Answer:
1.31 cM
Explanation:
Total offspring = 2205
Since two genes are involved, F1 progeny should have four types of combination. Out of them two are 17 and 12 which definitely means they are in lesser number. Since recombinants are always less than parental progeny in linkage, the given two types are recombinants.
Recombination frequency = (Number of recombinants / Total progeny) * 100
= [ ( 17 + 12 ) / 2205 ] * 100
= ( 29 / 2205 ) * 100
= 1.31 %
Map distance = Recombination frequency
Hence, distance between two genes = 1.31 cM
Most likely two dominant recessive genes. For example - pp or one parent could have Pp
Plants absorb ammonium and nitrate during the assimilation process, after which they are converted into nitrogen-containing organic molecules, such as amino acids and DNA
The period you’ve described would be the G2 phase which is considered a “checkpoint” that determines whether or not a cell can enter the mitosis phase
