I will say Large piscivorous fish since i figure that it eats it.
Amoebas are single-celled organisms, which means that they are composed of just one cell. Each amoeba is a cell capable of performing all living functions by itself. They can reproduce asexually. They are protozoans with no fixed shape. Most have no hard parts and look like blobs of jelly.
Answer:
b. Foraging birds eat left over rice and trample the ground making trilling unnecessary.
Explanation:
The California ricelands Habitat Partnership is a partnership between rice farmer and conservation group that will promote long term sustainability of wetland species.
This farming method is beneficial to both the farmers and the species. The wetland species such as foraging birds will benefit by coming to eat the left over rice after the farmers have harvested the rice.
As the birds eat the rice, they trample the grounds making trilling unnecessary hence benefiting the rice farmers.
At two points close together toward the upper right of the diagram, since X and Y chromosomes are more advanced.
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
