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CHAPTER 15 TEXTBOOK PROBLEMS
1-14 (pages 303-304)
1.
father is XhY
daughter is XHXh (she had to get a Xh from her father)
husband of daughter is XHY
XH
Xh
XH
XHXH
XHXh
Y
XHY
XhY
H = normal
H = hemophilia
Their chances of having a daughter who has hemophlia is 0%.
A son has a 50% chance of being a hemophiliac.
If they have 4 sons, the chance of all 4 having hemophilia is:
1/2 x 1/2 x 1/2 x 1/2 = 1/16
2.
The disease is sex-linked because it is only in boys. It is recessive because both
parents were normal. See Punnett square in question #1. This disease causes death
in the early teens so affected boys do not live long enough to pass on the abnormal
allele.
3.
Colorblind man is XhY
Woman is XHXh
XH
Xh
Xh
XHXh
XhXh
Y
XHY
XhY
H = normal
H = colorblind
Chance of having a colorblind daughter is 1/4.
Chance of having first son who is colorblind is 1/2. (Already know it is a boy)
4.
wildtype fruit fly heterozygous for gray body and normal wings is BbVv
black fly with vestigial wings is bbvv
Key
What is the recombination frequency? (Genes are linked)
B = gray
b = black
V = normal wings
v= vestigial wings
Offspring were:
wildtype (B_V_) = 778
black vesitigal (bbvv) = 785
black normal (bbV_) = 158
gray normal (B_vv) = 162
bv
BV
BbVv
bv
bbvv
This is the cross if these genes are linked
and have no cross over. To get black
normal and gray normal, there was a cross
over. So, add to find total offspring born
(1883). Add to find total cross over offspring
(320). Take crossover offspring divided by
total and multiply by 100 to get
recombination frequency.
320/1883
x 100 = 17%
1
5.
wildtype fruit fly heterozygous for gray body and red eyes = BbRr
black fruit fly with purple eyes = bbrr
Key
B = gray
b = black
R = red
r = purple
Cross is BbRr x bbrr:
BR
br
BbRr
Br
br
Bbrr
br
bbrr
bR
bbRr
This box shows correct linkage
– half wild type and half black
purple.
Genes are not linked like this
because this box does not give
right results.
Offspring:
721 wildtype
751 black purple
49 gray purple
45 black red
The gray purple and black red are crossovers.
Add total offspring. Add total crossovers.
94/1566 x 100 = 6% recombination frequency between body color and eye color
From problem 4: 17% recombination frequency between body color and wing type.
To find the recombination frequency between eye color and wing type, mate a wildtype
red eyed normal wing (RrVv) with a purple eyed vestigial wing (rrvv).
2
6.
b+ = B = gray
b = b = black
vg+ = G = normal wing
vg = g = vestigial wing
Cross = b+b+vgvg x bbvg+vg+ or BBgg x bbGG
Parent cross:
BBgg x bbGG
Parent cross generates the
following F1 generation:
All BbGg
Testcross for BbGg x bbgg (chromosomes drawn in Punnett square)
The last two boxes show offspring resulting from crossovers and the first two boxes show
offspringwithout crossovers.
If genes are 17 map units apart this means they have a 17% crossover frequency. Each of the two crossover
genotypes should occur approximately 8.5% of the time. The two noncrossover genotypes should each occur
41.5% of the time.
 41.5% gray vestigial
 41.5%black normal
 8.5% gray normal
 8.5% black normal
3
7.
The disorder would always be inherited from the mother because mitochondria come
from the egg cell– not the sperm cell.
8.
A female with 2 Barr bodies and aneuploidy = XXX (one X is active, the other two are
Barr bodies). Microscopy should show the 3 Barr bodies.
9.
A-C = 28
A-D = 25
B-C = 20
B-D = 33
20
C
8
B
25
A
D
28
33
10.
A and B linked = 50 map units
50 map units = 50% crossover or recombination frequency
If you have a 50% recombination frequency, this would look like a cross with genes that
are not linked!
11.
Cross tall heterozygote with antennae (TtAa) with a test cross (ttaa)
Offspring:
42 dwarf no antennae
46 tall antennae
7 dwarf antennae
5 tall no antennae
ta
TA
ta
TtAa
ttaa
Crossover total is 12.
12/100 x 100 = 12% between A and T
AaSS x aass
47antennae up-snout
48 no antennae down-snout
2 antennae down-snout
3 no antennae up-snout
as
Crossover total is 5.
5/100 x 100 = 5% between A and S
4
AS
as
AaSs
aass
12.
TtSs x ttss
ts
40 tall up-snout
42 dwarf down-snout
9 tall down-snout
9 dwarf up-snout
TS
ts
TtSs
ttss
Crossover total is 18.
18/100 x 100 = 18% between T and S
13.
Parent cross BBrr x bbRR yields all BbRr in F1.
Cross the F1 with homozygous white-oval: BbRr x bbrr
The 10% recombination frequency means that 100
out of the 1000 offspring should be recombinants
(resulting from crossovers). The B and r are linked
and the b and R are linked as seen in the parent
cross. The recombinant sex cells are BR and br.
Br
br Bbrr
bR
bbRr
BR
BbRr
Key
B = blue
b = white
R = round
r = oval
br
bbrr
Bbrr = 450 (blue oval)
bbRr = 450 (white round)
BbRr = 50 (blue round)
bbrr = 50 (white oval)
14.
Gene a – vestigial wings = 14%
Gene a – brown eyes = 26%
Gene “a” is located between the wing and
brown eyed genes. The numbers do not
added up perfectly, but 14+26 is 40 (or
approximately 37.5)
37.5
wing----------------------------eye
--------a-------------------14
26
5