Answer:
Carrying capacity is the maximum populations size of species that can be sustained continuously.
Explanation:
Answer:
0.2404
Explanation:
The genes R/r and E/e are linked and there is 4% recombination between them.
<u>The possible genotypes and phenotypes are:</u>
- RR or Rr: Rh+ blood type
- rr: Rh- blood type
- EE or Ee: elliptocytosis
- ee: normal red blood cells
Tom and Terri each have elliptocytosis (they are E_), and each is Rh+ (they are R_).
Tom's mother has elliptocytosis (E_) and is Rh- (rr), so she has the genotype Er/_r. His father is healthy (ee) and has Rh+ (R_), so he has the genotype eR/e_. Tom must have inherited his E allele from his mother and his R allele from his father, so he has the genotype eR/Er.
Terri's father is Rh+ (R_) and has elliptocytosis (E_), while Terri's mother is Rh- (rr) and is healthy (ee) with the genotype er/er. Terry could only receive the chromosome <em>er </em>from her mother, and because she is heterozygous for both genes the dominant alleles were both received from her father. Terri's genotype is ER/er.
The frequency of recombination is 4%, so 4% of the produced gametes will be recombinant. There are two possible recombinant gametes, so each will appear 2% of the times (a frequency of 0.02).
<u />
<u>Tom will produce the following gametes:</u>
- eR, parental (0.48)
- Er, parental (0.48)
- er, recombinant (0.02)
- ER (recombinant (0.02)
<u>Terri will produce the following gametes:</u>
- ER, parental (0.48)
- er, parental (0.48)
- Er, recombinant (0.02)
- eR, recombinant (0.02)
A child Rh- with elliptocytosis has the genotype rrE_. This can happen from the independent combination of the following gametes from Tom and Terri respectively:
- Er (0.48) × er (0.48) = 0.2304 Er/er
- Er (0.48) × Er (0.02) = 0.0096 Er/Er
- er (0.02) × Er (0.02) = 0.0004 er/Er
And the total probability of having a rrE_ child will be 0.2304 + 0.0096 + 0.0004 = 0.2404
Both process involve changes in energy and matter
That is true, enzymes are specific.
