They must both be OB (heterozygous). * OB produces B type * BB produces B type * OO produces O type We project that 1/4 of offspring will be BB, 1/2 will be OB, and 1/4 will be OO. Since BB and OB have the same phenotype (blood type B), statistically 3/4 will be type B and 1/4 will be type O.
We are looking for the possible blood types of a baby.
Parental information:
HOWEVER: There is more to ABO blood typing that just the ABO gene.
There is also an inhibitory gene that will change any genotype into the phenotype O.
Therefore a person with genetically AB blood can be tested as having Type O.
If the Type O parent has the inhibitory gene affecting his Type A or AB,then a Type A baby is definitely possible.
There are two possibilities.... either group 'O' - or group 'B'. Also, each group can be rhesus positive or negative - making a total of four combinations.
So, if one parent is Aa (heterozygous) and the other parent is aa (homozygous recessive) the punnett square would look like this: ___|_A__|__a_ _a_|_Aa_|_aa_ _a_|_Aa_|_aa_ The genotypes of the offspring 50% heterozygous and 50% homozygous recessive
The genotype each offspring has is determined by the parents. They can be both homozygous dominant or are they heterozygous and homozygous(dominant or recessive)
There are 3 probabilities: dominant homozygous, recessive homozygous, or heterozygous.
Assuming there is no co-dominance or partial dominance, the result would be that 100% of the offspring would be blue, heterozygous flowers with the phenotype Bb.
Alright, I suppose I will do your homework for you.. Here is your punnet square: F F F FF FF f Ff ff Therefore, 3/4, or 75%, offspring will have the phenotype of having freckles, and 1/4, or 25% will have the phenotype of no freckles. And 2/4, or 50%, of the offspring will have the genotype for homozygous for freckles, 1/4, or 25%, of the offspring will carry a heterozygous trait for freckles, and 1/4, or 25%, of the offspring will have the phenotype for homozygous no freckles.
There only certain crosses that will produce heterozygous offspring. These are heterozygous vs heterozygous, homozygous vs homozygous and heterozygous vs homozygous.
So, if one parent is Aa (heterozygous) and the other parent is aa (homozygous recessive) the punnett square would look like this: ___|_A__|__a_ _a_|_Aa_|_aa_ _a_|_Aa_|_aa_ The genotypes of the offspring 50% heterozygous and 50% homozygous recessive
The genotype each offspring has is determined by the parents. They can be both homozygous dominant or are they heterozygous and homozygous(dominant or recessive)
There are 3 probabilities: dominant homozygous, recessive homozygous, or heterozygous.
If the parent generation consisted of a homozygous dominant parent and a homozygous recessive parent, then the F1 generation would be 100% heterozygous.
Assuming there is no co-dominance or partial dominance, the result would be that 100% of the offspring would be blue, heterozygous flowers with the phenotype Bb.
You can perform test crosses with individuals of known genotypes and analyse the offspring you get.
The probability is 50%. There are four probabilities: dominant homozygous, recessive homozygous, or heterozygous.
There are two forms of Homozygous inheritance: Homozygous Dominant, and Homozygous Recessive. In order for two parents that are Homozygous to produce a Heterozygous offspring, one of them MUST be Homozygous Dominant, and the other MUST be Homozygous Recessive.
Alright, I suppose I will do your homework for you.. Here is your punnet square: F F F FF FF f Ff ff Therefore, 3/4, or 75%, offspring will have the phenotype of having freckles, and 1/4, or 25% will have the phenotype of no freckles. And 2/4, or 50%, of the offspring will have the genotype for homozygous for freckles, 1/4, or 25%, of the offspring will carry a heterozygous trait for freckles, and 1/4, or 25%, of the offspring will have the phenotype for homozygous no freckles.
Impossible. You can only be heterozygous or homozygous, not both.
You get one homozygous dominant (TT), one homozygous recessive (tt), and two heterozygous (Tt).