Genetics Probability Problem solving!!!!?

Dark haired M in humans is dominant over blond hair m. Freckles F are dominant over no freckles f. If a dark haired, freckled man whose mother was blond and had no freckles marries a woman with dark hair and no freckles whose mother was also blond., what proportion of their children would have the following phenotypes? Blond with no freckles? Blond with freckles? Dark haired with freckles? Dark haired with no freckles?

I know that 's a lot and I'm sorry but I just don't get how to solve this... :( We CANNOT USE PUNNET SQUARES!!!! I think the man's genotype would be MmFf and the womans Mmff???? I don't know!!! Please help!!!! Show all work and explain every step! Thank you soooo much!!!!!

Genetics Probability Problem solving!!!!?

kinda funny. im doing the same stuff. ur in ap bio? i got a test on this tomorrow. first thing u gotta know the rules of multiplication in genetics. U r for the mans genotype. in case that was a guess, this is a bc his mother was a blond. regardless, he has the m allele. (his father could be MM or Mm. MM x mm equals all Mm. Mm x mm = 1/2 Mm 1/2 mm) we know the father has dark hair, this means he cant be mm; he must be Mm. This applies the same way for freckles and the womans dark hair. so what u got is MmFf x Mmff. Mm x Mm= 1/4 MM, 1/4 mm, 1/2 Mm. Ff x ff= 1/2 Ff, and 1/2 ff.

ok so a blond with no freckles is mmff. since mm= 1/4 and ff= 1/2, 1/2 x 1/4 = 1/8.

Blond hair w/ freckles = mm x F_. From our rules of multiplication and dominance, we know freckles is 3/4 (since freckles r dom, any allele that has it will be expressed, so 1/2 (which is form Ff) + 1/4 (from FF) = 3/4). mm= 1/4. 1/2 x 3/4 = 3/8.

the rest is:

M_F_= 3/4 x 1/2= 3/8

M_ff= 3/4 x 1/2= 3/8

hope i explained it well.