Normally when one studies an enzymatic reaction you need to do the following:
1. The initial rate of the reaction - or initial velocity - Vo - (when you've just combined the enzyme and the substrate) - you measure the amount of product made per unit time right at the beginning of the reaction.
2. You have to do this for each concentration of substrate - this would be a range of substrate concentrations.
3. You plot these on a graph with (substrate, Vo) as your (x,y)
What you will see is a graph that rises from 0,0 and eventually plateaus.
The plateau indicates that the enzyme is saturated with the substrate and is working as fast as it can. This is the maximum velocity or Vmax.
Sun, grass, grass-hopper, bird, snake,hawk, wolf.
Is very necessary in the development of multiceular organisms. This is so because <span>the complexity of animals and plants depends on a remarkable feature of the genetic control system. Cells have a memory: the genes a cell expresses and the way it behaves depend on the cell's past as well as its present environment. Hope this info has given you the answer you need </span>
If we name the gene for hands and fingers with A, the possible genotypes are Aa and AA for malformed hands with shortened fingers and aa for normal hands and fingers.
If we name the gene for hair with B, the possible genotypes are Bb and BB for woolly hair and bb for normal hair.
A woman with normal hands and non-woolly hair has a genotype aabb while her husband who has malformed hands and woolly hair might have AaBb, AaBB, AABb or AABB genotypes.
Since their first child has normal hands and non-woolly hair:aabb we can conclude that he inherited half of recessive alleles from mother, and other half from father. That means that father in his genotype contain both recessive alleles: AaBb
P: aabb x AaBb
F1: (table)
4/16 or ¼ is the possibility that one child will be with normal hands and woolly hair and ¼* 1/4 is the possibility that both of them will be with normal hands and woolly hair