Learning Genetics Can Be Fun - Bio 121
1. A pea plant with round seeds is crossed with a pea plant that has wrinkled seeds. For the cross, indicate each of the following:
a) the genotype of each of the parents if the round seed plant is heterozygous.
b) the gametes produced by the parents
c) the genotypes and phenotypes of the F1 generation
d) the F2 generation if two round plants from the F1 generation were crossed
2. A plant of genotype TTYy was crossed with one of genotype ttYy. If the cross yielded 800 seeds, what number of what phenotypes probably resulted? What were their genotypes?
3. In summer squash, white fruit color is dominant. Yellow is recessive. A squash plant that is homozygous for white is crossed with a homozygous yellow one. Predict the appearance of
a) the F1 generation
b) the F2
c) the offspring of a cross between an F1 individual and a homozygous white individual.
4. For Dalmatian dogs, the spotted condition is dominant to non-spotted.
a) Using a Punnett square, show the cross between two heterozygous parents.
b) A spotted female Dalmatian dog is mated to an unknown male. From the appearance of the pups, the owner concludes that the unknown male was a Dalmatian. The owner notes that the female had six pups, three spotted and three non-spotted. What are the genotype and phenotype of the unknown male?
5. In horses, the trotter characteristic is dominant to the pacer characteristic. A male trotter mates with three different females, and each female produces a foal. The first female, a pacer, gives birth to a foal that is a pacer. The second female, also a pacer, gives birth to a foal that is a trotter. The third female, a trotter, gives birth to a foal that is a pacer. Determine the genotypes of the male, all three females, and the three foals sired.
6. Imagine for hair color that B gives brown hair and b gives blonde hair. Use a Punnett square to determine the following in a cross of two heterozygous parents.
a) What are the chances of the offspring being homozygous brown haired?
b) What are the chances of the offspring having blonde hair?
c) What are the chances of the offspring being heterozygous brown haired?
d) What is the genotypic ratio?
e) What is the phenotypic ratio?
f) Is there a heterozygous blonde haired offspring? Why?
g) If curly hair is dominant to straight hair, what letters will we use to show these genes?
h) A heterozygous curly haired male marries a straight haired female. What are their genotypes?
i) What would be the gametes for the male parent?
j) What would be the gametes for the female parent?
k) What are the chances of the offspring being homozygous curly haired?
l) What are the chances of the offspring having straight hair?
m) What are the chances of the offspring being heterozygous curly haired?
n) What is the genotypic ratio?
o) What is the phenotypic ratio?
p) Is there a heterozygous straight haired offspring? Why?
7. Blonde hair is due to a recessive gene, while brown hair is dominant. Agnes has brown hair, but
her mother was a blonde. Ralph, her husband, has brown hair also, and his father was blonde. Is it
possible for Ralph and Agnes to have blonde children? If so, what percentage?
8. Thalassemia is a serious human genetic disorder that causes severe anemia. The homozygous condition (TmTm) leads to severe anemia. People with thalassemia die before sexual maturity. The heterozygous condition (TmTn) causes a less serious form of anemia. The genotype TnTn causes no symptoms of the disease. Indicate the possible genotypes and phenotypes of the offspring if a male with the genotype TmTn marries a female of the same genotype.
9. For guinea pigs, black fur is dominant to white fur color. Short hair is dominant to long hair. A guinea pig that is homozygous for white and homozygous for short hair is mated with a guinea pig that is homozygous for black and homozygous for long hair. Indicate the phenotype(s) of the F1 generation. If two hybrids from the F1, generation are mated, determine the phenotypic ratio of the F2 generation.
10 Black coat color (B) in cocker spaniels is dominant to white coat color (b). Solid coat pattern (S) is dominant to spotted pattern (s). The pattern arrangement is located on a different chromosome than the one for color, and its gene segregates independently of the color gene. A male that is black with a solid pattern mates with three females. The mating with female A, which is white, solid, produces four pups: two black, solid, and two white, solid. The mating with female B, which is black, solid, produces a single pup, which is white, spotted. The mating with female C, which is white, spotted, produces four pups: one white, solid; one white spotted; one black, solid; one black, spotted. Indicate the genotypes of the parents.
11. For human blood type, the alleles for types A and B are codominant, but both are dominant over the type O allele. The Rh factor is separate from the ABO blood group and is located on a separate chromosome. The Rh+ allele is dominant to Rh-. Indicate the possible phenotypes from the mating of a woman, type,O, Rh-, with a man, type A, Rh+.
12. In a disputed paternity case, a woman with blood type B has a child with type O, and she claimed that it had been fathered by a man with type A. What can be proved from these facts?
13. Multiple alleles control the intensity of pigment in mice. The gene D1 designates full color, D2 designates dilute color and D3 is deadly when homozygous. The order of dominance is D1 > D2 > D3. When a full color male is mated to a dilute color female, the offspring are produced in the following ratio: two full color to one dilute to one dead. Indicate the genotypes of the parents.
14. Multiple alleles control the coat color of rabbits. A gray color is produced by the dominant allele C. The Cch allele produces a silver-gray color when present in the homozygous condition, CchCch, called chinchilla. When Cch is present with a recessive gene, a light silver-gray color is produced. The allele Ch is recessive to both the full color allele and the chinchilla allele. The Ch allele produces a white color with black extremities. This coloration pattern is called Himalayan. An allele Ca is recessive to all the other alleles. The Ca allele results in a lack of pigment, called albino. The dominance hierarchy is C > Cch > Ch > Ca. The table below provides the possible genotypes and phenotypes for coat color in rabbits. Notice that four genotypes are possible for full color but only one for albino.
Phenotypes Genotypes
Full color CC, CCch, CCh, CCa
Chinchilla CchCch
Light gray CchCh, CchCa
Himalaya ChCh, ChCa
Albino CaCa
a) Indicate the genotypes and phenotypes of the F1 generation from the mating of a heterozygous Himalayan coat rabbit with an albino coat rabbit.
b) The mating of a full color rabbit with a light gray rabbit produces two full color offspring, one light gray offspring, and one albino offspring. Indicate the genotypes of the parents.
c) A chinchilla color rabbit is mated with a light gray rabbit. The breeder knows that the light grey rabbit had an albino mother. Indicate the genotypes and phenotypes of the F1 generation from this mating.
d) A test cross is performed with light gray rabbit, and the following offspring are noted: five Himalayan color rabbits and five light gray rabbits. Indicate the genotype of the light-gray rabbit.
15. A geneticist notes that crossing a round shaped radish with a long shaped radish produces oval shape radishes. If oval radishes are crossed with oval radishes, the following phenotypes are noted: 100 long, 200 oval, and 100 round radishes. For the cross between a pure-breeding round and a pure-breeding long radish, use symbols to show the F1 and F2 generations.
16. Palomino horses are known to be caused by the interaction of two different genes. The allele Cr in the homozygous condition produces a chestnut, or reddish color, horse. The allele Cm produces a very pale cream color, called cremello, in the homozygous condition. The palomino color is caused by the interaction of both the chestnut and cremello alleles. Indicate the expected ratios in the F1 generation from mating a palomino with a cremello.
17 In mice, the gene C causes pigment to be produced, while the recessive gene c makes it impossible to produce pigment. Individuals without pigment are albino. Another gene, B, located on a different chromosome, causes a chemical reaction with the pigment and produces a black coat color. The recessive gene, b, causes an incomplete breakdown of the pigment, and a tan, or light-brown, color is produced. The genes that produce black or tan coat color rely on the gene C, which produces pigment, but are independent of it. Indicate the phenotypes of the parents and provide the genotypic and phenotypic ratios of the F1 generation from the following crosses:
a) CCBB x Ccbb (b) ccBB x CcBb (c) CcBb x ccbb (d) CcBb x CcBb
18. For ABO blood groups, the A and B genes are codominant but both A and B are dominant over type O. Indicate the blood types possible from the mating of a male who is blood type O with a female of blood type AB. Could a female with blood type AB ever produce a child with blood type AB? Could she ever have a child with blood type O?
19. Baldness (HB) is dominant in males but recessive in females. The normal gene (Hn) is dominant in females, but recessive in males. Explain how a bald offspring can he produced from the mating of a normal female with a normal male. Could these parents ever produce a bald girl? Explain your answer.
20. Use the phenotype chart (pedigree) at the right to answer the following questions.
a) How many children do the parents A and B have?
b) Indicate the genotypes of the parents.
c) Give the genotypes of M and N.
21. For the family outlined, draw a pedigree
a) Generation I : A male with blood type O marries a female with blood type A.
b) Generation II : They have 4 children. The first born is a male with A type blood, the second born is a girl with O type blood, and the last two children were identical twin boys.
c) Generation III : The last boy married a woman with type B blood. They had two girls, one with AB blood type and one with B blood type.