<span> Mitosis involves Prophase, Metaphase, Anaphase, and Telophase.
So you have
13 + 12 + 3 + 2 = 30 cells in mitosis
and
90 + 30 = 120 cells in total
Therefore your ratio of cells in mitosis to total cells is 30 / 120 = 1 / 4. That means mitosis should take up roughly 1/4 of the total cell cycle length. Since the completely cycle takes 24 hours, mitosis would then take (1/4)*24 = 6 hours</span>
Atomic number shows the number of protons in the nucleus of an atom.
Atomic weight shows the is an another term for atomic mass, which means <span>It is approximately equivalent to the number of protons and neutrons in the atom (the mass number) or to the average number allowing for the relative abundances of different isotopes.</span>
Answer:
The ability to bring water into the gills when the mouth is closed would benefit a bottom dwelling shark or ray, because it would allow them to be more discreet when in the process of catching prey and they would be able to stop swimming. Most sharks have to continuously swim or else they would sink, while being able to bring water into the gills when the mouth is closed would highly benefit them and allow them to stop swimming for a moment.
Explanation:
I am not 100% sure this is the correct answer, but I am currently doing a lab on this, so I am using the knowledge I previously had and online resources. I am pretty confident though.
Answer:
See the answer below
Explanation:
Let the disorder be represented by the allele a.
Since the disease is an autosomal recessive one, affected individuals will have the genotype aa and normal individuals will have the genotype Aa or AA.
Since the four adults are carriers, their genotypes would be Aa.
Aa x Aa
Progeny: AA 2Aa aa
Probability of being affected = 1/4
Probability of being a carrier = 1/2
Probability of not being affected = 3/4
(a) The chance that the child second child of Mary and Frank will have alkaptonuria = 1/2
(b) The chance that the third child of Sara and James will be free of the condition = 3/4
(c)
(d) If someone has no family history of the disorder, their genotype would be AA.
AA x aa
4 Aa
<em>The chance that a child with alkaptonuria will have an offspring with alkaptonuria if the child's mate has no family history </em>= 0
(e)
(f) <em>The chance that a child with alkaptonuria will have an offspring with alkaptonuria if the child's mate has no family history</em> = 0
