Oh, an albino has a special genotype that gives them their unique appearance. You see, albinism is typically caused by a mutation in the genes responsible for producing melanin, the pigment that gives color to our hair, skin, and eyes. So, the genotype of an albino would usually involve inheriting two copies of the mutated gene that leads to a lack of melanin production.
For the mother to have a non-albino offspring, she would need to have at least one dominant allele for the trait, so her genotype could be either Aa or AA. The father must have two recessive alleles (aa) to pass on albinism to the offspring. The albino offspring's genotype would be aa.
The albino woman has to be AA. So the man has to be either AA or Aa in order to be normal. So if they have a baby and its albino it has to contain the AA gene. a a A Aa | Aa ---------------- A Aa | Aa doesn't work.. All outcomes are normal. a a A Aa | Aa ----------------- a AA | AA these combination outcomes 50% chance of an albino baby. This means the man is Aa Heterozygous dominant.. the mother is AA homozygous recessive.. and the baby is the same as the mother. AA...
Both parents would each have one dominant gene for normal pigmintation and each would have one recessive albino gene. Say P is the dominant gene and q is the albino gene. Then the parents genotype would both be Pq. Below is a Punnet Square. The child's genotype is qq P q ------------------------------------- P / PP / Pq (carrier) / / / / ///////////////////////////////////////// q / Pq (carrier) / qq (albino) / / / / ////////////////////////////////////////
Albinism is a sex-linked trait. Accordingly, the description of "heterozygous male" doesn't make sense. The sex chromosomes for males are XY, and usually genetic disorders are passed down by the X chromosome (which is the case for albinism). So in the problem, we aren't sure whether the male is XaY or XAY. It would be more helpful to know if the male was albino or not; we would then immediately know his genotype. We can, however, figure out the genotype of the female. Because albinism is recessive, we can deduce that her genotyple is XaXa. But without knowing the male's genotype or phenotype, the offspring percentages cannot be calculated.
Albinism is a recessive trait, meaning that an albino person has the genotype AA, while a person who isn't albino has the trait Aa or AA. Two albino people (AA x AA) will have all albino children. Aa x AA will have a 50% probability of albino children, and 50% normal children who are carriers (Aa). AA x AA will have all normal children who are carriers (Aa). Aa x Aa will have 25% AA (normal/non carriers), 25% AA (albino), and 50% Aa (normal/carriers) offspring. AA x AA will have all normal children who are not carriers (AA).*A person who has the genotype Aa is a carrier because they carry the allele (a) for albinism but they are not albino. The allele (A) is dominant so it covers the (a) allele.
For the mother to have a non-albino offspring, she would need to have at least one dominant allele for the trait, so her genotype could be either Aa or AA. The father must have two recessive alleles (aa) to pass on albinism to the offspring. The albino offspring's genotype would be aa.
The albino woman has to be AA. So the man has to be either AA or Aa in order to be normal. So if they have a baby and its albino it has to contain the AA gene. a a A Aa | Aa ---------------- A Aa | Aa doesn't work.. All outcomes are normal. a a A Aa | Aa ----------------- a AA | AA these combination outcomes 50% chance of an albino baby. This means the man is Aa Heterozygous dominant.. the mother is AA homozygous recessive.. and the baby is the same as the mother. AA...
Each parent is heterozygous for the gene.
Both parents would each have one dominant gene for normal pigmintation and each would have one recessive albino gene. Say P is the dominant gene and q is the albino gene. Then the parents genotype would both be Pq. Below is a Punnet Square. The child's genotype is qq P q ------------------------------------- P / PP / Pq (carrier) / / / / ///////////////////////////////////////// q / Pq (carrier) / qq (albino) / / / / ////////////////////////////////////////
Albinism is a sex-linked trait. Accordingly, the description of "heterozygous male" doesn't make sense. The sex chromosomes for males are XY, and usually genetic disorders are passed down by the X chromosome (which is the case for albinism). So in the problem, we aren't sure whether the male is XaY or XAY. It would be more helpful to know if the male was albino or not; we would then immediately know his genotype. We can, however, figure out the genotype of the female. Because albinism is recessive, we can deduce that her genotyple is XaXa. But without knowing the male's genotype or phenotype, the offspring percentages cannot be calculated.
There is a very low percentage of having a albino child if you yourself are not albino ==you can have an albino baby==
Temperature affects the C gene codes for the enzyme tyrosinase, the first step in pigment production. The recessive alleles determine whether a cat is a complete albino or a temperature sensitive albino. These genes affect the color of the cats eyes and coat.
Albinism is a recessive trait, meaning that an albino person has the genotype AA, while a person who isn't albino has the trait Aa or AA. Two albino people (AA x AA) will have all albino children. Aa x AA will have a 50% probability of albino children, and 50% normal children who are carriers (Aa). AA x AA will have all normal children who are carriers (Aa). Aa x Aa will have 25% AA (normal/non carriers), 25% AA (albino), and 50% Aa (normal/carriers) offspring. AA x AA will have all normal children who are not carriers (AA).*A person who has the genotype Aa is a carrier because they carry the allele (a) for albinism but they are not albino. The allele (A) is dominant so it covers the (a) allele.
Are you Albino?
I think it's genotype...
Because their albino....
a shark thats an albino