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Mendelian Problems
Just a quick reminder - when solving genetic problems, always choose the first letter of the allele that is
dominant. For instance, if the dominant trait is curly hair over straight hair, you would use CC to
represent a homozygous curly haired person and cc to represent someone with straight hair.
1. (a) A homozygous violet flowered pea plant is crossed with a homozygous white flowered
pea plant. If the violet colour is dominant (V), what is the genotype and phenotype of
the F1 generation.
P1 = VV
P2 = vv
P1/P2 V
V
v
Vv
Vv
v
Vv
Vv
Genotype of F1 = 100% Vv
Phenotype for F1 = 100% violet
(b) If 2 members of the F1 are crossed, what ratio of the F2 generation will have the same
phenotype as the parents of the F1 and how many will have the same phenotype?
F1 = Vv
F1 = Vv
P1/P2 V
V
VV
v
Vv
v
Vv
vv
Ratio of F2 generation with same phenotyoe: 3:1 or 75%
Ratio of F2 generation with same genotype: 2:2 or 50%
2. Blue eyes are recessive to brown eyes. Show the expected children in a marriage between a
blue eyed woman and a brown eyed man who had a blue eyed mother.
P1 = bb
P2 = BB or Bb –
Parents of P2:
Woman: bb
Man: Bb or BB
P1/P2 B
B
b
Bb
Bb
b
Bb
Bb
Therefore all children Bb, phenotype 100% brown eyes
Or
P1/P2 B
b
b
Bb
bb
b
Bb
bb
Therefore children either Bb or bb – only Bb will produce brown eyes
Therefore P2 = Bb
F1 generation:
P1/P2 B
b
b
Bb
bb
b
Bb
bb
F1 generation:
50% brown eyes Bb
50% blue eyes bb
3. Two white sheep produce a black offspring. What must the parents’ genotypes for colour
be? What is the probability that the next offspring will be black? (assume white is
dominant - W)
Assuming that white is dominant to black, possibly genotypes for parents are WW or Ww
Cross #1: Parents WW and WW
P1/P2 W
W
W
WW WW
W
WW WW
100% WW, phenotype 100% white, therefore not black.
Cross #2: Parents Ww and WW
P1/P2 W
w
W
WW Ww
W
WW Ww
50% WW = white
50% Ww = white, therefore not black
Cross #3: Parents Ww and Ww
P1/P2 W
w
W
WW Ww
w
Ww ww
25% WW = white
50% Ww = white
25% ww = black
Therefore the genotypes of the parents must be Ww and Ww.
The ratio of white to black is 3:1, and the possibility that the next offspring will be black is
25%.
4. In guineas pigs, black coat colour is dominant to white. Suppose you had a black guinea
pig whose parents were unknown to you, how could you find out whether the guinea pig
was homozygous or heterozygous for black?
Possible genotypes for my guinea pig: Bb or BB
If I cross my guinea pig with a known heterozygous guinea pig, Bb then I will get the following
crosses:
Cross #1: Bb known, and Bb unknown
P1/P2 B
b
B
BB
Bb
b
Bb
bb
75% black, 25% white
Cross #2: Bb known, and BB unknown.
P1/P2 B
b
B
BB
B
BB
100% black
Bb
Bb
If I cross my guinea pig with a known heterozygote, then if I get 100% black offspring, I know
it was homozygous, and if I get 75% black and 25% white, I know it’s heterozygous.
5. Suppose you were raising what you thought was a purebred line of cats, but found that
some of the cats had spots of white fur at the tips of their tails, while others did not. How
could you determine whether this trait is recessive or dominant?
Cross a white tailed cat with a non-white tailed cat
Possible crosses:
Cross #1 WW x ww
P1/P2 W
W
w
Ww Ww
w
Ww Ww
100% Ww – which ever is dominant, 100% that colour
Cross #2: Ww x ww
P1/P2 W
w
w
Ww ww
w
Ww ww
50% Ww – dominant colour
50% ww – non dominant colour
Therefore if you cross a white tailed cat with a non-white tailed, depending on the ratio of
kittens, you can determine whether or not it’s dominant
6. An extra finger in man is rare but is due to a dominant gene. When one parent is normal
and the other parent has an extra finger, but is heterozygous for the trait, what is the
probability that their first child will be normal? Their second child?
P1 = FF
P2 = Ff
P1/P2 F
F
F
FF
FF
f
Ff
Ff
50% FF – dominant
50% Ff – extra finger
50% chance their child will be normal, 50% chance their second child will be normal
7. Determine the probability that a plant of genotype CcWw will be produced from parental
plants of the genotypes CcWw and Ccww.
P1/P2
Cw
Cw
cw
cw
CW
CCWw
CCWw
CcWw
CcWw
cW
CcWw
CcWw
ccWw
ccWw
Cw
CCww
CCww
Ccww
Ccww
cw
Ccww
Ccww
Ccww
ccww
Probability of CcWw = 4/16 or 25%
8. If you thought we would stop at dihybrid crosses, you were wrong. Trihybrid crosses are
those that involve 3 inherited characteristics where tri in Latin means “three”. What are
the possible gametes for the following cross between AABBCC and aabbcc. What are the
possible gametes of the F1 generation?
Trihydrid cross: AaBbCc x AaBbCc
P1/P2 ABC
ABc
AbC
ABC
AABBCC AABBCc AABbCC
ABc
AABBCc AABBcc AABbCc
AbC
AABbCC AABbCc AAbbCC
Abc
AABbCc AABbcc AAbbCc
aBC
AaBBCC AaBBCc AaBbCC
aBc
AaBBCc AaBBcc AaBbCc
abC
AaBbCC AaBbCc AabbCC
abc
AaBbCc AaBbcc AabbCc
Abc
AABbCc
AABbcc
AAbbCc
AAbbcc
AaBbCc
AaBbcc
AabbCc
Aabbcc
aBC
AaBBCC
AaBBCc
AaBbCC
AaBbCc
aaBBCC
aaBBCc
aaBbCC
aaBbCc
aBc
AaBBCc
AaBBcc
AaBbCc
AaBbcc
aaBBCc
aaBBcc
aaBbCC
aaBbcc
abC
AaBbCC
AaBbCc
AabbCc
AabbCc
aaBbCc
aaBbCc
aabbCC
aabbCc
abc
AaBbCc
AaBbcc
Aabbcc
Aabbcc
aaBbCc
aaBbcc
aabbCc
aabbcc
9. Albinism (nn) in humans is inherited as a simple recessive trait where N represents normal.
For the following families, determine the genotypes of the parents and offspring. When
two genotypes are possible, list both.
(a) Two normal parents have five children, four normal and one albino.
(b) A normal male and an albino female have six children, all normal.
(c )A normal male and an albino female have six children, three normal and three
albino.
a)
4:1 sounds like 3:1, which is the ratio for two heterozygous parents
P1/P2 N
n
N
NN
Nn
n
Nn
nn
This cross shows a 3:1 ratio for normal to albino.
b)
Possible genotypes for normal parent: Nn or NN
P1/P2 N
n
n
Nn
nn
n
Nn
nn
50% normal, 50% albino, therefore not correct.
Parents must be NN and nn.
c)
Possible genotypes for normal parent: Nn or NN
P1/P2 N
n
n
Nn
nn
n
Nn
nn
50% normal
50% albino
Therefore the genotype of the parents are Nn and nn.
10. In human families it is often observed that certain characteristics may “skip” a generation
or two and then reappear. How would you explain this in light of the facts learned in this
unit?
Phenotypes can “skip” a generation when a trait is dominant, and the middle generation is
heterozygous. Then the dominant trait “covers” the recessive part of the gene, and then when
the middle generation mates with someone else that is also heterozygous, the recessive trait can
possibly show up when an offspring is made homozygous recessive.
Rating scale: Assessment FOR learning
1
2
Ability to write
the correct geno
and phenotype
(Knowledge)
Ability to write
complete
conclusion
sentence stating
their answer in
words
(Communication)
Ability to generate
the correct crosses
(Application)
Ability to
formulate the
correct genotypes
without being
given them
(Thinking/Inquiy)
3
4