1. Galactosemia is a disease that will only be expressed when a person is<span><span><span> homozygotic recessive for that trait. It's the same as saying it </span> has</span> 2 recessive alleles.
Dominant allele-</span><span> G
recessive allele- g
</span>
Homozygotic dominant: GG <span><span>(doesn't express the disease)
</span>Heterozygotic : Gg (doesn't express the disease)
Homozygotic recessive: gg (expresses it)</span>
2.
-Mary has this genotype: G_ . This means it can be GG or Gg
-The exercise already says that justin's mother is GG (<span>Homozygotic dominant)
</span>- If his mother is GG, one of these G's is going to be passed to Justin. So, his genotype is either GG or Gg. Since we are not sure we write as: G_.
Justin's genotype: G_
3.
-Justin's uncle has galactosemia so his genotype is: gg
-If the uncle was able to receive two recessive alleles it means the mother had one to pass, and so did the father. However, in the diagram, it's not pointed out that they have a disease so it only leaves one possible genotype: Gg. Justin's grandparents are both Gg.
4. The last person to analyze is Justin's father.
If we crossed the grandparents (Gg x Gg) we could obtain these genotypes: GG, Gg, gg.
Justin's father doesn't express the trait, so it's not gg. That leaves us with either GG or Gg. Since we can't know for sure, onece again we write as G_
Justin's father: G_
Answer:there is a cell division as it occurs every 20min
I think it would be 50% because it could land on heads just as much as it lands on tails. its a even chance it could land on whichever it chooses to.
The heads of phospholipids are hydrophilic while the tails are hydrophobic
