Answer:
a. M_D_
: 9/16 Black furred individuals
b. M__dd
: 3/16 gray furred individuals (1/16 MMdd + 2/16 Mmdd)
c. mmD_
: 3/16 Brow-furred individuals (1/16 mmDD + 2/16 mmDd)
d. mmdd: 1/16 gray furred individuals.
Explanation:
Available data:
- Dense pigment gene, D, codes for melanophilin.
- D allele dominant over d.
- Gene M/m is responsible for the production of the pigment
- M is dominant over m
- If D is present, M/m is expressed
- If dd is present, M/m is not expressed
- DDMM, DdMM, DDMm, DdMm --->Black fur
- DDmm, Ddmm ---> Brown fur
- ddMM, ddMm -----> Light gray fur
Cross: between two dihybrids
Parental) DdMm x DdMm
Phenotype) Black-furred Black-furred
Gametes) DM, Dm, dM, dm DM, Dm, dM, dm
Punnet Square) DM Dm dM dm
DM DDMM DDMm DdMM DdMm
Dm DDMm DDmm DdMm Ddmm
dM DdMM DdMm ddMM ddMm
dm DdMm Ddmm ddMm ddmm
F1) 9/16 Black-furred individuals, D-M- (1/16 DDMM + 2/16 DDMm + 2/16
DdMM + 4/16 DdMm)
3/16 Brown-furred individuals, D-mm ( 1/16 DDmm + 2/16 Ddmm)
4/16 Gray-furred individuals, dd-- ( 1/16ddMM + 1/16 ddmm + 2/16 ddMm)
Answer:
it is simple. the combine two graphs into one basically by combining two graphs for grey and brown species the red line in between separates green and brown and just take the reading
Explanation:
8=10
12=12
14=1
note equals means frequency
Answer:
photosynthesis is the answer
Active transport moves from low to high
Answer:
low level of calcium would result in fewer signals sent between pre and post synaptic cells.
Explanation:
The calcium has a main role in signal transmission mechanism and here is the mechanism :
1-the impulse reaches the synapse (depolarization which is a voltage change).
2-the voltage gated calcium channels open and let calcium flow into the cell.
3-the calcium causes the synaptic vesicles to rupture and release chemical transmitters which cross the synaptic cleft and finally bind to receptors on the post synaptic cell (transmitting the signal)
so it's obvious that if calcium concentration decreases, the signal transmission also decreases.
