16.18 Past Paper Practise
Question 1:
Mendel’s first law states that ‘only one of a pair of contrasting characters may be represented in a single gamete’.
His second law states that ‘either of a pair of contracting characters may be combined with either of another pair.’
Explain how meiosis illustrates these laws.
Law 1. [4]
Law 2. [4]
- Law 1:
- ref to chromosomes in homologous pairs;
- only one of the pair passes into a gamete during meiosis;
- the alleles of a contrasting pair are situated on different but homologous chromosomes;
- thus only one allele of the pair can go to a gamete (dependent mark);
- Law 2:
- chromosomes/chromatids assort independently during meiosis;
- one chromosome/chromatid of the pair goes to one pole and the other to the other pole but which goes either way is random;
- the two genes (allelic pairs) occupy different homologous chromosomes;
- thus it is purely random which of the alleles of one gene becomes combined with which of the alleles of the other gene; (dependent mark)
Question 2:
a)
There are 12 chromosomes in the somatic cells of the bean, Vicia.
(i) How many chromosomes does the offspring receive from the female parent via the egg cell nucleus? [1]
6;
(ii) How many chromosomes will be present in the primary endosperm nucleus? [1]
18 (it is triploid);
(iii) A female sex chromosome is donated by X and a male sex chromosome by Y. Specify the sex chromosome content in the primary endosperm nucleus. [1]
XXY/XXX (formed from 2 female nuclei and one male nucleus);
b)
The diploid chromosome number of a donkey is 66 chromosomes and that of a horse is 60 chromosomes.
(i) A mule is the offspring of a cross between a female horse and a male donkey. Calculate the number of chromosomes to be found in the somatic cells of the mule. Show your working. [2]
(ii) Bivalent formation occurs in prophase 1 of meiosis. What is the theoretical maximum number of bivalents that could form in the mule? [1]
31;
(iii) Suggest why mules are infertile. [2]
- chromosomes of horse and donkey are different shapes and numbers;
- thus exact pairing/synapsis to form bivalents cannot occur;
- meiosis/gamete production fails;
Question 3:
The Brindled Beauty (Lycia hirtaria) is a moth whose colour is controlled by a single gene with a pair of alleles.
a)
A homozygous brown and white individual was allowed to breed with a heterozygous individual.
(i) Use the symbol B to represent the dominant allele for black (melanic) colour and b to represent the recessive allele for brown and white colour. Write down genotypes and phenotypes of the parents and offspring and the genotypes of the gametes.
Parents ………….. x …………. [1]
Gametes ……………. x ……………. [1]
F1 …………. [1]
(ii) The black offspring of this cross were allowed to interbreed together. State the genotypes and phenotypes of the offspring and their ratio.
Genotypes [1]
Phenotypes [1]
Ratio [1]
b)
Describe how you could find out whether black individuals were homozygous or heterozygous for the colour alleles. [5]
- test cross black moths with the double recessive/brown and white moths;
- if all offspring black then test parent is probably BB/homozygous;
- if some offspring brown and white then test parent is Bb/heterozygous;
- relevant complete genetic diagrams to show this;
- comment on necessity to hatch eggs, rear larvae and pupae to adult before crosses can be performed;
Question 4:
Alkaptonuria is a harmless, rare autosomal genetic defect in humans. The family tree below shows the pedigree of a family affected by alkaptonuria. (Individuals are numbered 1 to 14).
a)
(i) Is the condition dominant or recessive? Explain your answer. [2]
- recessive;
- because obvious heterozygotes/carriers don’t show the condition/the alletes in 6/8/13 must have come from the parents who do not show the condition;
(ii) State the numbers of all the individuals that are certain to be heterozygous for this gene. [3]
1 + 2; 3 + 4; 10 + 11; (lose 1 mark for each incorrect)
(iii) What is the probability that individual 14 is heterozygous for this gene? [1]
2/3 rds;
b)
Alkaptonuria occurs due to point mutation of the gene. State two other point mutation diseases of humans. [2]
sickle cell anaemia;
phenylketonuria/albinism/any other valid defect;
Question 5:
Alleles of the ABO blood group system are usually shown as IA, IB and IO. I represents the gene locus and A, B and O represent the three alleles of the ABO gene.
(a) What is meant by the term ‘gene locus’? [1]
the specific position of the alleles on the (specific) homologous chromosomes;
(b) By reference to the inheritance of the ABO blood group system explain what is meant by (i) multiple alleles, (ii) codominance and (iii) a recessive allele.
(i) Multiple alleles [3]
(ii) Codominance [2]
(iii) Recessive allele [2]
(i) when a gene has more than two alleles;
thus more variations of the character can be shown;
such as the four blood groups A, B, AB and O/credit any relevant examples;
(ii) when two alleles both exhibit an effect in the phenotype;
correct example/shown in group AB where allele A causes antigen A to be formed and allele B causes antigen B to be
formed;
(iii) only shown in the homozygote/ in the absence of a dominant allele;
correct example/the recessive O allele can only exert its effect to cause group O in the double recessive/OO condition;
c) By means of a genetic diagram show the possible results of a cross between individuals of genotypes AO and BO.
Question 6:
(a) Explain the meaning of the term ‘gene’. [3]
a length of DNA which contains the genetic code to enable the synthesis of a specific polypeptide;
situated at a specific locus/position on a specific chromosome;
consists of two or more variants called alleles;
(b) Define each of the following terms.
(i) genotype [1]
(ii) phenotype [1]
(iii) heterozygous [1]
(i) the actual genes/alleles an individual organism/species possesses; 1
(ii) the actual appearance of the individual/species which is a result of the effects of the genotype and the environment; 1
(iii) where an organism receives different alleles for the gene from each parent;
(c) Kerry type cattle with normal leg length are produced by a homozygous dominant genotype DD. Short legged Dexter type cattle are produced by the genotype Dd. The homozygous recessive genotype dd produces bulldog calves which are always stillborn.
Give complete genetic diagrams to illustrate the following crosses:
(i) Kerry cattle x Dexter cattle.[3]
(ii) Dexter cattle x Dexter cattle.[3]
(ii) Which of these crosses would be the better breeding programme for the farmer to use in order to obtain Dexter calves? Explain your answer. [2]
Kerry x Dexter;
other cross produces bulldog calves which is uneconomical;
Question 7:
In pigs erect ears is controlled by the dominant allele E and flop ears by the recessive allele e. Black coat is controlled by the dominant allele B whilst red coat is controlled by the recessive allele b. The pairs of alleles are not linked.
(a) (i) With the aid of a genetic diagram show the genotypes and phenotypes of a cross between two black erect eared heterozygous pigs and state the ratio of the phenotypes obtained. [6]
3 marks for correct punnet square with phenotypes associated with genotypes;;;
(1 mark penalty for each error/omission)
Ratio – 9 black erect : 3 black flop : 3 red erect : 1 red flop;
(ii) How do these results illustrate Mendel’s second law? [2]
either colour allele can be associated with either ear allele;
Mendel’s second law states that either of a pair of contrasting characters can be combined with either of another pair;
(b)With the aid of a genetic diagram show the genotypes, phenotypes and ratio of offspring produced by crossing a black erect eared heterozygous pig with a red flop eared pig. In this case assume that the genes are linked, in adjacent association on the same bivalent. Comment on the result obtained. [6]
Accept answers that show a small stated proportion of recombinants.
gametes Be and bE are unlikely to form since B is next to E and b next to e on the same chromosome pair;
could only form if chiasmata/crossovers occur and separate the gene loci;
Question 8:
Rats have a dominant allele B which gives black coat colour and a recessive allele b which gives a cream coat colour. This gene acts in association with a randomly assorting epistatic gene which possesses two alleles, dominant E which allows the development of colour and the recessive e which suppresses colour development, resulting in albino rats which have white coats and pink eyes.
(a) Write down the possible genotypes of the following rat phenotypes.
(i) a black rat [1]
(ii) a cream rat [1]
(iii) an albino rat [1]
(i) BBEE, BBEe, BbEE, BbEe;
(ii) bbEE. bbEe;
(iii) BBee, Bbee, bbee; (all genotypes required for the marks)
(b)What do you understand by the term ‘randomly assorting epistatic gene’? [3]
an epistatic gene is one that influences the expression of another gene ;
randomly assorting means that the gene is on a different homologous pair of chromosomes to the gene it is influencing ;
thus segregates independently in meiosis/not linked/behaves in a Mendelian manner ;
(c) With the aid of a genetic diagram show the genotypes and phenotypes which could result from a cross between a cream individual of genotype bbEe and a black rat heterozygous for both characters. Indicate the ratio of phenotypes possible. [4]
2 marks for punnet square with correct phenotypes linked to genotypes;;
(1 mark penalty for each error/omission)
Ratio – 3 black : 3 cream : 2 albino;
Question 9:
The genetic disease ‘progressive retinal atrophy’ is caused by a mutant recessive gene in Irish Setter dogs. If the dog inherits the double recessive genotype then symptoms will develop by 10 weeks of age. Although the dogs become blind they are otherwise normal. The Irish Setters Breeders Society has run a breeding programme in an attempt to eliminate the disease. Use the symbol R to represent the normal condition and the symbol r to represent the disease condition. Dogs of genotype Rr do not show the disease but act as carriers.
(a) Describe and explain a suitable breeding programme which could be used to eliminate the blindness gene. [6]
- never breed from blind dogs (except in a test cross);
- test cross normal dogs with blind dogs;
- check puppies at 10 weeks for signs of disease;
- if disease shows then normal parent is heterozygous/Rr;
- thus must not be used for breeding;
- if the litter contains at least six normal pups and no abnormal assume the normal parent is homozygous/RR;
- if the litter only contains two or three normal dogs then must repeat the testcross since normal heterozygote may not have
- passed on the recessive gene;
(b)Would such a breeding programme completely eliminate the gene causing blindness? Explain your answer. [2]
no ;
the normal gene will continue to mutate into the disease causing gene;
Question 10:
The family tree below shows the inheritance of Duchenne muscular dystrophy in a family. This disease is caused by a recessive allele of an X-linked gene.
(a) (i) Indicate below the possible genotypes of individuals 1 to 10. Use D for the normal gene and d for the Duchenne gene and X and Y for the sex chromosomes. [10]
l. XDY;
2. XDXd ;
3. XDXD or XDXd;
4. XdY ;
5. XdY ;
6. XDXd ;
7. XDY ;
8. XDXD or XDXd ;
9. XdY ;
10. XdY ;
(ii) What would a genetic counsellor say to parents 6 and 7 when explaining what the probability would be of their next child suffering from Duchenne muscular dystrophy? [4]
- Duchenne allele is on the X chromosome in the part which does not have a corresponding allele on the Y chromosome;
- in females is masked by the dominant allele on the homologous locus so such individuals act as carriers;
- not covered in male so they manifest the recessive gene if they receive it;
- 1 in 4 chance of having an affected male child if mother is a carrier;
- mother has a 1 in 2 chance of being a carrier;
- [max 2 if just shown as a cross with ratios]
(b) Name two other X-linked diseases of humans. [2]
haemophilia;
red green colour blindness;
Question 11:
(a) The multiple allele inheritance system of the ABO blood group system contains the codominant alleles A and B and the recessive allele O. The table below shows possible crosses between people of certain ABO groups, the alleles they can contribute and the phenotypes of offspring which could not be produced as a result of the cross. Complete the table by writing in the alleles and blood group phenotypes in the empty boxes. [9]
(b) A husband suspects that his wife’s third child has been fathered by another man. His own first two children possess blood groups O and AB. The third, suspect child, is blood group B. Are the husband’s suspicions justified? Explain your answer. [4]
since lst child has group O both parents must possess the O allele ;
since second child has group AB one parent must have A allele and the other the B allele ;
thus could give rise to a group B child of genotype BO ;
thus the man’s claim is not justified on this evidence ;
Question 12:
The graph below represents the distribution of height in a pea plant population and in a human population.
(a) (i) What types of variation are shown by the pea plants and humans in relation to stature? [2]
peas – discontinuous;
humans – continuous;
(ii) With reference to the genetic mechanisms involved explain the types of distribution of height in:
peas [5]
and humans. [5]
peas –
stature in peas is regulated by one gene which has only two alleles ;
the dominant allele causes tall plants to be produced and the recessive allele causes short plants to be produced;
thus shortness can only be shown in the double recessive state;
breeding by heterozygotes would produce the monohybrid ratio of 75 % tall to 25 % short;
there is no overlap in the expression of the two alleles;
thus two populations develop in relation to stature; max 5
humans –
stature in humans is regulated by several genes/ref polygenic;
each of these genes may have many alleles/multiple alleles;
no clear dominant or recessive alleles;
each particular allelic variant (of the gene) may exert a slightly different effect in the phenotype;
thus heterozygotes could contain any two alleles for the gene out of possible hundreds;
this would give a wide variation/ continuous variation of phenotypes;
(b) State three precautions which would be taken when obtaining the human data. [3]
- select individuals of:
- same age ;
- same sex ;
- same state of health/nutrition/race;
Question 13:
Coat colour in horses is governed by a pair of alleles. The allele CC governs the development of a cream coat and the allele CH governs the development of a chestnut (brown) coat. If a pure bred cream horse is crossed with a pure bred chestnut horse then the resulting offspring is golden in colour with a white mane and this coat colour is referred to as palomino.
(a) (i) What name is given to this type of genetic result? [1]
codominance;
(ii) Write down a genetic diagram to show the cross between a pure bred cream horse and a pure bred chestnut horse. Show all genotypes, phenotypes and gametes. [3]
(iii) What would be the probabilities of obtaining a palomino, cream or chestnut horse when two palominos are interbred?. Give a genetic diagram to explain your answer.
Probabilities [1]
Genetic diagram [3]
(b) The Manx cat is a variety that has no tail. The trait is caused by a dominant allele M. In the homozygous condition this allele is lethal and results in stillborn kittens. The recessive allele, m, results in the tailed.
condition. How can breeders of Manx cats avoid getting stillborn kittens? Explain your answer with genetic diagrams.
How avoided [1]
Genetic diagrams [6]
