CHAPTER 20
EXTENDING YOUR IDEAS
1.
A–B = 20%
A–C = 8%
Thus
_____________________________________________
B
20
A
8
C
C = 28%
or
A
2.
_______________________________
8
C
12
B
Tall dominant
Yellow dominant
Smooth dominant
a.
T
Y
S
TtYySs
tys
x
i.e. recombination B–C = 12%
dwarf
green
wrinkled
t
y
s
ttyyss
TYS
TYs
TyS
tYS
Tys
tYs
tyS
tys
TtYySs
TtYyss
TtyySs
ttYySs
Ttyyss
ttYyss
ttyySs
ttyyss
tall yellow smooth (TtYySs)
tall yellow wrinkled (TtYyss)
tall green smooth (TtyySs)
tall green wrinkled (Ttyyss)
dwarf yellow smooth (ttYySs)
dwarf yellow wrinkled (ttYyss)
dwarf green smooth (ttyySs)
dwarf green wrinkled (ttyyss)
b.
i.e. recombination B–
TtyySs
x
1
1
1
1
1
1
1
1
ttYySs
TyS
Tys
tyS
tys
tYS
TtYySS
TtYySs
ttYySS
ttYySs
tYs
TtYySs
TtYyss
ttYySs
ttYyss
tyS
TtyySS
TtyySs
ttyySS
ttyySs
tys
TtyySs
Ttyyss
ttyySs
ttyyss
tall yellow smooth
tall yellow wrinkled
tall green smooth
tall green wrinkled
dwarf yellow smooth
dwarf yellow wrinkled
dwarf green smooth
dwarf green wrinkled
3
1
3
1
3
1
3
1
TtYySs; TtYySS
TtYyss
TtyySS; TtyySs
TtYyss
ttYySS; ttYySs
ttYyss
ttyySS; ttyySs
ttyyss
3.
Red (R) dominant to yellow (r).
Round (P) dominant to pear-shaped (p).
Tall (T) dominant to dwarf (t).
a.
TTPPRR
x
F1 all TtPpRr
b.
ttpprr
i.e. tall, round and red
F2
TPR
TPr
TpR
Tpr
tPR
tPr
tpR
tpr
TPR
TTPPRR
TTPPRr
TTPpRR
TTPpRr
TtPPRR
TtPPRr
TtPpRR
TtPpRr
TPr
TTPPRr
TTPPrr
TTPpRr
TTPprr
TtPPRr
TtPPrr
TtPpRr
TtPprr
TpR
TTPpRR
TTPpRr
TTppRR
TTppRr
TtPpRR
TtPpRr
TtppRR
TtppRr
Tpr
TTPpRr
TTPprr
TTppRr
TTpprr
TtPpRr
TtPprr
TtppRr
Ttpprr
tPR
TtPPRR
TtPPRr
TtPpRR
TtPpRr
ttPPRR
ttPPRr
ttPpRR
ttPpRr
tPr
TtPPRr
TtPPrr
TtPpRr
TtPprr
ttPPRr
ttPPrr
ttPpRr
ttPprr
tpR
TtPpRR
TtPpRr
TtppRR
TtppRr
ttPpRR
ttPpRr
ttppRR
ttppRr
tpr
TtPpRr
TtPprr
TtppRr
Ttpprr
ttPpRr
ttPprr
ttppRr
ttpprr
Thus
4.
tall round red (TPR)
tall round yellow (TPr)
tall pear red (TpR)
tall pear yellow (Tpr)
dwarf round red (tPR)
dwarf round yellow (tPr)
dwarf pear red (tpR)
dwarf pear yellow (tpr)
a. Presence of
27
9
9
3
9
3
3
1
three dominant alleles (WWW)
two dominant alleles (WW)
one dominant allele (W)
no dominant allele
22% amylase
20.4%
18.4%
0.0%
Thus for wax to be present at least one dominant allele must be present. Further,
approximately 1.6% was deposited for each extra dominant allele. E.g. predict:
WxWxWxWx
would have approximately 23.6% amylase
WxWxWxWxWx
would have approximately 25.2%
Case of continuous variation rather than simple dominance. Each dominant allele causes
production of an amount of the specific enzyme, amylase.
5.
b.
A gene determines, through protein synthesis, the presence or absence of specific
characteristics.
a.
i. 0.1 x 0.1 = 0.001 per thousand.
ii. 0.04 x 5 (average) x 0.4 = 0.08 per thousand.
Individuals that are heterozygous for a recessively inherited disease do not exhibit the
symptoms and thus may be unaware that they carry the disease (e.g. cystic fibrosis),
whereas diseases that are inherited through the dominant allele are expressed and the
individual is unlikely to reach adulthood (e.g. neurofibromatosis), or may elect not to
have children (e.g. myotonic dystrophy).
b.
c.
d.
In both myotonic and muscular dystrophy there is progressive muscular weakness. This
affects not only the skeletal muscles but also the cardiac and smooth muscles. Thus
whilst there will be a decrease in locomotion and ventilation, there will also be
deterioration of peristalsis and cardiovascular activity. The individual will take in less
oxygen and be less able to distribute this to the organs, particularly the brain cells. There
would probably also be an accumulation of wastes as the circulatory system would be
slower in their removal to the appropriate organs. The overall result would be poor
metabolic activities, which would cause deterioration of other organs and systems.
Because these diseases are carried on the X chromosome, there is no masking effect of a
dominant allele in males and thus the disease is more commonly expressed in them.
6.
Barking dominant
B
Erect ears dominant
E
Want pure breeding
BBee
Select only dogs which bark and have drooping ears.
Mate to drooping eared silent dogs – if any progeny are silent (i.e. barker was heterozygous
Bb), then cull that barker from breeding.
7.
Short dominant S
Black dominant B
Long recessive
Black and tan recessive
s
b
BBss
x
bbSS
All offspring will be BbSs – short haired and black.
Proportion F2 being long haired as well as black and tan (bbSS or bbSs) = 3/16 (Dihybrid
ratio).
8.
a.
RRPp
rP
x
rrPp
RP
Rp
RrPP
Rrpp
Thus
rp
RrPp
Rrpp
3 walnut (RrPP and RrPp) : 1 rose (Rrpp)
b.
rrPP
x
RrPp
rP
Thus
c.
RP
RrPP
walnut
Rp
RrPp
walnut
rP
rrPP
pea
rp
rrPp
1 walnut : 1 pea comb
pea
RrPp
x
Rp
rp
Thus
Rrpp
RP
Rp
rP
rp
RRPp
walnut
RRpp
rose
RrPp
walnut
Rrpp
rose
RrPp
Rrpp
rrPp
walnut
rose
pea
3 walnut : 3 rose : 1 pea : 1 normal
rrpp
normal
9.
X Sc
X sc
Cv V
X sc
x
cv v
cv V
Y
X Sc
X sc
cv V
Y
Cv V
X sc
cv v
X Sc Cv V
X sc cv V
X sc
X sc
cv v
X Sc
Y
X sc
Y
cv v
Cv V
cv V
Thus:
1 wild type female : 1 scute crossveinless female : 1 wild type male : 1 scute crossveinless
vermillion male.
But the original female may have the following genotypes on her X chromosomes:
Sc cv V
sc Cv v
sc Cv V
Sc cv v
sc cv V
Sc Cv v
Sc Cv v
sc cv V
Sc cv v
sc CV V
in which case the genotypes and phenotypes of the offspring will differ. E.g.
X Sc
X sc
cv V
Y
cv V
X sc
Cv v
X Sc cv V
X sc cv V
X sc Cv v
X sc cv V
X Sc
Y
X sc
Y
cv V
Cv v
i.e. 1 crossveinless female : 1 scute female : 1 crossveinless male : 1 scute vermillion
male
10.
Text error: the cross should read:
y xn lz
×
y+ sn+ Lz+
–
y sn lz
Possibilities for crossing over assuming y, sn and lz are equidistant on the chromosome:
a.
between y and sn
b.
between sn and lz
c.
between y and sn and between sn and lz
Thus with female
y+ sn+ lz+
y
sn lz
Male progeny:
No crossing over y+ sn+ lz+
Y
y sn lz
Y
Crossover a
Crossover b
wild type
yellow, singed, lozenge
y sn+ lz+
Y
y+ sn lz
Y
yellow
y+ sn+ lz
Y
y sn lz+
Y
lozenge
singed, lozenge
yellow, singed
Crossover c
y sn+ lz
Y
+
y sn lz+
Y
yellow, lozenge
singed
11.
No. It is possible that the female was homozygous for a dominant allele whose only effect
was to suppress the effect of the bar gene in males.
12.
If A and B are 10 crossover units apart then you would get 10% recombination between A and
B.
Thus the proportion of gametes and possible combinations will be:
9 AB
9 ab
1 aB
1 Ab
9 AB
81 AB//AB
81 AB//ab
9 AB//aB
9AB//Ab
9 ab
81 AB//ab
81 ab//ab
9 aB//ab
9Ab//ab
1 aB
9 AB//aB
9 aB//ab
1 aB//aB
1 Ab//aB
1 Ab
9 AB//Ab
9 Ab//ab
1 Ab//aB
1 Ab//Ab
i.e. 81 AB//AB : 162 AB//ab : 81 ab//ab : 18 AB//aB : 18 AB//Ab : 18 aB//ab: Ab//ab: 2Ab//aB :
1aB//aB : 1 Ab//Ab
13
h–c
Xg–h
Xg–c
12 units
46 units
34 units
Thus:
Xg
_________________________________
34
c
12
h