Cambridge International A Level Biology Revision Guide

Cambridge International A Level Biology
Answers to self-assessment questions
21
22
Gametes
from
other
parent
23a
b
c
show one of the recessive characteristics,
then the ‘unknown’ genotype must be
heterozygous for that characteristic.
iiCC and iiCc
Parental pink white
phenotypes
Parental AAbb aaBB
genotype
Gametes Ab aB
F1 genotypes
AaBb
F1 phenotypes
purple
Although a pure-breeding white-flowered
variety could be either aaBB or aabb the
latter would not give purple flowers in the
and so is the incorrect choice in this case.
F1 × F1 AaBb AaBb
Gametes AB Ab aB ab AB Ab aB
F2 genotypes and phenotypes:
Gametes from one parent
AB Ab aB ab
AB AABB AABb AaBB AaBb
the
purple purple purple purple
Ab AABb AAbb AaBb Aabb
purple pink purple pink
aB AaBB AaBb aaBB aaBb
purple purple white white
ab AaBb Aabb aaBb aabb
purple pink white white
1 : 1 : 1 : 1
Linkage; that is the two loci are on the same
chromosome. The alleles for grey body
and straight wings are on one homologous
chromosome in the heterozygote and the
alleles for ebony body and curled wings are
on the other homologous chtromosome.
30 + 29
113 + 30 + 29 + 115 × 100% = 20.6% d The curled wing locus is further away from
the ebony locus (cross-over value = 21%)
than is the aristopedia locus (cross-over
value = 12%).
24 The expected ratio would be 9 grey long : 3
grey short : 3 white long : 1 white short.
The total number of offspring is 80, so we
would expect 9 ÷ 16 of these to be grey long,
and so on.
Expected numbers:
9 ÷ 16 × 80 = 45 grey long
3 ÷ 16 × 80 = 15 grey short
3 ÷ 16 × 80 = 15 white long
1 ÷ 16 × 80 = 5 white short
The table below shows how χ 2 is
calculated.
Now look at Table 19.3 on page 000. There
are 4 classes of data, so there are 3 degrees
of freedom. The value for χ 2 is much greater
than any of the numbers in this row so
there is a significant difference between the
observed and expected results.
Phenotypes of animals
grey, grey, white, white,
long short long short
Observed
54 4 4 18
number (O)
Expected ratio 9 3 3 1
Expected
45 15 15 5
number (E)
– E +9 –11 –11 +13
– E) 2 81 121 121 169
– E) 2 / E 1.8 8.1 8.1 33.8
Σ(O – E) 2 / E = 51.8
= 51.8
This is a huge value for χ 2
F1
ab
O
(O
(O
χ 2
Cambridge International AS and A Level Biology © Cambridge University Press 2014