Blake is correct about the genetics in my post. Sorry about my error.
Let’s take from the top: Everyone normally has two copies of each gene (except males for those genes on the sex chromosomes), one of which is inherited from the mother and the other from the father.
Assume that there are three genes for skin color in human, let’s call them A, B, and C.
Each of these genes has two forms, or alleles, one of which produces little melanin (let’s call it the “light” allele) and the other of which produces a lot of melanin (let’s call it the “dark” allele). Let’s indicate the dark allele with a capital letter (A,B.C), and the light allele with a small letter (a,b,c).
Unlike the main genes for eye color, these alleles do not show a dominant /recessive pattern of inheritance, in which you can have either brown eyes (which are dominant, so that it only takes one brown allele to produce brown eyes, even if the other allele is for blue), or blue eyes (which are recessive, so that you must have two blue alleles in order to produce blue eyes), but not an intermediate condition. (I am ignoring some of the complexities here that can produce hazel or green eyes.) Instead, the skin color genes show what is called co-dominance, in which those individuals with differemt alleles show an intermediate condition.
AA = dark
aa = light
Aa = medium-colored, intermediate between the two extremes
If we assume that the effects of the three different genes are additive, we end up with seven different possible degrees of skin color:
6 dark alleles (AABBCC) = darkest
5 dark (e.g. AaBBCC and other possibilities) = next darkest
4 dark (AaBbCC, etc) = slightly darker than average
3 dark (AaBbCc, etc) = intermediate between the extremes
2 dark (aaBbCc, etc) = slightly lighter than average
1 dark (aabbCc, etc) = slightly darker than the lightest
0 dark (aabbcc) = lightest
Right. If both are aabbcc, then there are no dark alleles present, so it’s impossible for them to have a darker child.
If you have a pairing of an aabbcc (white) with an Aabbcc (lightest skinned mulatto), the only possible results are aabbcc (white) and Aabbcc (same color as mulatto parent). The offspring cannot be darker than the mulatto parent. This is true of any case in white one parent is aabbcc. Any dark alleles that are present must come from the mulatto parent, so that the offspring cannot be darker than the mulatto parent.
As an example, let’s take two very light skinned mulattos, both Aabbcc. There are then three possible genotypes for offspring:
aabbcc = white
Aabbcc = lightest mulatto
AAbbcc = darker mulatto, darker than either parent. This individual has obtained one dark allele from each of the parents.
Taking the example of two intermediate mulattos of genotype AaBbCc, they can produce all possible genotypes between them, including AABBCC (darkest), aabbcc (lightest) and all intermediate combinations. This is the point of the second quote in astro’s post.
Most likely, unless her kid really really does look like the milkman. 