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:
0.05 mg/mL ( B )
Explanation:
Given data:
20 mg/ml starch
2% solution = 2g of solute is in 100g of solvent
<u>Determine the new concentration in mg/ml </u>
Dilution equation = C1V1 = C2V2
new concentration ; applying the dilution factor
dilution factor = 1 : 400 ; ( 2 /400 )g = 0.005 g of solute is present in every 100 mL
∴ new concentration = 0.00005 g / 1 mL * ( 1000 mg / 1g ) = 0.05 mg/mL
The land underneath would slowly decay
