Genetics Assignment: Human Twin Studies and Variation

This document presents a comprehensive solution to a Human Biology assignment focusing on genetics and variation. The assignment covers a range of topics, including discontinuous and continuous variation, exemplified by ABO blood groups and height, respectively. It explores the genetic control, environmental influences, and phenotypic ranges of these variations. The solution analyzes twin studies, examining concordance rates in monozygotic and dizygotic twins to assess the influence of genetic factors on height. It delves into the concepts of mutation and meiosis, explaining how these processes contribute to genetic variation. The assignment includes an information leaflet evaluating the benefits and dilemmas of human twin studies. Furthermore, the document addresses Mendelian genetics, family trees, Punnett squares, and genetic diagrams to solve problems related to eye color, pea plant crosses, and guinea pig hair traits. The solution also examines haemophilia inheritance patterns, including autosomal and sex-linked inheritance, and calculates probabilities of carrier status and disease transmission. Finally, the document addresses the ABO blood grouping system and provides genetic diagrams for blood group inheritance.

Running head: LEAFLET
Information Leaflet on Human Twin Studies
Name of the Student
Name of the University
Author Note

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this assignment is my own work and contains no plagiarised content. All sources
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Learner Signature: Date:
Provider: The Manchester College Pathway: Health
Diploma Title: Access to Higher Education Diploma (Health and Physiotherapy)
Module: Human Biology Group: 1 and 2
Unit Title:Genetics. Unit Code:GA33BIO06
Tutor/Assessor: Dr. Beth Tunstall Internal Verifier: Aqsa Hussain
Assignment Title:Lifestyle choices and health
Type of Assignment: SAQ
Word Count: (outlined in assessment per section)
Time allowed for supervised assessment: NA
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assessment evidence, extensions, resubmissions and referrals.

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Task 1
Assessment criteria 1.1
Q1. (a) ABO blood groups in human are an example of discontinuous variation,
whereas height is an example of continuous variation. Describe how these two
examples differ in terms of:
(i) Genetic control (i.e. the number of genes involved).
Discontinuous variations, like height, are controlled by the effects of a single gene or
a very small set of genes that control the phenotype of the characteristic (Mather and
Jinks 2013). Continuous variations, like height differences, are controlled by the
effects of multiple genes, which are known as the polygenic inheritance (Boyd and
Silk 2014).
(ii) The effect of the environment on each characteristic.
Continuous variations are much likely to be influenced by the effects of environment

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on the particular trait which provides intermediate variants. Discontinuous variations
are not likely to be affected by the environmental influence and provide distinct
phenotypes (Reichenberg et al 2016).
(iii) The range of phenotypes.
The phenotypes of discontinuous variations have two or more distinct forms which
arises from presence or absence of a gene. However, the continuous variations
exhibit a vast range of unbroken phenotypes which is not equally distributed in the
population effected by environmental factors and multiple genes (Mather and Jinks
2013).
(b) Give one other example of continuous variation and one other example of
discontinuous variation.
 Continuous: Another example of continuous variation is weight of human
beings is different throughout the population (Goodrich et al 2014).
 Discontinuous: A different example of discontinuous variation observed in
human population is the variation in finger prints (Bbc.co.uk, 2018).
Assessment criteria 1.2
Q2. The table shows concordance for height between monozygotic and dizygotic
twin pairs from birth to the age of eight years. A concordance of 1 indicates that the
twins are identical in height.
Concordance for height between
Age monozygotic twin pairs dizygotic twin pairs of the same
sex
Birth
3 months
6 months
12 months
24 months
3 years
0.62
0.78
0.80
0.86
0.89
0.93
0.79
0.72
0.67
0.66
0.54
0.56

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5 years
8 years
0.94
0.94
0.51
0.49
(a) If height were entirely genetically controlled, what concordance would you
expect between monozygotic twin pairs? Explain your answer.
Complete genetic control would provide a perfect 1 probability score, but it is not the
case in case of continuous variation is subjected to environmental effect (Hallmayer
et al 2011).
(b) Does data for eight-year-old twins suggest that height is largely controlled
by genetic factors? Explain you answer.
The eight year old twins’ data show that concordance is maintained for monozygotic
twins and for fraternal dizygotic twins the score is as similar to normal siblings
(Joseph 2013).
(c) Suggest an explanation for the low concordance at birth for monozygotic
twins.
Certain modification take place in the zygote as a form of somatic variation, this is
the cause of a phenomenon called developmental noise. The difference in the
growth of twins may add to the low rate of concordance at birth (Bell and Spector
2011).

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Assessment criteria 1.3
Q3. (a) Explain the meaning of each of the following terms:
(i) Variation: genetic variation is defined as the diversification of genes across
a population which provides different phenotypes. These variations occur
naturally and provide multiple phenotypes. Natural phenotypic variations
do not cause to the individual.
(ii) Mutation: Mutation is the sudden or unintended change in the gene which
provides altered or abnormal phenotype. Mutations often cause harm to
the individual (1000 Genomes Project Consortium 2010).
a. Explain how mutation causes variation. Give examples.
Mutations are alterations of DNA that has the ability to alter the phenotype, either in
a small scale or a large scale. The changes of mutation determine the outcome and
its severity in an organism. Mutations that are harmful and cause diseases
manifestation are sickle cell anaemia, cystic fibrosis et cetera. Mutations, which are
beneficial to the organism, are naturally selected over the course of time by evolution
(1000 Genomes Project Consortium 2010). Those mutated organisms with the
beneficial mutation become a part of the variation. Example can be provided of the
genetic variant of the apolipoprotein called Apolipoprotein- Al Milano, which is more
efficient that than the normal apolipoprotein variants and can dissolve accumulated
plaques from the arteries, prevents inflammation in the pulmonary cavity along with
functioning as an antioxidant (Speidl et al 2010).
b. Explain how meiosis causes genetic variation in the gametes.
Meiosis is the process by which an individual receives one half of each of its parent’s
genetic material. The process occurs in cell division in the germ line cells, where the
each of the parental gametes are separated into two sister chromatid which is
independently assorted in the progeny and possible recombination give rise to
variation in the progeny (Giraut et al 2011). Homologous chromosomes undergo
crossing over which is inherited by the offspring and the phenotypic expression gives

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rise to an altered gene expression helping in the survival rate of the individual. These
variations in the gene can also result in the formation of mutated alleles which
expresses a disease condition.
Assessment criterion 1.4
The classical twin study is established as the definitive study design for investigating
the relative importance of genetic and environmental factors to traits and diseases in
human population.
Monozygotic (identical) twins share all of their genes, while Dizygotic(fraternal) twins
share only about 50 percent of them, the same as non-twin siblings. If a researcher
compares the similarity between sets of identical twins to that of fraternal twins for a
particular trait, then any excess likeness between the identical twins should be due
to genes rather than environment(Hallmayer et al 2011)..
Write an information leaflet evaluating the benefits and dilemmas in the use of
human twin studies to investigate the causes of variation.
You should aim for two sides of A4 with images to make the leaflet engaging.
Assessment criteria 2.1,2.2
Q4. Assume eye colour in humans is controlled by a pair of alleles of a gene where
the allele giving brown eyes is dominant to the allele giving blue eyes. Both parents
of a blue-eyed man, John, were brown-eyed. He married a brown-eyed woman,
Sara, whose father had brown eyes and mother blue eyes. Sara had a blue-eyed
sister. John and Sara had a brown-eyed child.

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1) Fill in the boxes and circles on the family tree below to show the genotype of
each individual.
B-brown(dominant)
b- blue(recessive)
Q6. When Mendel crossed a large number of tall pea plants with short pea plants, all
F1 plants were tall. The F2 generation was created by self-pollinating the F1 plants.
T= tall; t=short
(a) Complete a genetic cross of F2 to show the genotypes and phenotypes of the
offspring.
Phenotype of F2
Genotype of F2: 1Tt : 2 Tt : 1 tt
(b) State the ratio of phenotypes expected in the F2 offspring.
Phenotype of F2= 2 tall: 1 short
(c) State Mendel’s First Law of inheritance
The first law of inheritance by Mendel states that the “two alleles responsible for a
John
Bb Bb bb
bb Bb bb
Bb

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gene segregate from each other during gamete formation, where half of the genes
from each parent will be passed on to the progeny” (Bateson and Mendel 2013).
(d) State Mendel’s Second Law of inheritance
The second law of inheritance by Mendel states that “ the alleles of one gene sort
into gametes independently of alleles of another gene.” (Bateson and Mendel 2013).
Assessment criteria 2.3,2.4
Q7. Guinea pigs, which were homozygous for long, black hair were crossed with
ones which were homozygous for short, white hair. All the F1 offspring had short,
black hair.
Let the genotype for homozygous long be : LL (dominant)
Let the genotype for homozygous short be : ll (recessive)
Let the genotype for homozygous black be : BB (dominant)
Let the genotype for homozygous white be : bb (recessive)
(a) Using suitable symbols, draw a genetic diagram to explain this result.
llBB × LLbb
Gametes: lB × Lb
F1: BbLl

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(b) Complete the Punnett square to show the results of interbreeding the F1
offspring.
Gametes BL Bl bL bl
BL BBLL
Black-short
BBLl
Black-short
BbLL
Black-short
BbLl
Black-short
Bl BBLl
Black-short
BBll
Black-long
BbLl
Black-short
Bbll
Black-long
bL BbLL
Black-short
BbLl
Black-short
bbLL
white-short
bbLl
white-short
bl BbLl
Black-short
Bbll
Black-long
bbLl
white-short
bbll
white long

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(c) Complete the following table to show the different phenotypes you would
expect in the F2 and their ratio.
Phenotype Ratio
Black-short 9
Black-long 3
White-short 3
White-long 1
(d) State the ratio of phenotypes expected in the F2 offspring.
9 Black-short : 3 Black-long : 3 white-short : 1 white-long

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Assessment criteria 3.1,3.2, 3.3
Q8. Haemophiliacs possess a non-functional form of the gene responsible for the
production of blood clotting factors. Shown below is the occurrence of haemophilia
in one family.
Using the following symbols:
H =dominant allele h = recessive allele
1) State the genotypes of the following individuals.
Individual Genotype