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x = cx = aa+1 If x = cx, then c = 1 aa = a2 if a2 + 1 = x, then a2 = x-1 short of parametrization, this is the answer for this equation, but if you doing advanced maths, then let x= t (teR) a2 = -1 +t (teR)
14-a=714= 7+aa= 7
It depends on the shape in question. Clearly it is not a straightforward shape since AA - conventionally, the distance of the point A from itself - is 33 rather than 0.
Here are the four possibilities: AA Aa Aa aa Therefore there is a 25% chance of producing a homozygous dominant offspring (AA).
p represents the square root of the frequency of the homozygous genotype AA.
+/- 11
If K(equilibrium constant) is greater than Q(concentration constant at a prticular point) then the reaction will tend to the right. If Q is less that K the reverse reaction will occur and if they are equal the reaction is at equilibrium. Example: aA+bB<--->cC+dD K=1.5 if Q<1.5 the reaction is aA + bB ---> cC + dD if Q> 1.5 the reaction is aA + bB <--- cC + dD K= [C]c[D]d/ [A]a[B]b at any point Q=[C]c[D]d/ [A]a[B]b at a particular point in time
AA
Heterozygotes. If AA X AA, or AA X AA, is all the mating allowed, then Aa will lose frequency in the population.
x = cx = aa+1 If x = cx, then c = 1 aa = a2 if a2 + 1 = x, then a2 = x-1 short of parametrization, this is the answer for this equation, but if you doing advanced maths, then let x= t (teR) a2 = -1 +t (teR)
Molarity of products divided by reactants Keq=(products)/(reactants)
aa+2 is an expression, not an equation. An expression has no answer.
The Hardy-Weinberg Equilibrium equation: p2 + 2pq + q2 = 1 p is frequency of dominant allele A q is frequency of recessive allele a p + q always equals 1 pp or p2 is probability of AA occurring qq or q2 is probability of AA occurring 2pq is probability of Aa occurring (pq is probability of Aa, qp is probability of aA, so 2pq is probability of all heterozygotes Aa) These add up to 1 because they represent all possibilities. The frequency of the homozygous recessive genotype
Yes an No. You can have blood type AB in following situations: AO+BO AO+BB AA+BO AA+BB AB+BB AA+AB AB+AB So there's more possible ways to get type AB other than the combination of ABs.
The Hardy-Weinberg Equilibrium equation: p2 + 2pq + q2 = 1 p is frequency of dominant allele A q is frequency of recessive allele a p + q always equals 1 pp or p2 is probability of AA occurring qq or q2 is probability of AA occurring 2pq is probability of Aa occurring (pq is probability of Aa, qp is probability of aA, so 2pq is probability of all heterozygotes Aa) These add up to 1 because they represent all possibilities. The frequency of the homozygous recessive genotype
aa+bb=cbc The a would be a 9, the b would be a 2 and the c would be a 1
14-a=714= 7+aa= 7