Rr heterozygous
You can be a carrier of a recessive gene as part of your genotype.
She is a carrier of hemophilia but does not have the condition
If an AC genotype man and an AS genotype woman marry, each of their children has a 25% chance of being born with a sickle cell disease (SS genotype), a 50% chance of being a carrier like their parents (AS genotype), and a 25% chance of having a normal genotype (AA).
I think no. It's caused by a recessive gene so if a person has albinism their genotype can only be recessive, recessive ---> AA (small a small a or whatever you call it). No other genotype will mean that person has Albinism, like if it has at least one big A, that means it's just a carrier of albinism.
As long as a persons genotype consists of at least one recessive gene, they can pass it on to offspring to give them the disease, which makes them a carrier. Since this persons genes are both recessive, it is definite that they will pass on the recessive gene.
A genetic carrier has a dominant and a recessive version of an allele. Normally, the term genetic carrier is used in relation to genetic illnesses where two copies of the recessive allele cause that illness. Therefore, a carrier does not have the illness themself (as the dominant, non-disease allele is expressed over the recessive allele). However, they have the ability to create an offspring who has the double recessive genotype and therefore has the condition if they mate with another carrier or someone who is double recessive (who has the disease).
There is not really enough information in the question to answer it properly. You would need to know what trait the D and d stood for to give a proper answer.If the dog's genotype is Dd and it has the disease, this suggests that the disease is caused by a dominant allele. A healthy bitch would have the genotype dd and half of her pups would have the disease. There is no mate which would produce all heathly pups.If the dog is symptom free then the disease is caused by a recessive allele. In which case the dog should be matted with a dog with the genotype DD to have 100% symptom free pups. Mating Dd to Dd would give 75% symptom free pups and 25% sufferers. Mating the Dd dog to a dd dog would give 50% disease free pups and 50% sufferers.
If their genotype contains both a dominant and a recessive allele for a trait.
The host doesnt show symptoms and the carrier does
A Punnett square for cystic fibrosis would involve crossing two parents who are carriers of the recessive allele for the disease (denoted as "cf"). The square would show the probability of having a child with cystic fibrosis (25%), a carrier (50%), or unaffected by the disease (25%). Each parent would have a genotype of "Cf" (carrier) for the Punnett square.
This person is called a carrier. Some carriers do not show any signs of the disease but others do (as someone with a cold).