Algae, alligator, bacteria, bass, bear, bladderwort,bream,butterfly, butterfly larva, cattail, cute, crayfish, cricket, cypress tree, dandelion, Florida panther, fox, frog, fungi, Gambusa, garfish, gopher tortoise, grat, Kork, Heron, Killifish, Live oak tree, Millard duck, Manatee, mangrove trees, mice, mosquito, water, owl, pelican, Pine tree, rabbit, raccoon, Rattlesnake,
Answer:
12:3:1
Explanation:
<em>The typical F2 ratio in cases of dominant epistasis is 12:3:1.</em>
<u>The epistasis is a form of gene interaction in which an allele in one locus interacts with and modifies the effects of alleles in another locus</u>. There are different types of epistasis depending on the type of alleles that are interacting. These include:
- Dominant/simple epistasis: Here, a dominant allele on one locus suppresses the expression of both alleles on another locus irrespective of whether they are dominant or recessive. Instead of the Mendelian dihybrid F2 ratio of 9:3:3:1, what is obtained is 12:3:1. Examples of this type of gene interaction are found in seed coat color in barley, skin color in mice, etc.
- Other types of epistasis include <em>recessive epistasis (9:3:4), dominant inhibitory epistasis (13:3), duplicate recessive epistasis (9:7), duplicate dominant epistasis (15:1), and polymeric gene interaction (9:6:1).</em>
Answer:
I think it is D
Explanation:
Plants are all of the above (I,II,III) but fungi are heterotrophic, but other than that they have a cell wall in their eukaryotic cells and are multicellular, so it is most likely D.
Answer:
The answer is autotrophs
Explanation:
During the early days of the Earths life there was no oxygen but there was plenty of CO2 which the autotrophs took and used as food while discarding the excess which was the oxygen.
Answer: the cfu/g Gram-negative bacteria in the fecal sample is C = 3.0 × 10^3
Explanation:
We know that; Gram negative bacteria looks pale reddish in color under a light microscope from Gram staining.
therefore
There are 30 red bacterial colonies counted.
1 mL of from tube 1 was removed and added to tube with 99 mL saline (tube 2) dilution is 1/100.
transferred volume into the plate is 1 mL.
Now, we have to determine the cfu/g Gram-negative bacteria in the fecal sample
Formula to calculate CFU/g bacteria in fecal sample is expressed as;
C = n/(s×d )
where C is concentration (CFU/g)
, n is number of colonies
, s is volume transferred to plate
, d is dilution factor.
so we substitute
C = 30 / ((1/100) × 1)
C = 30 / 0.01
C = 3000
C = 3.0 × 10^3
THERFERE, the cfu/g Gram-negative bacteria in the fecal sample is C = 3.0 × 10^3
