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GA33BIO06 Genetics-Lifestyle Choices And Health

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  • Course Code: GA33BIO06
  • University: The Manchester College
  • Country: United Kingdom

Question:

Task 1

Assessment criteria 

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.

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.

                                                         

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

 

 

 

 

 

 

 

 

 

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