CAIE IGCSE Biology 17.4.18 Codominant & Sex-linked Genetic Crosses

Practise completing genetic diagrams for codominance and sex linkage and calculating offspring phenotype probabilities.

Exam points

write parental genotypes and gametes using correct codominant or X-linked notation
combine gametes to show every offspring genotype and corresponding phenotype
convert affected or named-phenotype outcomes into ratios, fractions or percentages

[Maximum number: 5]

The eye is adapted for focusing on near and distant objects.
Fig. 3.1 shows the parts of the eye involved in focusing. The eye is focused on a distant object.

Fig. 3.1

Colour blindness can be caused by a mutation in a gene. The gene is located on the X chromosome.

Fig. 3.3 is a pedigree diagram of a family which has several people who are colour-blind.

Fig. 3.3

Use the information in Fig. 3.3 to complete the genetic diagram to show the probability of person 3 and person 4 having another child with colour blindness.

Table for Question 17.4.18—Genetic diagrams to predict question 1 — CAIE IGCSE Biology

offspring
genotypes
offspring
phenotypes
probability of a child having colour blindness

1 correct genotype for person 3, i.e. $(X)^{A}(X)^{a}$ ;

2 correct genotype for person 4, i.e. $\mathrm{X}^{\mathrm{A}} \mathrm{Y}$ ;

3 correct gametes from the parental genotypes ;

4 correct offspring genotypes from their gametes / parental genotypes (in any order) ;

5 correct offspring correct probability ;

MP3 ecf from MP1 and MP2 MP4 ecf from MP3

MP5 ecf from their offspring genotype

expected answer:

genotype (person 3) $\mathrm{X}^{\mathrm{A}} \mathrm{X}^{\mathrm{a}} \quad \times\left(\right.$ person 4) $\mathrm{X}^{\mathrm{A}} \mathrm{Y}$ gametes $\quad \mathrm{X}^{\mathrm{A}}, \mathrm{X}^{\mathrm{a}}+\quad \mathrm{X}^{\mathrm{A}}, \mathrm{Y}$ offspring genotypes $X^{A} X^{A}, X^{A} X^{a}, X^{A} Y, X^{a} Y$

offspring phenotypes female with normal colour vision female with normal colour vision male with normal colour vision male with colour blindness)

probability $\quad 0.25$ / 25\% / 1 in 4 / $\frac{1}{4}$

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17.4.18—Genetic diagrams to predict question 1