Download MONOHYBRID PROBLEMS

GENETICS REVIEW (KEY)
MONOHYBRID PROBLEMS
1.
If in guinea pigs, B=black coat color and b = brown coat color, what would be the genotypic and phenotypic
results of this cross:
Bb X BB
offspring genotype probabilities: Bb 50% / BB 50%
offspring phenotype probabilities: BLACK 100%
2.
If in guinea pigs S=short hair and s=long hair, what would be the genotypic and phenotypic results of this
cross:
Ss X Ss
offspring genotype probabilities: SS 25% / Ss 50% / ss 25%
offspring phenotype probabilities: BLACK 75% / BROWN 25%
3.
Cross a brown coat color guinea pig with a heterozygous coat color guinea pig. (b = white, B = brown)
offspring genotype probabilities: Bb 50% / bb 50%
offspring phenotype probabilities: Brown 50% / White 50%
4.
In pea plants, green-pea pods is due to the dominant gene "G", while yellow-pea pods is due to its recessive
allele, "g". If you cross a homozygous green pea pod plant with a homozygous yellow one, what F1 results
(genotype and phenotype) would you expect? Crossing two F1 individuals would give the F2 generation.
What is the F2 genotype results (genotype and phenotype)?
Genotype: Gg
Phenotype: Green
Crossing two F1 individuals would give the F2 generation. What is the F2 genotype results (genotype and
phenotype)?
Genotypes: GG / Gg / gg
Phenotype: Green / Striped
DIHYBRID PROBLEMS
In rabbits, grey hair is dominant to white hair, and black eyes are dominant to red eyes. Write the phenotype for
each genotype below.
GG = grey
BB = black
Gg = grey
Bb = black
gg = white
bb = red
1.
What are the phenotypes of rabbits that have the following genotypes:
Ggbb ______grey, red_______
2.
ggBB ________white, black________
ggbb ______white, red______
GgBb ________grey, black_______
A male rabbit has the genotype GgBb. Determine the gametes produced by this rabbit (the sperm would
have these combinations of alleles) and describe how these sperm cells demonstrate the laws of segregation
and independent assortment.
GB, Gb, gB, gb
Each gamete only has one of the two alleles that the parent has, demonstrating the law of segregation. Any hair
allele can combine with any eye allele, demonstrating independent assortment.
3.
A female rabbit has the genotype ggBb. Determine the gametes (eggs) produced by this rabbit.
gB and gb
4.
Use the gametes from #3 and #4 to set up a Punnett square. Put the male's gametes on the top and the
female's gametes down the side. Then fill out the square and determine what kind of offspring would be
produced from this cross and in what proportion.
GB
Gb
gB
gb
GgBB
GgBb
ggBB
ggBb
gB
gB
GgBB
GgBb
ggBB
ggBb
gb
GgBb
Ggbb
ggBb
Ggbb
gb
GgBb
Ggbb
ggBb
ggbb
EXTRA PRACTICE….. (#5)
5.
A male rabbit homozygous dominant for hair and homozygous recessive for eyes is crossed with a female
rabbit heterozygous for both traits. Determine the phenotypes and proportions in the offspring.
GB
Gb
GGBb
Gb
GGBb
Gb
GGBb
Gb
GGBb
GGbb
GGbb
GGbb
GGbb
GgBb
GgBb
GgBb
GgBb
Ggbb
Ggbb
Ggbb
Ggbb
How many out of 16 are:
Grey, red eyed __8__
Gb
gB
gb
Grey, black eyed __8__
White, red eyed __0__
White, black eyed __0__
INCOMPLETE OR CODOMINANT PROBLEMS
1.
In a certain type of flower, color is incompletely dominant. Cross a pure red flower with a pure white
flower. What will be the phenotype of the offspring? (R = red, W = white) Pink
2.
In flowers, color is an incomplete dominance trait. (see above problem) When the genotype is
heterozygous, the phenotype is pink. Cross a pink flower with a pink flower. What will be the genotype
and the phenotype of the offspring?
Genotype: RR / RW / WW
Phenotype: Red / Pink / White
3.
When a pure red cow is bred with a pure white bull, the F1 offspring is a roan (this means that it has both
red and white hairs). What will be the results of a cross between a red bull and a roan cow?
RR x RW = Genotype: RR 50% / RW 50%
Phenotype: Red 50% / Roan 50%
Could a farmer raise a herd of roan cattle that would only produce roan offspring? Explain. (R = red hairs W
= white hairs) No, will always produce 25% Red, 50% Roan, and 25% White
4. In Guppy vultures, pure long legs when crossed with pure short legs cause medium legs. Cross a long
Guppy vulture male with: 1. a medium hen and 2. a short hen and give the phenotype of the offspring. (L =
long legs, S = short legs)
1) LL x LS = Long Legs / Medium Legs
2) LL x SS = Medium Legs
SEX-LINKED PROBLEMS
1.
Show, using a Punnett Square, how the sex of a child is determined. Does the egg from the mother or the
sperm from the father determine the sex of the offspring? Dad
X
Y
X
XX
XY
X
XX
XY
2.
Color blindness is a sex-linked recessive trait. What is the probability that a color blind woman and a colorblind man will have: a. a normal sighted son b. a color blind daughter
XcXc x XcY = a) 0% b) 100%
3.
What is the probability that two parents with normal color vision will have color blind sons or daughters if
the mother's father is color -blind? XCXc x XCY = 25% Colorblind Children
4.
Hemophilia is a sex-lined blood disorder. People with this disease can bleed to death from a small cut
because their blood does not clot. It is caused by a recessive allele. You are a normal (homozygous) female
who is about to marry a hemophiliac male. What are your chances of having a child with this blood
disorder? (XH= normal blood Xh= hemophilia) XHXH x XhY = 0% Hemophiliacs
MULTIPLE ALLELE PROBLEMS
1.
If a man with blood type AB marries a woman heterozygous for type A, what is the probability that their
child will be type B? IAIB x IAi = 25% Type B
2.
A mother has type A blood and a father has type B blood. If their baby has type O blood, what is the
genotype of the parents? IAi x IBi
3.
Use a Punnett square to show the possible genotypes and phenotypes for blood type of the offspring of two
parents, one with blood type O and one with blood type AB.
IAIB x ii = Genotypes: IAi / IBi
Phenotypes: Type A / Type B
4.
Two women gave birth to girls in the same hospital at the same time. The nurses think they may have
accidentally switched the babies' name tags and given the babies to the wrong parents. One baby, Jane, is
type O. The other baby, Mary, is blood type A. The father in one set of parents, the Reds, is blood type A,
and the mother is type B. The father in the other set of parents, the Greens, is blood type AB, and the
mother is type O. Figure out which baby belongs to which parents. Show your work and reasoning. Jane
cannot be Greens baby b/c father is type AB. Jane is a Red. Mary is a Green.
PEDIGREE PROBLEMS
1.
Determine the genotypes of each of the following 12 individuals (color blindness is sex-linked)?
B
A or B
B
C
C
D
D
B
C
B
C
E
A = XCXC
B = XCXc
C = XCY
D = Xc Y
E = XcXc
2.
The following is a pedigree of myopia (near-sightedness). This condition is caused by a recessive gene, but
is NOT sex linked. Determine whether the 8 individuals are A. Heterozygous B. Homozygous C. Not
enough sufficient data to determine genotype.
A
A
C
A
B
A
A
B
CHALLENGE PROBLEMS
1.
What kind of gametes (sperm) could a man who was heterozygous for long eye lashes, free earlobes, and
brown eyes produce? LEB, LeB, LEb, Leb, lEB, leB, lEb, leb
If he were to marry a woman of exactly the same genotype for these 3 traits, what size Punnett square
would be needed? DO NOT ACTUALLY MAKE THE CROSS!!!!!! 8 x 8
2. In cocker spaniels, white spotting is recessive to solid color. You want to know if your red solid-colored
cocker is homozygous or heterozygous for spotting. How could you find out?
You need to do a test cross with a homozygous recessive (spotted) dog.
If yours is homozygous, what results would you expect? 100% Red
If yours is heterozygous, what results would you expect from the cross? 50% Red / 50% Spotted
3. In poultry, feathered legs (F) are dominant over clean legs (f) and pea comb (P) are dominant over single
comb (p). Two cocks, A and B, are bred to two hens, C and D. All four birds are feather legged and pea
combed. Cock A with both hens produced offspring that are feathered and pea combed. Cock B with hen C
produced both feathered and clean, but all pea combed, but with hen D he produced all feathered, but part
pea combed and single. What are the genotypes of all four birds?
Cock A = FFPP, Cock B = FfPp , Hen C = FfPP, Hen D = FFPp