Answer:
The water and polar residues cannot bind to the hydrophobic ones, so in the lowest energy state the hydrophobic residues are pushed together in the middle.
Answer:
From this information, we know both parents are heterozygous. This means each parent has an O allele to produce a child with type O blood. A and B blood type alleles are dominant to the O allele which allows one to be heterozygous and express A and B blood type but still have an O allele. This makes that Alice's blood type would be AO. Additionally, the father genotype would have to be BO.
Answer: Option A.
nephron loop multiplies the concentration of salts in the interstitial fluid of the kidney medulla.
Explanation:
countercurrent multiplier is a system that utilize energy that create osmotic gradients and enable reabsorption of water thereby increase concentration of urine. The countercurrent multiplier uses electrolyte pumps there by making the nephron loop increase the concentration of salts in the interstitial fluid of the kidney medulla because it allows the nephron to reabsorb alot of water and and increase the concentration of urine for excretion while at the same time using as little energy as possible.
The expected phenotype ratio is 9:3:3:1 for a BbFf x BbFf cross.
Explanation:
Data given:
Black fur dominant = BB, short fur dominant FF
brown fur recessive = bb, long fur ff
genotype of parents:
BbFf and BbFf
The cross between these two genotype of parents is shown in the below punnet square:
BF Bf bF bf
BF BBFF BBFf BbFF BbFf
Bf BBFf BBff BbFf Bbff
bF BbFF BbFf bbFF bbFf
bf BbFf Bbff bbFf bbff
phenotype ratio:
black fur and short guinea pig = 9
black and long fur = 3
brown and long fur = 3
brown and short fur-1
The expected phenotypic ratio is 9:3:3:1
