The genotypes of this cross are:
The phenotypes of this cross are:
Well, if you use a punnent square, all you would have to do is put Aa on the top and AA on the bottom.
A a
a Aa AA
a Aa AA
So as you can see there is a 50 % chance the off spring will be Aa and a 50% chance the off spring will be AA.
Gamete possibilities from parent 1. aB AB
Gamete possibilities from parent 2. AB Ab aB ab
Genotypes AaBB AaBb aaBB AaBb AABB AABb AaBB AaBb
Ratio: 1 AABb 2 AaBb 1 AABB 1AABb 1AaBB 1aaBB 1aaBb
Phenotype: Since there is no information on what A or B represent
6 dominant appearance for both traits
2 recessive for trait A and dominant for trait B
Genotype: AA AA Aa Aa 50% Homodominate 50%Hetro
Phenotype: 100% Dominate charicteristics
The phenotypic ratio for the cross AaBb and AaBb is 1/4 AaAa, 1/2 AaBb, and 1/4 BbBb. This can be determined by using a Punnett Square.
AA:Aa:aa in the ratio of 1:2:1 in that order
1:2:1
100%
The expected genotypic ratios for a dihybrid cross is 9:3:3:1, and for a monohybrid cross is 1:2:1.
Gk, gk
If an individual is of the dominant phenotype, its genotype can be either AA or Aa (letter A picked for demonstration's sake). To do a good test cross, we have to be able to unambiguously tell which one of the genotypes the individual is. The best test cross would be conducted with a homozygous recessive (aa) individual. This way, if the cross generates 100% dominant phenotype, then the original individual was homozygous dominant. If the cross generates 50% dominant and 50% recessive phenotype, then the original individual was heterozygous.
Gregor Mendel devised the fundamental tool of the test cross. It is an experimental cross of an individual organism of dominant phenotype but unknown genotype and an organism with a homozygous recessive genotype .
To determine the genotype of an individual that shows the dominant phenotype you would cross that individual with one that is homozygous recessive. A monohybrid cross of two individuals that are heterozygous for a trait exhibiting complete dominance would probably result in a phenotype ratio is 3 dominant 1 recessive.
The phenotypic rationof a dihybrid cross is 9:3:3:1
RrYy
The expected genotypic ratios for a dihybrid cross is 9:3:3:1, and for a monohybrid cross is 1:2:1.
A test cross.
possible mendelian ratios for monohybrid cross genotype is 1:2:1 and phenotype is 3:1
normal sexual cross pollination - each parent plant contributes traits from its genotype (which are expressed as the phenotype)
A punnett square is the diagram used to determine the expected genotypic ratios for the offspring.A dihybrid cross is a cross involving two different traits. For example RrDd X RrDd would be a dihybrid cross. You could use a punnett square to determine the expected ratios for this cross:RDRdrDrdRDRRDDRRDdRrDDRrDdRdRRDdRRddRrDdRrddrDRrDDRrDdrrDDrrDdrdRrDdRrddrrDdrrddOne parent's genotype is shown across the top, the other down the side (both in bold).
Gk, gk
For monohybrid cross the genotype ratio in f2 generation would be 1:2:1 and phenotype ratio would be 3: 1
If an individual is of the dominant phenotype, its genotype can be either AA or Aa (letter A picked for demonstration's sake). To do a good test cross, we have to be able to unambiguously tell which one of the genotypes the individual is. The best test cross would be conducted with a homozygous recessive (aa) individual. This way, if the cross generates 100% dominant phenotype, then the original individual was homozygous dominant. If the cross generates 50% dominant and 50% recessive phenotype, then the original individual was heterozygous.
This question is worded weird, because if someone had a dominant phenotype, as in Brown eyes dominant, than they would be BB or Bb. The exact genotype would need to have a key to what one of their parents' genotype was. Example: If a man with brown eyes (dominant) had a mom with blue eyes (recessive) then what would the man's genotype be? the answer would be Bb because that means that the man's dad would have BB and mom would be bb and when using punnent square, his genotype is Bb (only possible genotype is Bb)
Try using a Punnett Square.