Part 1 involved MeOH addition to form the key tetrahedral intermediate. Part 2 involves loss of H2O from the tetrahedral intermediate to form the ester. Part 2 of 2: Intermediate species (from Part 1):

For all remaining steps in the mechanism: 1) Draw each species (organic and inorganic) resulting from the previous step. Each step will include all of, and only, the atoms given in the first step (i. e., do not use extra reagent molecules/ions, and do not omit spectator ions).

(Depending on how you drew the mechanism, the drawing area below might not be needed. Do not draw the final products.)

the child will have heterozygous dominant chromosomes as characters of his father are dominant over the characters of his mother.

the child is heterozygous

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally