Answer:
The probability that their first child will both be ab/ab = 41%.
Explanation:
Available data:
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 recombination in total Â
              = % aB + % Ab Â
              = 9% aB + 9% Ab
    100% - 18% = 82% of parental in total Â
              = % of AB + % ab Â
              = 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) Â Â AB ( 41%) Â Â Â Ab (9%) Â Â Â aB (9%) Â Â Â 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
