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In the F2 generation, the possible gametes produced would be the result of the random assortment of alleles from the parents. Each parent can produce two types of gametes based on their genotype. For example, if the parents are AaBb and AaBb, the possible gametes would be AB, Ab, aB, and ab.
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.
There are four different gametes for XXHhuU: XHU, XHu, XhU, and Xhu.
No gametes are formed in asexual reproduction because they are only formed in sexual reproduction. Many cells can come from one cell in asexual reproduction which is really a cloning process, but not gametes.
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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.
The percentage of AB gametes produced by an AaBb parent is 25%. This is because during meiosis, the two alleles segregate independently, resulting in equal proportions of gametes with different combinations of alleles (AB, Ab, aB, ab). By calculating the possible combinations, we see that only 25% will be AB gametes.
In the F2 generation, the possible gametes produced would be the result of the random assortment of alleles from the parents. Each parent can produce two types of gametes based on their genotype. For example, if the parents are AaBb and AaBb, the possible gametes would be AB, Ab, aB, and ab.
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)
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Ab and ab There would be about a 50/50 ratio of each.
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
Four different phenotypes can be produced: AABB, AABb, AaBB, and AaBb. This is the result of different combinations of alleles from each parent in the offspring.
AABB was created in 1947.
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'
To determine the number of types of gametes each parent produces, you can use the formula ( 2^n ), where ( n ) is the number of heterozygous gene pairs. In problem number 1, if both parents are heterozygous for one trait (Aa), each will produce 2 types of gametes (A and a). In problem number 2, if each parent is heterozygous for two traits (AaBb), they will produce 4 types of gametes (AB, Ab, aB, ab).