The EMB (Eosin Methylene Blue) agar is a selective and differential agar medium. It contains sucrose and lactose as fermentable substrates together with Eosin Y and methylene blue dyes which in combination gives the agar its characteristic color when prepared and serves as a pH indicator as well as an inhibitor of growth of Gram-positive bacteria. Hence, it is primarily used to isolate Gram-negative fecal coliforms while some positive fecal coliforms such as Staphylococci are also able to grow.
These coliforms are of two types:
- Lactose or sucrose fermenting coliforms form metallic sheens on the agar as a result of acid production and the response of the indicator to the increased acidity.
- Non-lactose/sucrose fermenters are only able to produce acid by protein deamination. Hence they either form colorless or pinkish colonies on the agar
On the other hand, Mannitol salt agar is a selective and differential agar medium that is primarily used to isolate Staphylococcal bacteria. The presence of sodium chloride in the medium makes it a partial or complete inhibitor of other bacteria.
Hence, an unknown bacterium that forms colorless colonies on EMB will either be a non-lactose fermenting, Gram-negative coliform that will hardly grow on Mannitol salt agar, or a Gram-positive fecal Staphylococcus.
Staphylococci bacteria are of two types:
- Coagulase-positive will form yellow colonies on the Mannitol salt agar.
- Coagulase-negative will form red colonies on the Mannitol salt agar.
More about isolating media can be found here: https://brainly.in/question/7238948
Nope. Nonpoint source comes from a Specific <span>locations and is easy to trace.
In short, Your Answer would be "False"
Hope this helps!</span>
<span>In Drosophila + indicates wild-type allele for any gene, m is mahogany and e is ebony.
Female parents are m+/m+ and males are +e/+e.
F1 are m+/+e, all wild type. F1 females are crossed with me/me males - the test cross.
Offspring will be : non recombinant m+/me, mahogany wild type or +e/me wild type ebony. OR
recombinant me/me mahogany ebony or ++/++ wild type.
As the two genes are 25 map units apart, the percentage of recombinants will be 25% and therefore percentage parental types will be 75%.
75% 1000 is 750. There are two parental types, so you would expect 375 of each. Therefore, you would expect 375 m+/me and 375 +e/me.
25% of 1000 is 250 split between two recombinants =125 of each. Therefore you would expect 135 me/me and 125 ++/++</span>
Answer:
43 monosaccharides will be joined together to make the complex carbohydrate.
Explanation:
The complex carbohydrates are synthesized by the joining of monosaccharide residues by glycosidic linkages.One glycosidic linkage joins two monosaccharide residues by the elimination of one molecule of water.
In other words it can be said that elimination of one water molecule joins two monosaccharides.By the same way elimination of 42 molecules of water will result in the joining of 43 monosaccaride residues.
