Ab and ab
There would be about a 50/50 ratio of each.
one type of gamete will be formed having genotype Ab
2 can
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%
Each time a gamete is formed, one allele from each gene (i.e. either A or a) is included.This means for an individual with AaBbCc, there are 8 different gametes:ABCABcAbCAbcaBCabCaBcabc
The most accurate description of an organism with genotype AaBb is heterozygous. A homozygous genotype is aaBB and AA.
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
2 can
law of independent assortment b
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.
If the genotype is AaBb then there will be 4 gametes produced: AB, Ab, aB, ab. Therefore the percentage of ab gametes is 25%
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.
Each time a gamete is formed, one allele from each gene (i.e. either A or a) is included.This means for an individual with AaBbCc, there are 8 different gametes:ABCABcAbCAbcaBCabCaBcabc
4
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)
if a man were non- hemophiliac and he marries a woman whois homozygus for nan-hemophilia,give the possible genotypes of the children
A gamete is haploid (1N) so 'Aa' & 'AA' are diploid and during cell division (mitosis) gametes are formed and then 2 gametes merge together to make a diploid (think of sperm and egg, each is haploid or 1N, when fertilization occurs the egg and sperm form 1 cell that is 2N or diploid). So the possible gametes for 'Aa' would be 'A' & 'a' while for 'AA' the only gametes possible are 'A' If the question is asking what the possible gametes are for 'AaBB' the haploid (gamete) can be 'AB' or 'aB'
The most accurate description of an organism with genotype AaBb is heterozygous. A homozygous genotype is aaBB and AA.
The two types of gametes that could result from the AABb allele combination are AB and Ab. This is because during meiosis, homologous chromosomes separate and randomly assort, leading to different combinations of alleles in gametes.