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AaBb usually, but can also be AAbb or aaBB depending on what they ask you is heterozygous. To be heterozygous for one trait, it's AaBb. To have a heterozygous genotype, unless specified, it can be any of the above.
This is called a dihybrid cross in which both parents are heterogeneous.
if a man were non- hemophiliac and he marries a woman whois homozygus for nan-hemophilia,give the possible genotypes of the children
Generally, if the parents are heterozygous and one allele is dominant over the other there are only 2 phenotypes and 3 genotypes. Parents Aa can produce AA, Aa and aa offspring. If the heterozygous individuals have an intermediate phenotype, then three genotypes and 3 phenotypes are possible. If 2 traits are being studied using heterozygous parents AaBb then the possible Genotypes are AABB, AABb, AAbb, AaBB, AaBb, Aabb, aaBb, aaBB, aabb which is nine genotypes. But there are 4 phenotypes. AABB AABb AaBB AaBb are phenotypically the same. aaBb, aaBB are phenotypically the same. Aabb, AAbb are phenotypically the same. aabb
AABB was created in 1947.
2 can
AaBb usually, but can also be AAbb or aaBB depending on what they ask you is heterozygous. To be heterozygous for one trait, it's AaBb. To have a heterozygous genotype, unless specified, it can be any of the above.
law of independent assortment b
If the genotype is AaBb then there will be 4 gametes produced: AB, Ab, aB, ab. Therefore the percentage of ab gametes is 25%
The most accurate description of an organism with genotype AaBb is heterozygous. A homozygous genotype is aaBB and AA.
This is called a dihybrid cross in which both parents are heterogeneous.
Ab and ab There would be about a 50/50 ratio of each.
The genotypes of this cross are:AA - 25%Aa - 50%aa - 25%The phenotypes of this cross are:Dominant trait (A) - 75%Recessive trait (a) - 25%A ratio of dominant to recessive phenotypes - 3:1
if a man were non- hemophiliac and he marries a woman whois homozygus for nan-hemophilia,give the possible genotypes of the children
Generally, if the parents are heterozygous and one allele is dominant over the other there are only 2 phenotypes and 3 genotypes. Parents Aa can produce AA, Aa and aa offspring. If the heterozygous individuals have an intermediate phenotype, then three genotypes and 3 phenotypes are possible. If 2 traits are being studied using heterozygous parents AaBb then the possible Genotypes are AABB, AABb, AAbb, AaBB, AaBb, Aabb, aaBb, aaBB, aabb which is nine genotypes. But there are 4 phenotypes. AABB AABb AaBB AaBb are phenotypically the same. aaBb, aaBB are phenotypically the same. Aabb, AAbb are phenotypically the same. aabb
The 9/3/3/1 ration is the ratio of phenotypes that are the result of a dihybrid cross. Consider two genes, A and B, that reside on different chromosomes (so that they independently assort). Assume each gene has two alleles. For A, A is dominant and a is recessive, while for the B gene, B is dominant and b is recessive. Now consider a cross between two individuals that are heterozygous for both genes (this is called a dihybrid cross): AaBb X AaBb There are only 4 possible gametes that each individual can produce (in equal proportion): AB Ab aB ab So if we cross the two we get 16 combinations. This will result in 9 possible genotypes: AABB AABb AAbb AaBB AaBb Aabb aaBB aaBb aabb However, there are only 4 possible phenotypes (with proportion in parentheses): Dominant A and B (9/16) (AABB, AABb, AaBB, AaBb) Dominant A, Recessive B (3/16) (AAbb, Aabb) Recessive A, Dominant B (3/16) (aaBB, aaBb) Recessive A, Recessive B (1/16) (aabb)
You need to make a Punnet Square Put A and a on top and B and B on the left side. These represent the possible gametes. Your results should be: AB AB Ba (aB) Ba (aB) So 50% of the genotype have the A allele.