Answer:
The near-UV CD range (>250 nm) of proteins delivers info on the define configuration. The indications found within the 250–300 nm section are because of the engagement, dipole alignment and also the environment of the encircling atmosphere of the essential amino acid, tyrosine, aminoalkanoic acid (or S-S disulfide associations) and essential amino acids. In contrast to far-UV CD, the near-UV CD variation can't be allotted to any specific 3D configuration. Relatively, near-UV CD varieties offer essential info on the character of the prosthetic teams in proteins, e.g., the pigment teams in hemoprotein and cytochrome.
The answer is “The Nervous System”
Cells need energy for various purposes
Answer:
Because it is important to support the body to function properly
Answer: mother: XX^aa, father: X^YAa, son: X^YAa, daughter: X^X^aa.
Explanation:
Color blindness is a genetic disorder that affects the ability to distinguish colors. It is hereditary and is transmitted by an X-linked recessive allele. If a male inherits an X chromosome with the altered allele he will be color blind. In contrast, females, who have two X chromosomes, will only be colorblind if both of their X chromosomes have the altered allele. This is because <u>males have one X chromosome and one Y chromosome, while females have two X chromosomes</u>.
If the woman has normal vision, that means she cannot have both chromosomes affected. She can only have one affected chromosome (be a carrier) or none at all. Also, if she has blue eyes, which is a recessive trait, then both alleles are recessive. But the eye color is not on the X chromosome. For example, her eye color genotype can only be aa, because if she had at least one dominant allele she would have brown eye color. As for the other trait, she can be XX^, with X^ being an affected (carrier) allele or XX, i.e. both normal. So in summary, her genotype can be XXaa or XX^aa
If she has a brown-eyed male child who is also colour blind, he has inherited the allele for colour blindness from his mother, since the father does not pass on an X chromosome to the male children, only the Y. With this we can now rule out the mother's XXaa genotype since she had to have passed on her affected X^ chromosome. Then the genotype of the mother is XX^aa. And since her mother can only pass on one allele to (recessive) because she does not have allele A, the dominant that determines her brown eye color can only come from the father. So the genotype of this son is X^YAa. The female daughter has color blindness and blue eyes. So she had to inherit the affected X^ chromosome from the mother (which we already know she has) and an affected X^ chromosome from the father, because the daughter needs to inherit both affected X^ chromosomes to develop the disease. And if she also has blue eyes, she had to have inherited a recessive allele from the mother and another from the father. So with this information we can say that the father's genotype can only be X^YAa. Because the father must have both A and A alleles of the same eye color, because he passed the dominant one to the son and the recessive one to the daughter. At last, the genotype of the daughter is X^X^aa.
