Examples Of Variation And Twin Studies For Investigating Genetic Factors

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 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.

(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. 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).

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 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.

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 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.

(b) Complete the Punnett square to show the results of interbreeding the F1 offspring.

(c) Complete the following table to show the different phenotypes you would expect in the F2 and their ratio.

(d) State the ratio of phenotypes expected in the F2 offspring.                                         

9 Black-short : 3 Black-long : 3 white-short : 1 white-long

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.

2) On the basis of the information provided, is the inheritance of haemophilia:

(i) autosomal or sex-linked?

(ii) dominant or recessive?

3) State the probability of individual 8 being a carrier of haemophilia.- 33%

4) Explain why only females can be carriers of haemophilia- Since women have two pairs of X chromosomes, if either one of those pairs is a diseases allele, then the effect of the diseased gene will be suppressed by lyonization*.Q9. (i)  Complete the following genetic diagram to show how parents who did not suffer from haemophilia, could have a son with haemophilia but also other children who did not suffer from haemophilia.                

(i) What is the probability of the couple having a daughter with haemophilia?       

In this particular scenario none of the daughters will have haemophilia as the male parent does not have haemophilia. The diseases is prominent in females when they acquire both recessive genes for heamophilia. Therefore, all the daughter will be either carrier or normal.

(ii) What is the probability of the couple having another son with haemophilia?

33%     

Q10.   In the ABOblood grouping system, a single gene with three alleles controls the production of the antigens that determine an individual’s blood group.

(a) State the possible genotypes for an individual who is:

Let the genotype for “O” be – O, the genotype for “A” be –A and the genotype for “B” blood group be – B

Blood group A : AO / AA

Blood group AB : AB

(b)  In a particular family, the father is blood group A and the mother is blood group B.

They have four children, each with a different blood group.

Draw a genetic diagram below to show how it is possible for the parents to have four children all with different blood groups.   

AO  ×  BO

A, O, B

AO, OO, BO, AB

Q11.   

1) What is meant by epistasis?

The phenomenon by which non allelic genes intereact to mask on or the other’s phenotype is called epistasis.

The natural colouration of wild mice is called agouti and is produced from banded hairs. Two genes are involved, each with a dominant (A and B) and a recessive allele (a and b). The allele A codes for the ability to produce hair pigment: AA and Aa mice have pigmented hairs but all aa individuals are albinos. The B allele codes for the ability to make hair with grduated colouration: BB and Bb mice have graduated hair, bb mice have hair that is all one colour which is black.

2) Two agouti mice, genotypes AaBb, are bred together. What phenotypic ratio would you expect in the next generation? Set out the crosses using a Punnett square.  

Ratio: 9 agouti : 3 black : 4 albino