Answer:
C
Explanation:
Add radio labeled actin sub units to a mixture of actin filaments in which conditions are favorable for polymerization. This would enable you to identify the plus end of actin filaments.
Actin filaments are linear polymer of globular actin sub units Actin filaments have polarity. This means that the two ends can be identified due to structural differences.
Answer:
C
Explanation:
Since the dark green phenotype allows the insects' to blend in with their surroundings, there will be generally more of the same dark green phenotype on the remote island. Since they are blending in with their surroundings, they survive easier, and they are able to stay alive long enough to reproduce. Let me know if this helped!
Answer:
Kettlewell thought that if natural selection caused the change in the moth population, the following must be true: Heavily polluted forests will have mostly dark peppered moths. Clean forests will have mostly light peppered moths. Dark moths resting on light trees are more likely than light moths to be eaten by birds.
Answer: Amino acids are absorbed via a Sodium cotransporter, in a similar mechanism to the monosaccharides.
Explanation: Amino acids are absorbed via a Sodium cotransporter, in a similar mechanism to the monosaccharides. They are then transported across the alabaster membrane via facilitated diffusion. Di and tripeptides are absorbed via separate H+ dependent cotransporters and once inside the cell are hydrolyzed to amino acids.
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
