Answer:
BRYAN!!! stop cheating this is your teacher if you are caught again you will be getting a message home and a suspension
Explanation:
So we’ll just use “R” and “r” for this example. If the mother AND father are heterozygous, then both of their genotypes are “Rr” if you work out the lumber square or use the foil method, the box would look like this: RR on top left, Rr on top right, Rr on bottom left, and rr on bottom right. So the genetic probabilities, using four as the sum would be 1:2:1
Answer:
D) B/b;S/s (x) b/b;s/s
Explanation:
Parent 1 : belted syndactylous sow
Since it is showing dominant phenotype for both the traits, it can either be BBSS or BbSs
Parent 2: unbelted cloven-hoofed
Since it is showing recessive phenotype for both the traits, it can only have bbss genotype
If we assume parent 1 to be BBSS all the resulting progeny with bbss will have dominant phenotype which is not the case.
If we assume parent 1 to be BbSs:
BbSs X bbss =
BbSs : belted syndactylous
bbSs : unbelted syndactylous
Bbss : belted cloven
bbss : unbelted cloven
The progeny will be produced in 1:1:1:1 ratio which means that each of them will make 25% of the population.
Hence, parent 1 will have BbSs genotype and parent 2 will have bbss genotype
Answer:
the diagram represents DNA
They differ in single trait
