Four genes, M N O and P are linked on the same chromosome.

The basic assumption about crossing over is that the greater percentage of crossing over, the farther apart the genes are. In this case:

The closest genes are M-O (which is 8%) and O-P(which is 10%).
We also know that M-P is 18%.

So from this, we can conclude that the genes are lined up as:
MOPN/NPOM or NMOP/POMN.

We also know that N-O (which is 28%) is farther apart than N-M (which is 20%)

Hence, the only order possible is: NMOP / POMN

The linear order of the genes on the chromosome is, therefore, C.

For the first question it can't be B

m and o are closest:

M-O

p is close to o and must be on the far end of o from m (not between o and m) because their is more recombination between m and p then between m and o s:

M-O-P

This eliminates answer C and D. Answer A and B are the same answer because genes are ordered in relation to each other, there isn't a first or last, just one end, middle genes, other end (unless you are including information about the centromere which this problem doesn't). C could also be correct if this was a very small piece of circular DNA but again, this question doesn't allow for that and furthermore it wouldn't be a better answer then A & B.

Thank you so much !

That's not correct, Arkadion. The question is not talking about circular DNA. The question specifies linked genes on a single chromosome. The question further specifies that it is asking for the linear order of the genes.

The only answer that can be reasonably inferred from the crossover data (and I know we can argue for any of the choices by assuming a variety of inhibitory mechanisms/regions/etc...) is choice C. It has the N-M-O-P sequence. None of the other choices have this.

Good call, bioboi.