It performs all functions for the organism.
Hope this helps!
Answer:
Don't have sex. 5 points isnt a lot so I am not spending alot of time on it.
Answer:
beginner at this what he proves to be true
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:
b. public containers is the answer
