Answer:

Explanation:
As we know that the formula for pH is
![pH = -[log H+]\\](https://tex.z-dn.net/?f=pH%20%3D%20-%5Blog%20H%2B%5D%5C%5C)
It is also given in the question that one pH unit represents the 10 fold change in the concentration of hydrogen ion.
Now for lemon juice whose pH is 2, hydrogen ion concentration is equal to

And for pure water whose pH is
, hydrogen ion concentration is equal to

Thus, the H+ ion concentration in lemon juice is
times greater than the pure water.
Mycelium-Fungi
<span>
Heterotrophic-Both</span>
<span>
Pseudopods-Protists</span>
<span>
Contain cell walls with chitin-Fungi </span>
<span>
Fruiting body-Fungi</span>
<span>
Flagellum-Protists
I hope this helps out alot. </span>
Answer:
Seen on the explanation.
Explanation:
Glandular tissues are a mixture of both endocrine (ductless, hormones are secreted directly into the blood) and exocrine (have ducts, hormones are secreted into surfaces) glands.
The pancreas and salivary glands are both exocrine glands although the pancreas has an endocrine function via the islets of Langerhans, both aid in the digestion of food.
The pancreas lacks striated ducts and the intralobular duct is similar to the intercalated duct of salivary glands. It often has a collapsed lumens and the cells are cuboidal with little cytoplasmic staining in contrast to acinar cells.
The pancreas can also be distinguished from the parotid gland by the presence of the pancreatic islet (islets of langerharns) which has an endocrine function.
Answer:
2% of the progeny will be double crossovers for the trihybrid test cross
Explanation:
By knowing the positions of genes, we can estimate the distances in MU between them per region.
- Genes A and B are 10 map units apart (Region I)
- Genes B and C are 20 map units apart (Region II)
- Genes A and C are 30 map units apart
----A-------10MU--------B-------------20MU-------------C---
Region I Region II
We can estimate the recombination frequencies by dividing each distance by 100.
• recombination frequency of A-B region = 10MU / 100 = 0.10
• recombination frequency of B-C region = 20MU / 100 = 0.20
Now that we know the recombination frequencies in each region, we can calculate the expected double recombinant frequency, EDRF, like this:
EDRF = recombination frequency in region I x recombination frequency in region II.
EDRF = 0.10 x 0.20 = 0.02
2% of the progeny will be double crossovers for the trihybrid test cross