Answer:
2 out of 4
Explanation:
Let the allele for cleft chin be represented by A, while the alternate recessive version would be a.
A parent that is heterozygous for cleft chin would have the genotype Aa.
A parent without cleft chin would have the genotype aa.
Aa x aa
Aa Aa aa aa
Aa - cleft chinned = 2/4
aa - normal chin = 2/4
Thus, two out of every four (2/4) of their offspring would have cleft chin.
Animal cells do not have cell walls because they do not need them. Cell walls, which are found in plant cells, maintain cell shape, almost as if each cell has its own exoskeleton.
Answer:
1 x 10^13 stadiums will be needed in this scenario
Explanation:
We are told that
1 stadium holds = 1 × 10^5 people and
Number of iron atoms = 1 × 10^18 atoms
If the stadium carries an equivalent number of atoms as that of people.
We can infer that 1 stadium will carry 1 × 10^5 atoms.
The calculation to determine the number of stadiums would then be 1 × 10^18 divided by 10^5 atoms/stadium which was gotten by dividing the total number of atoms by the number of atoms per stadium.
Number of stadiums = Total number of atoms ÷ Number of atoms per stadium
= 1 × 10^18 atoms ÷ 1 × 10^5 atoms/stadium
= 1 × 10^13 Stadiums
This means that 1 × 10^18 atoms would occupy 1 × 10^13 stadiums
