Monohybrid (single-trait inheritance) Problems
1. a) Several black guinea pigs of the same genotype were mated and produced 29 black and 9 white
offspring. If black is the dominant allele and white is the recessive allele, what would you predict
the genotype of the black parent guinea pigs to be? Explain your choice.
Both parents are most likely heterozygous (black)
b) If a black female guinea pig is test-crossed and produces at least one white offspring, determine…
i)
the genotype and phenotype of the sire (male parent) which produced the offspring.
the male parent must be homozygous recessive (white)
ii)
the genotype of the female parent
female parent must be heterozygous
c) Heterozygous black guinea pigs are mated to white guinea pigs, predict the genotypic and
phenotypic ratios expected from crossing the F1 offspring to…
i)
the black parent
Black parent x heterozygous (Bb) offspring
F2 genotypes
¼ homozygous dominant
½ heterozygous
¼ homozygous recessive
F2 phenotypes
¾ black
¼ white
Black parent x homozygous recessive (bb) offspring
F2 genotypes
½ heterozygous
½ homozygous recessive
F2 phenotypes
½ black
½ white
ii)
the white parent
white parent x heterozygous (Bb) offspring
F2 genotypes
½ heterozygous
½ homozygous recessive
F2 phenotypes
½ black
½ white
white parent x homozygous recessive (bb) offspring
F2 genotypes
all homozygous recessive
F2 phenotypes
all white
2. In Drosophila, sepia coloured eyes are due to a recessive allele e and wild-type (red eye colour) to
its dominant allele E. If sepia eyed females are crossed to pure-breeding wild-type males, what
phenotypic and genotypic ratios are expected if the F1 males are mated to the sepia eyed parental
females?
F1 genotypes: all heterozygous
F1 phenotypes: all wild-type eyes
F2 genotypes: ½ heterozygous
½ homozygous recessive
F2 phenotypes: ½ wild-type eyes
½ sepia
3. The lack of pigmentation, called albinism, in man is the result of a recessive allele (a) and normal
pigmentation is the result of its dominant allele (A). Two normal parents have an albino child.
Determine the probability that
both parents must be heterozygous (Aa)
i) the next child is albino ¼ or 25%
ii) the next two children are albino 1/16
iii) the couple have two more children, the first child has albinism and the second is normal 3/16
4. Short hair is due to a dominant gene (L) in rabbits, and long hair to its recessive allele (l). A cross
between a short haired female and a long haired male produced a litter of 1 long haired and 7
short haired bunnies.
a) what are the genotypes of the parents?
female must be heterozygous (Ll); male must be homozygous recessive (ll)
b) what phenotypic ratio was expected in the offspring generation?
F1 expected phenotypes:
½ short hair
½ long hair
c) How many of the 8 bunnies were expected to be long haired? 4
5. A dominant gene H produces wired-haired texture in dogs; its recessive allele h produces smooth
hair. A group of heterozygous wire-haired individuals are crossed and their F1 offspring are testcrossed. Determine the expected genotypic and phenotypic ratios among the test-cross offspring.
F1 genotypes:
¼ homozygous dominant
½ heterozygous
¼ homozygous recessive
F1 phenotype:
¾ wired-haired
¼ smooth haired
6. In foxes, silver-black coat colour is governed by a recessive allele b and red colour by its dominant
allele B. Determine the genotypic and phenotypic ratios expected from the following matings…
a) pure-breeding red x heterozygous red
F1 genotypes:
½ homozygous dominant
½ heterozygous
F1 phenotypes:
All red coat
b) heterozygous red x silver-black
F1 genotypes:
½ homozygous recessive
½ heterozygous
F1 phenotypes:
½ silver-black coat
½ red coat
c) pure-breeding red x silver-black
F1 genotypes:
All heterozygous
F1 phenotypes:
All red coat