ONE GENE CROSSES:
1.
R = red, r = white a.
The parental cross: RR x rr b.
The gametes: RR can produce Ⓡ or Ⓡ, rr can produce ⓡ or ⓡ c.
Punnett Square:
R R r r
Rr Rr
Rr Rr d.
Genotypic ratio: 0:4:0 (100% heterozygous) e.
Phenotypic ratio: 4:0 (100% red)
2.
R = red, r = white a.
The F
1
cross: Rr x Rr b.
The gametes: Rr can produce Ⓡ or ⓡ, Rr can produce Ⓡ or ⓡ c.
Punnett Square:
R r
R RR Rr r Rr rr d.
Genotypic ratio: 1:2:1 (25% homozygous dom., 50% heterozygous, 25% homozygous rec.) e.
Phenotypic ratio: 3:1 (75% red, 25% white)
3.
This is an example of a TEST CROSS , R = red, r = white a.
The cross: Rr x rr b.
The gametes: Rr can produce Ⓡ or ⓡ, rr can produce ⓡ or ⓡ c.
Punnett Square:
R r r Rr rr r Rr rr d.
Genotypic ratio: 0:2:2 (0% homozygous dom., 50% heterozygous, 50% homozygous rec.) e.
Phenotypic ratio: 2:2 (50% red, 50% white)
Although in the problem you were given that the red parent was heterozygous, this parent would appear the same as a homozygous dominant parent (i.e. they would both be red). The fact that some of the offspring could be white when the red parent is crossed with the white parent provides evidence that the red parent is heterozygous if any of the offspring are, in fact, white. If all offspring are red, the red parent could be either homozygous dominant or heterozygous.
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4.
G = gray-bodied, g = black-bodied a.
The cross: Gg x gg b.
The gametes: Gg can produce Ⓖ or ⓖ, gg can produce ⓖ or ⓖ c.
Punnett Square:
G g g Gg gg g Gg gg d.
Genotypic ratio: 0:2:2 (0% homozygous dom., 50% heterozygous, 50% homozygous rec.) e.
Phenotypic ratio: 2:2 (50% gray-bodied, 50% black-bodied) f.
Of 100 offspring, 50 are expected to have black bodies (50% of 100 offspring)
5.
The brown parent is heterozygous (Bb) and the albino parent is homozygous recessive (bb). a.
The cross: Bb x bb b.
The gametes: Bb can produce Ⓑ or ⓑ, bb can produce ⓑ or ⓑ c.
Punnett Square:
B b b Bb bb b Bb bb d.
Genotypic ratio: 0:2:2 (0% homozygous dom., 50% heterozygous, 50% homozygous rec.) e.
Phenotypic ratio: 2:2 (50% brown, 50% albino) f.
This is the only cross between brown and albino parents that can result in some of the offspring being brown and other offspring being albino phenotype. The albino parent must be homozygous recessive (that is the only way the recessive trait appears). The brown parent could have been homozygous dominant or heterozygous. However, if this parent were homozygous dominant, only brown offspring are possible. Since the problem states that some offspring were brown and others were albino, the brown parent must be heterozygous. This problem provides an example of a TEST CROSS.
6.
R = can roll tongue, r = cannot roll tongue a.
Both parents must be heterozygous (Rr) for tongue-rolling and their son must be homozygous recessive (rr).
The cross Rr x Rr is the only combination of two parents with the dominant phenotype (i.e. both can roll their tongues) where their offspring could show the recessive phenotype.
Punnett Square:
R r
Phenotypic ratio: 3:1
(75% tongue-roller, 25% non-tongue roller)
R RR Rr r Rr rr b.
The probability of the next child (or any subsequent children) being a non-tongue roller is
25% , as predicted by the Punnett square, above.
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7.
This problem illustrates a form of intermediate inheritance called incomplete dominance. The intermediate trait (in this case short-tailed) is a blend of the two extreme traits (long-tailed and tailless) and results from the inheritance of the two contrasting alleles for the extreme traits.
L = allele for long-tailed
N = allele for tailless
Note that two different, capital letters are used to represent the alleles since neither is dominant over the other.
A cat with the genotype: LL has a long tail
LN has a short tail
NN has no tail a.
The parental cross: LL x NN b.
The gametes: LL can produce Ⓛ or Ⓛ, NN can produce Ⓝ or Ⓝ c.
Punnett Square:
L L
N LN LN
N LN LN d.
As predicted by the Punnett square, above, the problem states that all of the F
1
offspring have short tails. e.
The F
1
cross: LN x LN f.
The gametes: LN can produce Ⓛ or Ⓝ, LN can produce Ⓛ or Ⓝ g.
Punnett Square:
L N
L LL LN
N LN NN h.
According to the Punnett square, above, we expect that 25% of the kittens will have long tails,
50% will have short tails, and 25% will be tailless. The problem states that kittens of each of these phenotypes resulted from the F
1
cross.
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TWO GENE CROSSES
Keys for two gene crosses (the actual letters used to represent alleles will vary with each problem)
A = Dominant allele for the first trait considered, a = recessive allele for the first trait considered
B = Dominant allele for the second trait considered, b = recessive allele for the second trait considered
Genotypic ratio: AABB : AABb : AaBB : AaBb : AAbb : Aabb : aaBB :aaBb : aabb
Phenotypic ratio: Dominant/Dominant : Dominant/recessive : recessive/Dominant : recessive/recessive
8.
B = Brown, b = blonde, C = Curly, c = straight a.
The parental cross: BBCC x bbcc
The possible gametes produced by each parent are determined using the principle of independent assortment (FOIL): BBCC can produce BC or BC or BC or BC bbcc can produce bc or bc or bc or bc
Punnett Square:
BC BC BC BC bc BbCc BbCc BbCc BbCc bc BbCc BbCc BbCc BbCc bc BbCc BbCc BbCc BbCc bc BbCc BbCc BbCc BbCc
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Genotypic ratio: 0:0:0:16:0:0:0:0:0 (100% dihybrid)
Phenotypic ratio: 16:0:0:0 (100% Brown/Curly) b.
The parental cross: BBcc x bbCC
The possible gametes produced by each parent are determined using the principle of independent assortment (FOIL): BBcc can produce Bc or Bc or Bc or Bc bbCC can produce bC or bC or bC or bC
Punnett Square:
Bc Bc Bc Bc bC BbCc BbCc BbCc BbCc bC BbCc BbCc BbCc BbCc bC BbCc BbCc BbCc BbCc bC BbCc BbCc BbCc BbCc
Genotypic ratio: 0:0:0:16:0:0:0:0:0 (100% dihybrid)
Phenotypic ratio: 16:0:0:0 (100% Brown/Curly)
Continued on page 5…

Question 8 continued: c.
The parental cross: BbCc x BbCc This is a DIHYBRID CROSS
The possible gametes produced by each parent are determined using the principle of independent assortment (FOIL): BbCc can produce BC or Bc or bC or bc
BbCc can produce BC or Bc or bC or bc
Punnett Square:
BC Bc bC bc
BC BBCC BBCc BbCC BbCc
Bc BBCc BBcc BbCc Bbcc bC BbCC BbCc bbCC bbCc bc BbCc Bbcc bbCc bbcc
Genotypic ratio: 1:2:2:4:1:2:1:2:1
Phenotypic ratio: 9:3:3:1 d.
The parental cross: BbCc x bbcc This is a dihybrid TEST CROSS
The possible gametes produced by each parent are determined using the principle of independent assortment (FOIL): BbCc can produce BC or Bc or bC or bc
Punnett Square: bbcc can produce bc or bc or bc or bc
BC Bc bC bc bc BbCc Bbcc bbCc bbcc bc BbCc Bbcc bbCc bbcc bc BbCc Bbcc bbCc bbcc bc BbCc Bbcc bbCc bbcc
Genotypic ratio: 0:0:0:4:0:4:0:4:4
Phenotypic ratio: 4:4:4: 4 For a dihybrid test cross, If all 4 possible phenotype combinations occur in offspring, then the parent showing the dominant phenotype for both traits that you are looking at is heterozygous for both traits.
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9.
T = Tall, t = short, Y = Yellow seed, y = green seed a.
The parental cross: TTYY x ttyy
The possible gametes produced by each parent are determined using the principle of independent assortment (FOIL): TTYY can produce TY or TY or TY or TY ttyy can produce ty or ty or ty or ty
Punnett Square:
TY TY TY TY ty TtYy TtYy TtYy TtYy ty TtYy TtYy TtYy TtYy ty TtYy TtYy TtYy TtYy ty TtYy TtYy TtYy TtYy
Genotypic ratio: 0:0:0:16:0:0:0:0:0 (100% dihybrid) b.
The F
1
cross: TtYy x TtYy This is a DIHYBRID CROSS
The possible gametes produced by each parent are determined using the principle of independent assortment (FOIL): TtYy can produce TY or Ty or tY or ty
TtYy can produce TY or Ty or tY or ty
Punnett Square:
TY Ty tY ty
TY TTYY TTYy TtYY TtYy
Ty TTYy TTyy TtYy Ttyy tY TtYY TtYy ttYY ttYy ty TtYy Ttyy ttYy ttyy
Phenotypic ratio: 9:3:3:1 9/16 or 56.25% Tall/Yellow-seeded
3/16 or 18.75% Tall/green-seeded
3/16 or 18.75% short/Yellow-seeded
1/16 or 6.25% short/green-seeded
Assuming the ideal ratio: If 90 are Tall/Yellow-seeded, then there are 160 offspring total (90 is 9/16 of 160). That means that we should expect that 10 offspring will be short/green-seeded (10 is 1/16 of
160).
10.
B = Black fur, b = white fur, E = Black eyes, e = pink eyes a.
The parental cross must be: bbee X BbEe b.
The father of this litter must be dihybrid (BbEe) because this is the only genotype that, when crossed with bbee (the mother’s genotype) can result in all the phenotypic combinations displayed by the offspring.
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11.
T = Long tail, t = short tail, C R = Red coat, C W = White coat (R and r, as the problem states, are not preferred since this trait provides an example of codominance, not complete dominance). If you used R and r, no problem, but correct this in the future. For codominance, we use a capital letter with different capital letter superscripts to represent different alleles for the codominant trait.
The long-tailed/roan bull is TtC R C W . We know that he is heterozygous for tail length because his cross with cow A resulted in a calf with a short tail. a.
Cow A: ttC b.
Cow B: ttC
W C
R c.
Cow C: TtC
C
R
W
W
C R
, the calf of cow A: ttC R C W .
, the calf of cow B: TtC R C R .
, the calf of cow A: ttC R C R .
12.
Follow the product rule of probability. Start by determining the probability of exhibiting the dominant phenotype for each separate trait using Punnett squares. For each trait, the probability of the dominant phenotype is: a.
Aa x Aa probability: 3/4 b.
Bb x bb probability: 2/4 c.
Cc x CC probability: 4/4 d.
dd x Dd probability: 2/4 e.
Ee x EE probability: 4/4 f.
Multiply these five fractions to determine the probability of all five events occurring simultaneously: 192/1024 = 3/16 = 18.75%
LINKED GENES
13.
For linked genes, treat the letters with the bar above them as if they are one symbol. Because the alleles represented by these letters are on the same chromosome, they cannot assort independently. a.
The parental cross: BW BW x bw bw
BW BW can produce BW or BW The gametes: bw bw can produce bw or bw
Punnett Square:
____
BW
____
BW
___ bw
___ bw
___ ___
BW bw
___ ___
BW bw
___ ___
BW bw
___ ___
BW bw
Phenotypic Ratio Key: Gray/Normal : Gray/stumpy : black/Normal : black/stumpy
Phenotypic ratio: 4:0:0:0 (100% Gray-bodied/Normal wings)
The remaining crosses are solved in the same manner. b.
Phenotypic ratio: 3:0:0:1 c.
Phenotypic ratio: 2:0:0:2
(75% Gray/Normal, 25% black/stumpy)
(50% Gray/Normal, 50% black/stumpy) d.
Phenotypic ratio: 4:0:0:0 (100% Gray-bodied/Normal wings)
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14.
The order of alleles on the chromosome is: B-C-A-D (or D-A-C-B)
B 10% C 15% A 15% D
GENE INTERACTIONS
15.
This problem serves as one example of EPISTASIS - one gene alters the expression of another gene.
The problem indicates that the genes for pigmentation and pigment distribution are linked (note the line over the gametes). a.
The parental cross must have been: Cs Cs x cS cS b.
The offspring of this cross (the F c.
The F
1
cross: Cs cS x Cs cS d.
Punnett Square:
1
’s) would all be: Cs cS (100% striped, as stated)
____
Cs
____ cS
___
Cs
___ cS
___ ___
Cs Cs
___ ___
Cs cS
___ ___
Cs cS
___ ___ cS cS
Phenotypic Ratio Key: Black stripes : Black : white e.
Phenotypic ratio: 2:1:1 (50% Black stripes, 25% Black, 25% white)
16.
This problem serves as one example of a type of EPISTASIS called COMPLIMENTARY GENES .
In this case, the cellular products of two separate genes work together to contribute to the same trait. a.
The F
1
cross: CcPp x CcPp This is a DIHYBRID CROSS
The possible gametes produced by each parent are determined using the principle of independent assortment (FOIL): CcPp can produce CP or Cp or cP or cp
CcPp can produce CP or Cp or cP or cp
Punnett Square:
CP Cp cP cp
CP CCPP CCPp CcPP CcPp
Cp CCPp CCpp CcPp Ccpp cP CcPP CcPp ccPP ccPp cp CcPp Ccpp ccPp ccpp
Phenotypic ratio: 9:7 9/16 or 56.25% Purple
7/16 or 43.75% white
Remember: the plant must have a dominant C and a dominant P in order to be purple.
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17.
This problem serves as one example of a type of EPISTASIS called COLLABORATION - two genes interact to give four distinct phenotypes depending on the combination inherited. a.
The parental cross: Bbyy x bbYy
The possible gametes produced by each parent are determined using the principle of independent assortment (FOIL): Bbyy can produce By or By or by or by bbYy can produce bY or by or bY or by
Punnett Square:
By By by by bY BbYy BbYy bbYy bbYy by Bbyy Bbyy bbyy bbyy bY BbYy BbYy bbYy bbYy by Bbyy Bbyy bbyy bbyy
Phenotypic Ratio Key: Blue : Yellow : Green : White b.
Phenotypic ratio: 4:4:4:4 (25% Blue, 25% Yellow, 25% Green, 25% White)
18.
This problem serves as a simple example of POLYGENIC INHERITANCE- an additive effect of two or more separate genes on a single phenotype. a.
The parental cross: AaBb x AaBb This is a DIHYBRID CROSS
The possible gametes produced by each parent are determined using the principle of independent assortment (FOIL): AaBb can produce AB or Ab or aB or ab
AaBb can produce AB or Ab or aB or ab
Punnett Square:
AB Ab aB ab
AB AABB AABb AaBB AaBb
Ab AABb AAbb AaBb Aabb aB AaBB AaBb aaBB aaBb ab AaBb Aabb aaBb aabb
Phenotypic Ratio Key: Red : Dark : Medium : Light : White b.
Phenotypic ratio: 1:4:6:4:1 1/16 or 6.25% Red
4/16 or 25% Dark
6/16 or 37.5 % Medium
4/16 or 25% Light
1/16 or 6.25% White
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ALLELE INTERACTIONS: Multiple Alleles and Intermediate Inheritance
19.
Although the representation of alleles in the problem set will work, the preferred way to represent the alleles for human blood groups is I A , I B , and i, since this trait shows BOTH codominance and multiple alleles. a.
The parental cross: I A i x I B i
Punnett Square:
I B I A
I A
I B I i
B i i I A i ii
Each child of this couple could have type A, type B, type AB, or type O blood. b.
It is possible for a type A woman and a type B man can have a child with type O blood if both the man and woman are heterozygous for their respective blood types. This is demonstrated in the Punnett square, above. However, if the man’s parents were both type AB, then the man must be homozygous B (I B I B ) since his parents do not possess a recessive allele (i) to pass on to him.
In this case, the man could not have a child with type O blood. Case dismissed.
20.
C B = Black allele, C c = chocolate allele, c g = golden allele
The parental cross: C B c g x C c c g a.
C B c g is a black lab, C c b.
Punnett square: c g is a chocolate lab
Genotypic Ratio Key:
Genotypic ratio:
Phenotypic Ratio Key:
Phenotypic ratio:
C B c g
C c C B C c C c c g c g C B c g c g c g
C B C B : C B C c : C B c g : C c C c : C c c g : c g c g
0 : 1 : 1 : 0 : 1 : 1
Black : Chocolate : Golden
2 : 1 : 1 (50% Black, 25% Choc., 25% Golden)
SEX LINKAGE
21.
If a sex-linked trait is due to a recessive allele, a female will express the recessive phenotype only if she is homozygous recessive, X r X r . A cross between a heterozygous (carrier) female and a recessive male can produce female offspring with the recessive phenotype, demonstrating that the trait is recessive.
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22.
X B = normal X-chromosome, X a.
The parental cross: X b
B
= defective X-chromosome,Y = normal Y-chromosome
Y x X b X b b.
Punnett square:
X b X b
X B X B X b X B X b
Y X b Y X b Y c.
The probability of their son(s) being colorblind is 100% d.
The probability of their daughter(s) being colorblind is 0%
AUTOSOMAL AND SEX-LINKED RECESSIVE TWO GENE CROSSES
23.
C = curly, c = straight,
X B = normal X-chromosome, X b = defective X-chromosome,Y = normal Y-chromosome
The cross between the now happily married man and woman: CcX b Y x ccX B X b
Punnett Square: cX B cX b cX B cX b
CX b CcX B X b CcX b X b CcX B X b CcX b X b
CY CcX B Y CcX b Y CcX B Y CcX b Y cX b ccX B X b ccX b X b ccX B X b ccX b X b cY ccX B Y ccX b Y ccX B Y ccX b Y a.
Probability of a curly-haired/normal visioned child: 4/16 (25%) b.
Probability of a straight-haired/colorblind child: 4/16 (25%) c.
Probability of a curly-haired/colorblind daughter: 2/16 (12.5%)
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24.
A = Blind from aniridia, a = normal vision,
X H = normal X-chromosome, X h = defective X-chromosome,Y = normal Y-chromosome
The cross between the couple:
Punnett Square:
AaX H Y x aaX H X h aX H aX h aX H aX h
AX H AaX H X H AaX H X h AaX H X H AaX H X h
AY AaX H Y AaX h Y AaX H Y AaX h Y aX H aaX H X H aaX H X h aaX H X H aaX H X h aY aaX H Y aaX h Y aaX H Y aaX h Y
There is a 25% chance that each son will exhibit blindness with hemophilia.
There is a 25% chance that each son will exhibit blindness without hemophilia.
There is a 25% chance that each son will exhibit normal vision with hemophilia.
There is a 25% chance that each son will exhibit normal vision withouts hemophilia.
25.
The genotypes (blood group; colorblindness) for the people mentioned in the problem are: a.
John: b.
Mary’s mother:
Heterozygous AB; Afflicted male
Homozygous recessive; Heterozygous (carrier female) iiX iiX c
C
Y
X c
Homozygous recessive; Afflicted male
Homozygous recessive; Heterozygous (carrier female) c.
Mary’s father: d.
Mary:
I A I B iiX C
X c
X c
Y
I A I B X c Y x iiX C X c Cross between John and Mary:
Punnett Square: iX C iX c iX C iX c
I A X c I A iX C X c I A iX c X c I A iX C X c I A iX c X c
I A Y I A iX C Y I A iX c Y I A iX C Y I A iX c Y
I B X c I B iX C X c I B iX c X c I B iX C X c I B iX c X c
I B Y I B iX C Y I B iX c Y I B iX C Y I B iX c Y
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Phenotypic ratios:
2/16 Type A/normal color vision female (12.5%) 2/16 Type A/normal color vision male (12.5%)
2/16 Type A/colorblind female (12.5%) 2/16 Type A/colorblind male (12.5%)
2/16 Type B/normal color vision female (12.5%) 2/16 Type B/normal color vision male (12.5%)
2/16 Type B/colorblind female (12.5%) 2/16 Type B/colorblind male (12.5%)