Answer:
The probability of their having a child with normal tooth enamel and color blindness is 3/8 or 0.375
Explanation:
Let the alleles for normal and defective tooth enamel be T and t respectively.
Color blindness is on the X-Chromosome. Let the allele normal vision and colour blindness be Xn and Xc respectively.
The woman had a colour blind father (XcY) and a normal vision mother, so her allelic pair for vision will be XnXc.
Her father had normal tooth enamel while her mother had a defective tooth enamel, therefore, she will have the allelic pair, Tt.
Her genotype for vision and tooth enamel will be XnXcTt.
Her husband who is color blind with normal tooth enamel had a father with defective tooth enamel, tt. Therefore, her husband's genotype for vision and tooth enamel will be XcYTt.
From the dihybrid cross of XnXcTt and XcYTt, the following offspring are obtained:
XnXcTT, XnXcTt, XnYTT, XnYTt
XnXcTt, XnXctt, XnYTt, XnYtt
XcXcTT, XcXcTt, XcYTT, XcYTt
XcXcTt, XcXctt, XcYTt, XcYtt
From the results of the cross, a child with normal tooth enamel and color blindness will have the following genotypes: XcXcTT, XcXcTt, XcYTT, XcYTt, XcXcTt, XcYTt.
Probability of their having a child with normal tooth enamel and color blindness = 6/16 = 3/8 or 0.375