Discontinuous And Continuous Variation In Biology

Genetic Control of Discontinuous and Continuous Variation

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).

(ii) The effect of the environment on each characteristic.

(iii) The range of phenotypes.

(b) Give one other example of continuous variation and one other example of discontinuous variation.

· Continuous

· Discontinuous

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 5 years 8 years	0.62 0.78 0.80 0.86 0.89 0.93 0.94 0.94	0.79 0.72 0.67 0.66 0.54 0.56 0.51 0.49

(a) If height were entirely genetically controlled, what concordance would you expect between monozygotic twin pairs? Explain your answer.

(b) Does data for eight-year-old twins suggest that height is largely controlled by genetic factors? Explain you answer.

(c) Suggest an explanation for the low concordance at birth for monozygotic twins.

Q3. (a) Explain the meaning of each of the following terms:

(i) Variation

(ii) Mutation 

a. Explain how mutation causes variation. Give examples.

b. Explain how meiosis causes genetic variation in the gametes.

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.

                                                         

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.

(a) Complete a genetic cross of F2 to show the genotypes and phenotypes of the offspring.

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

(c) State Mendel’s First Law of inheritance

(d) State Mendel’s Second Law of inheritance

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.

(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.              

Gametes

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

Phenotype

Ratio

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

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
1
2
3
5
6
9

2) State the probability of individual 8 being a carrier of haemophilia.

3) Explain why only females can be carriers of haemophilia.

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. 

Phenotype of parents	Genotype of parents	Genotype of gametes
Normal male
Normal female
Genotype of offspring
Phenotype of offspring

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

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

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:

Blood group A

Blood group 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.   

Q1.1) What is meant by 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.  

Gametes

Genetic Control of Discontinuous and Continuous Variation

1. a) i. 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. 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 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) 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).

2. a) 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) 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) 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).

3. 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) 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).

Effect of the Environment on Discontinuous and Continuous Variation

b) Mees 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.

                               

6. T= tall; t=short

a) Genotype of F2: 1Tt : 2 Tt : 1 tt

b) Phenotype of F2=  2 tall: 1 short

c) 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) 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).

7. 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) llBB × LLbb

Gametes:                                                             lB   × Lb

F1:                                                                        BbLl

b)

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

c) 

Phenotype	Ratio
Black-short	9
Black-long	3
White-short	3
White-long	1

d) 9 Black-short : 3 Black-long : 3 white-short : 1 white-long.

8. 1)

Individual	Genotype
1	HH
2	Hh
3	HH
5	hh
6	HH/Hh
9	Hh

2) 33%

3) 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*.

9. XY  ×  X^hX

X^hX, XX, X^hY, XY

Phenotype of parents	Genotype of parents	Genotype of gametes
Normal male	XY	X & Y
Normal female	XX / XhX	X & Xh
Genotype of offspring	XhX, XX, XhY, XY
Phenotype of offspring	Normal carrier female, normal female, diseased male and normal male

i) 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. 33%     

10. a) 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) 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

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

2) 

Gametes	AB	Ab	aB	ab
AB	AABB agouti	AABb agouti	AaBB agouti	AaBb agouti
Ab	AABb agouti	AAbb black	AaBb agouti	Aabb black
aB	AaBB agouti	AaBb agouti	aaBB albino	aaBb albino
ab	AaBb agouti	Aabb black	aaBb white	Aabb white

References

1000 Genomes Project Consortium, 2010. A map of human genome variation from population-scale sequencing. Nature, 467(7319), p.1061.

Bateson, W. and Mendel, G., 2013. Mendel's principles of heredity. Courier Corporation.

Bbc.co.uk. (2018). BBC - Standard Grade Bitesize Biology - Variation : Revision, Page 4. [online] Available at: https://www.bbc.co.uk/bitesize/standard/biology/inheritance/variation/revision/4/ [Accessed 3 May 2018].

Bell, J.T. and Spector, T.D., 2011. A twin approach to unraveling epigenetics. Trends in Genetics, 27(3), pp.116-125.

Boyd, R. and Silk, J.B., 2014. How humans evolved. WW Norton & Company.. How humans evolved. WW Norton & Company.

Giraut, L., Falque, M., Drouaud, J., Pereira, L., Martin, O.C. and Mézard, C., 2011. Genome-wide crossover distribution in Arabidopsis thaliana meiosis reveals sex-specific patterns along chromosomes. PLoS genetics, 7(11), p.e1002354.

Goodrich, J.K., Waters, J.L., Poole, A.C., Sutter, J.L., Koren, O., Blekhman, R., Beaumont, M., Van Treuren, W., Knight, R., Bell, J.T. and Spector, T.D., 2014. Human genetics shape the gut microbiome. Cell, 159(4), pp.789-799.

Hallmayer, J., Cleveland, S., Torres, A., Phillips, J., Cohen, B., Torigoe, T., Miller, J., Fedele, A., Collins, J., Smith, K. and Lotspeich, L., 2011. Genetic heritability and shared environmental factors among twin pairs with autism. Archives of general psychiatry, 68(11), pp.1095-1102.

Joseph, J., 2013. The use of the classical twin method in the social and behavioral sciences: The fallacy continues. The Journal of Mind and Behavior, pp.1-39.

Mather, K. and Jinks, J.L., 2013. Biometrical genetics: The study of continuous variation. Springer.

Reichenberg, A., Cederlöf, M., McMillan, A., Trzaskowski, M., Kapra, O., Fruchter, E., Ginat, K., Davidson, M., Weiser, M., Larsson, H. and Plomin, R., 2016. Discontinuity in the genetic and environmental causes of the intellectual disability spectrum. Proceedings of the National Academy of sciences, 113(4), pp.1098-1103.

Speidl, W.S., Cimmino, G., Ibanez, B., Elmariah, S., Hutter, R., Garcia, M.J., Fuster, V., Goldman, M.E. and Badimon, J.J., 2010. Recombinant apolipoprotein AI Milano rapidly reverses aortic valve stenosis and decreases leaflet inflammation in an experimental rabbit model. European heart journal, 31(16), pp.2049-2057.