons.
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Reading frame one: UAC GCA AGU CGA UGA AAU CUC GCU UAG G
Reading frame two: U ACG CAA GUC GAU GAA AUC UCG CUU AGG
Reading frame three: UA CGC AAG UCG AUG AAA UCU CGC UUA GG
d. What feature of the genetic code is revealed by these sequences? Identify a specific example.
The first reading frame consists of the stop codon of transcription meaning the sequence was
taken from the end of a gene sequence.
UAC GCA AGU CGA UGA AAU CUC GCU UAG G - UGA is the stop codon
In genetic sequences, codons consisting of three bases are translated into amino acid.
U ACG CAA GUC GAU GAA AUC UCG CUU AGG
Thr, Gln, Val ,Asp, Glu, Ile, Ser, Leu, Arg
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 names of the amino acids
for the codons noted. (6 marks)
a. Could a single type of tRNA molecule be used in the translation of CAC and CAG?
No tRNA are amino acid specific and since these two codes code for different molecules, they
cannot be translated by the same tRNA
b. Could a single type of tRNA molecule be used to translate AUU and AUA?
Yes, this codes both stand for isoleucine
c. Could a single type of tRNA molecule be used to translate UCU and AGC?
Yes this codes both codes for serine
10. Do these descriptions support or refute the statement that a region of DNA can code for
more than one polypeptide? Explain. (6 marks)
a. Exons can be joined in different combinations - this supports that a region of DNA is
multifunctional and can code for more than one polypeptide.
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b. An intron can become an exon – the ability of an intron to become an exon means that thee
non-coding region for one gene is a coding region for another gene.
c. Both strands of the DNA can be used as different genes – these statements refute the ability of
DNA code to code for more than one gene.
11. Compare and contrast genetic heterogeneity and allelic disorders. (4 marks)
Genetic heterogeneity explains how a phenotypic characteristic could be the result of a number
of mutations of an allele or non-allele while allelic disorders are a resultant consequence of allele
heterogeneity that may cause a mutation.
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? Explain.
a. Missense – may result in a change of amino acid produced changing the whole protein. This
kind of mutation is adverse as a non-functional protein may be produced.
b. The point mutation in an intron – an intron for one gene is an exon for another hence the
change in an intron would affect one protein and have no effect on another its effect is not
adverse c. Deletion of three bases-mutations caused when three bases are deleted means that one
or more amino acids in a protein are affected. This may change the folding and function of the
protein. Its results are adverse changes
d. A translocation – this means there is a new base on a different spot and the lack of a base on
the old spot. This mutation may be adverse as there is a range of codons that can be affected
including the region of a start and stop codon
e. A nonsense mutation at the end of the gene these may change the stop codon for the gene that
may result in a new protein. (12 marks)
Biology 341: Human Genetics (Rev. C14) 3
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 (10 marks)
a. The frequency of the homozygous recessive genotype in the population.
5545-3607=1938
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1938/5545
0.3495
Q2=0.1221
b. The frequency of the recessive allele.
0.1221+0.4231=05452
1-0.5452
0.4548
c. The frequency of the dominant allele.
1=P+Q
1-q=p
p=1-0.3495
p=0.6505
p2=0.4231
d. The number of homozygous dominant people.
0.4231×100=42.31
42.31/100×5545
2346
e. The number of heterozygous people.
2346+1938=4284
5545-4284=1261
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, and specifically describe how they could explain the
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difference between the rural and urban populations. What type of information or research would
be used to determine which explanation/s is/are most likely? (10 marks)
1. The Founder effect- the population of Quebec is made up of a high population of
phenylketonuria allele who make up the offspring of the founders.
2. Insular breeding. - There is inbreeding between the people of Quebec and these have
contributed to the increase in numbers of the phenylketonuria allele
3. Immigration for treatment in the world-famous center for PKU treatment hence a high number
of people with the condition there.
4. Common genetic ancestry of the population- the people share common ancestors who in-bred
and most of the population resulted in a homologous dominant allele that produced more
offspring of a homologous dominant allele
There is a need for tracing the world largest phenylketonuria allele facility to be able to help with
data collection and analysis.
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 sapiens Homo habilis
Around the world sub-Saharan Africa
200,000 years ago – present 2.1 million to 1.5 million years ago.
Homo sapiens skull has a short base but a
high braincase, its back is rounded while its
face is reasonably smaller and has a
projecting nose bone. Their bodies are short
an adaptation to the tropical environment.
They have slender trunks and long limbs but
The Homo habilis species was taller than
its predecessor an average of 100cm ~
135cm long. They had a large brain but a
smaller face and teeth. Their arms were
much longer than homo sapiens
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the legs are relatively longer that arms. They
have a vertical looking face as a result of a
short jaw.
Live in different habitats as a result of
adaptation Inhabited forests as well as caves
Their culture is more advanced than that of
the previous species that it preceded it.
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References
Arthur P. Mange, (2010). Elaine Johansen Mange. Genetics: Human Aspects. Minnesota.
Sinauer Associates, 1990.
Lewis, R. (2009). Lewis, Human Genetics: Concepts and Applications © 2010 9e, Student
Edition (Reinforced Binding). McGraw-Hill Education.
Ricki, L. (2006). Human Genetics: Concepts and Applications. 10th Edition. Dubuque, IA.
William Brown.
Robert H. Tamarin, Ron W. Leavitt.(2010). Principles of genetics. Madison. Wm. C. Brown
Publishers, 1991.
Ricki, L. (2017). Human Genetics: The Basics. 10th Edition. New York. Routledge.
Lewis, R. (2008). Human genetics: concepts and applications. Wurtsburg,Germany. McGraw-
Hill/Higher Education, [VitalSource]. Retrieved from
https://online.vitalsource.com/#/books/0077658892/
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