iGCSE Biology (0610)-17.4 Monohybrid inheritance

Question

Huntington’s disease is a genetic disease caused by a mutation in a single gene.
The allele for Huntington’s disease is dominant and is represented by the letter H.
The allele for no Huntington’s disease is recessive and is represented by the letter h.
Fig. 3.1 is a pedigree diagram showing the inheritance of Huntington’s disease in one family.

(a) (i) State the number of males with Huntington’s disease in Fig. 3.1.
(ii) State the two possible genotypes for person 3 in Fig. 3.1.
(iii) State the evidence from Fig. 3.1 that suggests that the allele for Huntington’s disease is dominant.

(b) A person that is heterozygous for Huntington’s disease has a child with a person that is homozygous recessive.
Complete the Punnett square in Fig. 3.2 by writing in the gametes and offspring for this cross and calculate the percentage chance of the child inheriting Huntington’s disease.

Punnett Square

Fig. 3.2

Percentage chance of the child inheriting Huntington’s disease………………………

(c) Chromosomes contain genetic information in the form of genes.
(i) Define the term gene.
(ii) State where chromosomes are found in cells.
(iii) State the chromosomes involved in the inheritance of sex in humans.

▶️ Answer/Explanation

Solution

(a)(i) 2

Explanation: From the pedigree diagram, two males (individuals 1 and 4) are shaded, indicating they have Huntington’s disease.

(a)(ii) HH or Hh

Explanation: Person 3 has Huntington’s disease (shaded), so they must have at least one H allele. Their genotype could be homozygous dominant (HH) or heterozygous (Hh).

(a)(iii) Individual 5 would have to inherit the condition if it were recessive, but he does not, so it must be dominant.

Explanation: If Huntington’s were recessive, unaffected parents (3 and 4) could not produce an unaffected child (5). Since they do, the allele must be dominant.

(b)

Percentage chance of the child inheriting Huntington’s disease is 50%.

Explanation: The cross is Hh (heterozygous) × hh (homozygous recessive). The Punnett square shows 2 Hh (affected) and 2 hh (unaffected) offspring, giving a 50% chance of inheritance.

(c)(i) A length of DNA that codes for a protein.

(c)(ii) Nucleus.

(c)(iii) X and Y.

Question

(a) The melanin gene in humans codes for a pigment produced in the skin, hair and eyes.

(i) Describe what is meant by the term gene.

(ii) State the name of a cell structure in humans that contains genes.

(b) Albinism is a genetic condition where there is a mutation in the gene that produces melanin. People with albinism have very pale skin, hair and eyes. In a population, 1 in 18000 are people with albinism. Calculate how many people with albinism would be expected in a population of 270000 people.

(c) The allele for normal melanin production is dominant and is represented by the letter A. The allele for albinism is recessive and is represented by the letter a. Fig. 5.1 is a pedigree diagram showing the inheritance of albinism in one family.

(i) State the sex chromosomes for person 1 in Fig. 5.1.

(ii) State how many people in Fig. 5.1 have albinism.

(iii) Circle the correct description of the genotype of person 3 in Fig. 5.1.

heterozygous homozygous dominant homozygous recessive

(d) An unaffected person has a child with a person with albinism. Complete the genetic diagram to show the possible genotypes and phenotypes of this child also the ratio of unaffected people:people with albinism.

▶️ Answer/Explanation

Solution

Ans:

(a)(i) a length of DNA ; that codes for a protein

A gene is a segment of DNA that carries instructions for synthesizing a specific protein, determining traits like melanin production.

(a)(ii) nucleus

Genes are located on chromosomes, which are housed in the nucleus of eukaryotic cells.

(b) 15 (people)

Calculation: \( \frac{270000}{18000} = 15 \). In a population of 270,000, 15 individuals are expected to have albinism based on the given frequency.

(c)(i) XX

Person 1 is female (indicated by the circle in the pedigree diagram), so their sex chromosomes are XX.

(c)(ii) 4

Four individuals in the pedigree (shaded circles/squares) exhibit the phenotype for albinism.

(c)(iii) homozygous recessive

Person 3 has albinism (shaded square), so their genotype must be homozygous recessive (aa).

(d) parental gametes: A a × a a ;
offspring genotypes: Aa (Aa) aa (aa) ;
offspring phenotypes: unaffected (unaffected) albinism (albinism) ;
ratio 1:1

An unaffected parent (Aa) and a parent with albinism (aa) can produce offspring with either genotype Aa (unaffected) or aa (albinism) in a 1:1 ratio.

iGCSE Biology (0610)-17.4 Monohybrid inheritance – iGCSE Style Questions Paper 3

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