Answer:
The probability that their first child will both be ab/ab = 41%. 
Explanation:
<u>Available data:</u>
- Two linked genes  (A) and (B) ⇒ 18 centiMorgans apart
- Man ⇒  Aa Bb
- Man´s father ⇒ AA BB
- Woman ⇒ aabb
The recombination frequency is given by the distance between genes. We have to know that 1% of recombinations = 1 map unit = 1cm.
Recombination frequency = 0.18 
Man: AaBb
Gametes) AB Parental
                  ab Parental
                  Ab Recombinant
                  aB recombinant
18 centi morgan = 18 % of <u>recombination</u> in total  
                             = % aB + % Ab  
                             = 9% aB + 9% Ab
        100% - 18% = 82% of <u>parental</u> in total  
                            = % of AB + % ab  
                            = 41% AB + 41% ab
The frequency for each gamete is:
AB 41%
ab 41%
Ab 9%
aB 9%
Cross: man x woman
Parental)          AaBb           x          aabb
Gametes) AB, Ab, aB, ab        ab, ab, ab, ab
Punnett square)     AB ( 41%)       Ab (9%)      aB (9%)       ab (41%)
                    ab      AaBb               Aabb          aaBb            aabb 
                    ab      AaBb               Aabb          aaBb            aabb
                    ab      AaBb               Aabb          aaBb            aabb
                    ab      AaBb               Aabb          aaBb            aabb
F1) 41% of the progeny is expected to be AaBb
      9% of the progeny is expected to be Aabb
      9% of the progeny is expected to be aaBb
      41% of the progeny is expected to be aabb