The most appropriate answer is 3 !! and that infection is leukemia, a type of cancer , in which WBC increases !!
Answer:
ΔS will be positive.
Explanation:
Denaturation may be defined as the loss of the biological activity of protein by changing the physical conditions of protein. Protein can be denatured by increasing temperature, pH and salts.
Entropy may be defined as the randomness of molecule. The entropy increase with increase in the randomness of the substance. Since, protein has been denatured, the random coil confirmation increase. This means the randomness of protein has been increased, this increase the entropy and makes it more positive.
Thus, entropy becomes more positive.
Answer:
A. parents: Hh and Hh
B. chances of being a carrier: 50%
C. 50%
Explanation:
A. Since hemophilia is a recessive trait, both parents must carry the recessive alelle for it to show through in the child. However, we know both of them are phenotypically normal, meaning a dominant trait (not hemophiliac, H) is masking the recessive trait (hemophiliac, h)
B. The options the woman has when you create a punnet square with her parents genotypes are HH (doesn't have any hemophilia, because hemophilia is recessive, so represented with a lower case h) Hh (there is twice the likeliness she is an Hh since it occurs two times when you combine her parents genotypes. Hh means she is a carrier, but does not have the trait) and hh (meaning she definitely has it). So there is four options, two of which mean carrier, meaning she has a 50% chance of being a carrier.
C. genotypes of the parents:
mom: Hh
dad: hh
when you create their punnet square, you will get two occurrences of Hh and two occurrences of hh. Remember that hh means hemophilia. This means the child has a 50% chance of having hemophilia.
If you were talking about an adult human male, then they can<span> sometimes </span>produce<span> over 100 million </span>sperm<span> per day.</span>
Answer:
The genotypic frequency = 1:1
The phenotypic frequency = 1:1
Explanation:
Given that:
The allele → R = Red beetles
The allele → B = Blue beetles
Since the gene color shows a codominant allele
The Red Beetle = RR
The blue beetles will be = BB
The heterozygous beetle will be = RB
∴
The punnet square showing the crossing of RB × RR is:
R B
R RR RB
R RR RB
The result shows that we have two red beetles and two heterozygous beetles.
Hence;
The genotypic frequency = 1:1
The phenotypic frequency = 1:1
