You could try breeding it with a homozygous recessive partner (hh)
Lets assume that you breed the original mystery rabbit with an hh recessive partner, and they have 10 offspring.
If the original rabbit is homozygous dominant, it would be HH + hh, which would give all 10 the offspring Hh genotypes, which would give them the dominant hair color.
If it was heterozygous dominant, it would be Hh + hh, which would lead to either Hh or hh offspring. This means that in theory, 5 would be dominant colored while the other 5 would not be.
A breeder could determine if a rough-coated guinea pig is homozygous or heterozygous by conducting a test mating. Breeding the rough-coated guinea pig with a smooth-coated guinea pig would show if the rough coat is dominant or recessive. If all the offspring have rough coats, it indicates the rough-coated guinea pig is homozygous for the trait. If both rough and smooth-coated offspring are produced, then the rough-coated guinea pig is heterozygous.
If both parents are black-haired guinea pigs and the black coat color is dominant, they could either be homozygous (BB) or heterozygous (Bb) for the black coat gene. If both are heterozygous (Bb), the offspring ratio would typically be 25% homozygous black (BB), 50% heterozygous black (Bb), and 25% brown (bb). Therefore, among the 20 offspring, we would expect around 15 to be black (BB or Bb) and about 5 to be brown (bb) if the parents are Bb. If both parents are homozygous (BB), all offspring would be black.
The phenotype of the homozygous black guinea pig would be black hair. Since black is dominant to white in guinea pigs, having two copies of the black allele (BB) would result in the guinea pig displaying the black hair phenotype.
100% because BB is dominant over bb and all the crosses make Bb
No idea i am guessing it is kinda like people. if a black and a white cross it is a 50/50 chance the child will be black or white so its unpredictable it was a 50/50 chance the baby pigs would be blacks or albinos. hope this helps YOU ARE SO RACIST!!!!!
A breeder could determine if a rough-coated guinea pig is homozygous or heterozygous by conducting a test mating. Breeding the rough-coated guinea pig with a smooth-coated guinea pig would show if the rough coat is dominant or recessive. If all the offspring have rough coats, it indicates the rough-coated guinea pig is homozygous for the trait. If both rough and smooth-coated offspring are produced, then the rough-coated guinea pig is heterozygous.
To determine the hair color of the guinea pigs' offspring, we need to know the genotypes of the parents. If one parent is homozygous dominant (BB) and the other is homozygous recessive (bb), all offspring will be heterozygous (Bb) and will have black hair. If both parents are heterozygous (Bb), approximately 25% of the offspring are expected to be homozygous dominant (BB), 50% heterozygous (Bb), and 25% homozygous recessive (bb), resulting in a 75% chance of black hair and a 25% chance of white hair.
A test cross with a homozygous recessive guinea pig (bb) would reveal the genotype of the black guinea pig. If all offspring are black, then the black guinea pig is most likely homozygous dominant (BB). If both black and white offspring are produced, then the black guinea pig is likely heterozygous (Bb).
When a heterozygous guinea pig (Rr) is mated with a homozygous recessive guinea pig (rr), the possible genotypes of their offspring are 50% Rr (heterozygous, rough coat) and 50% rr (homozygous recessive, smooth coat). This results in a genotype distribution of 1:1, with half of the offspring expected to have rough coats and the other half having smooth coats.
If a homozygous black guinea pig (BB) is crossed with a homozygous white guinea pig (bb), all offspring will inherit one black allele (B) from the black parent and one white allele (b) from the white parent, resulting in heterozygous offspring (Bb). Since black fur is dominant over white fur, all offspring will have black fur. Therefore, the probability of an offspring having black fur is 100%.
The phenotype "brown fur" can be associated with two possible genotypes: homozygous dominant (BB) and heterozygous (Bb). The phenotype "white fur," however, can only be paired with the homozygous recessive genotype (bb). Therefore, any pairing that suggests a genotype of BB or Bb with white fur would be incorrectly matched.
If both parents are black-haired guinea pigs and the black coat color is dominant, they could either be homozygous (BB) or heterozygous (Bb) for the black coat gene. If both are heterozygous (Bb), the offspring ratio would typically be 25% homozygous black (BB), 50% heterozygous black (Bb), and 25% brown (bb). Therefore, among the 20 offspring, we would expect around 15 to be black (BB or Bb) and about 5 to be brown (bb) if the parents are Bb. If both parents are homozygous (BB), all offspring would be black.
here is my answer found using a Punnett Square: for one trait (lets use hair color) both guinea pigs are hazel. But, because they are heterozygous they also have the resessive gene for white. So, you cross Hh with Hh. You then have this genotypic ratio: 1 HH:2 Hh:1 hh This can be translated into a phenotypic ratio: 3 hazel: 1 white The final answer: You will have 25% homozygous dominant, 50% heterozygous (showing the dominant), and 25% homozygous recessive.
The side because they are the same. The top because they are different.
The phenotype of the homozygous black guinea pig would be black hair. Since black is dominant to white in guinea pigs, having two copies of the black allele (BB) would result in the guinea pig displaying the black hair phenotype.
The offspring will look more like the mom but they will have mid hair.
100% because BB is dominant over bb and all the crosses make Bb