Answer:
A. outside plasma membrane:
1. cell wall 6. flagella
B. outside of the cell
4. fimbriae 5. pilus
C. in cytoplasm
2. nucleoid 3. ribosome
Explanation:
Bacteria is one of the three domains of cellular organisms. Bacteria are prokaryotes, cells that lack a true nucleus and membrane-bound organelle.
The plasma membrane is the outer membrane that enclosed the cytoplasm and cytoplasmic substance. The cell wall is covering the present out of the cell membrane or plasma membrane. Flagella are attached to the plasma membrane but it is present outside of the plasma membrane.
Fimbriae and pilus are the structure present outside of the cell wall and help to attach it to other cells or any surface. Nucleoid and ribosomes are present in the cytoplasm of the bacterial cell.
Answer:
Bison have limited type of natural predators.
Explanation:
A Bison is a land animal which has a similar appearance to that of a buffalo. They are known to be fast,agile and strong.
A prairie habitat is an ecosystem that is considered part of the temperate grasslands, savannas, and shrublands biome by ecologists due to the similarities in the temperate climates, moderate rainfall and composition of grasses, herbs, and shrubs instead of trees as the main and common vegetation type.
This habitat consists of animals and reptiles such as bison, coyote, antelope, badger, elk, prairie dog, otters, foxes,lizards.
This represents a limited number of predators for the Bison
DNA polymerase is the answer. Hope it helps. Cheers mate
Answer:
Explanation:
using the formula p² + 2pq + 2pr + q²+ 2qr + r² = 1 where p is IA allele is 0.35 and q is IB allele is 0.15.
Thus, the expected frequency of people with type AB blood in this population will be (2pq) = 2 x 0.35 x 0.15 = 0.105
b. The expected frequency of people with type A blood will be p² + 2pr = where r is found using the formular p + q + r = 1 = 1 - (p+q) = 1 - (0.35+0.15) = 0.5
Thus, we have the expected frequency of people with type A blood to be (0.35² + 2x0.35x0.5) = 0.4725.
c. <u> p² + 2pr</u> + <u>2pq </u>+ <u>q² + 2qr</u> +<u> r²</u>
AA AO AB BB BO OO
1850 180 635 2335
frequency of IA- p = (1850 + 180/2) / 5000 = 0.388
frequency of IB- q = (635 + 180/2) / 5000 = 0.145
frequency of i- r = 2335/ 5000 = 0.467.
since r² = 2335, r = √2335 = 48.32
using the formula p + q + r = 1