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Cambridge International A Level Biology Revision Guide

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<strong>Cambridge</strong> <strong>International</strong> A <strong>Level</strong> <strong>Biology</strong><br />

Answers to self-assessment questions<br />

Genotypes<br />

of sperm<br />

X N<br />

Y<br />

Genotypes of eggs<br />

X N X N<br />

X N<br />

normal female<br />

X N Y<br />

normal male<br />

X N X n<br />

X n<br />

carrier female<br />

X n Y<br />

male with<br />

colour<br />

blindness<br />

This can happen if the woman is<br />

heterozygous. The affected child will be male.<br />

c Yes, if the mother has at least one allele for<br />

colour blindness, and the father has colour<br />

blindness.<br />

15a Male cats cannot be tortoiseshell, because a<br />

tortoiseshell cat has two alleles of this gene.<br />

As the gene is on the X chromosome, and<br />

male cats have one X chromosome and one<br />

Y chromosome, then they can only have one<br />

allele of the gene.<br />

b<br />

Parental<br />

phenotype<br />

Parental<br />

genotype<br />

tortoiseshell<br />

female<br />

X CO X CB<br />

orange<br />

male<br />

X CO Y<br />

Gametes X CO or X CB X CO or Y<br />

Genotypes of eggs<br />

X CO<br />

X CB<br />

chromosomes carrying the A or a alleles<br />

behave quite independently from those<br />

carrying the D or d alleles. This means that<br />

allele A can end up in a gamete with either<br />

D or d, and allele a can also end up in a<br />

gamete with either D or d. Thus independent<br />

assortment is responsible for the fact that<br />

each parent can produce four different types<br />

of gamete, AD, Ad, aD and ad.<br />

b Random fertilisation means that it is equally<br />

likely that any one male gamete will fuse with<br />

any female gamete. This results in the 16<br />

genotypes shown in the square. Thus random<br />

fertilisation is responsible for the different<br />

genotypes of the offspring and the ratios in<br />

which they are found.<br />

17a AaBb and Aabb in a ratio of 1:1<br />

b GgHh, Gghh, ggHh and gghh in a ratio of<br />

1 : 1 : 1 : 1<br />

c All TtYy<br />

d EeFf, Eeff, eeFf and eeff in a ratio of 1 : 1 : 1 : 1<br />

18a They would all have genotype GgTt and<br />

phenotype grey fur and long tail.<br />

b Grey long, grey short, white long, white short<br />

in a ratio of 9 : 3 : 3 : 1.<br />

19 Let T represent the allele for tall stem, t the<br />

allele for short stem, L G the allele for green<br />

leaves and L W the allele for white leaves.<br />

Parental<br />

genotype<br />

TTL G L G<br />

ttL G L W<br />

Genotypes<br />

of sperm<br />

X CO<br />

X CO X CO<br />

orange<br />

female<br />

X CB X CO<br />

tortoiseshell<br />

female<br />

Gametes TL G tL G or tL W<br />

Offspring<br />

TtL G L G and TtL G L W<br />

in a ratio of 1 : 1<br />

Y<br />

X CO Y<br />

orange<br />

male<br />

X CB Y<br />

black<br />

male<br />

The kittens would therefore be expected to be<br />

in the ratio of 1 orange female : 1 tortoiseshell<br />

female : 1 orange male : 1 black male.<br />

16a Independent assortment results from<br />

the random alignment of the pairs of<br />

homologous chromosomes on the equator<br />

during metaphase I. It ensures that the<br />

20a If B represents the allele for black eyes<br />

and b represents the allele for red eyes,<br />

L represents the allele for long fur and l<br />

represents the allele for short fur, then the<br />

four possible genotypes of an animal with<br />

black eyes and long fur are: BBLL, BbLL, BBLl<br />

and BbLl.<br />

b Perform a test cross – that is, breed the<br />

animal with an animal showing both<br />

recessive characteristics. If the offspring<br />

<strong>Cambridge</strong> <strong>International</strong> AS and A <strong>Level</strong> <strong>Biology</strong> © <strong>Cambridge</strong> University Press 2014

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