Answer:
a way to inform conservation efforts based on genetic information from species
Explanation:
Genetic studies in wildlife and captivity species provide significant information to understand aspects regarding the biology and ecology of species, especially endangered species. Many discoveries have been useful to solve several taxonomic issues, determine the risk of diseases, and explain reproductive problems, among other problems. What is even more important is that genetic knowledge helps to plan management programs and conservation. Together with other areas related to behavior and conservation, genetic knowledge provides the basis to carry out successful conservation strategies.
Single Recognition Particle (SRP) RNA is necessary for the targeting of proteins to the prokaryotic plasma membrane or to the eukaryotic endoplasmic reticulum membrane. Its job is to bind to the signal peptide of the membrane or secretory proteins coming from the ribosome at which time it forms a ribosome-nascent chain (RNC)-SRP complex.
SRP plays an important role in understsnding bacterial physiology, emphasizing the importance of proper membrane protein biogenesis, and demonstrates the ability of time-resolved quantitative proteomic analysis to provide new biological insights.
Answer:
After this treatment, the investigators should expect to get a mixture of the desired enzyme, plus fragments of the peptide used to desorb the enzyme in question.
This would be the result of using a peptide as a desorption solution when the desired protein is a protease,
Assuming that the protease retains its activity in the medium in question, and that the peptide can act as a substrate (which would make sense), as the peptide solution is added, it will interact with and bind to the antibody, but some molecules will also interact with the active site of the enzyme as it desorbs and passes through, culminating on the elution of the hydrolized part of the peptide along with the enzyme.
The correct answer is: (a) RNA polymerase (along with its sigma subunit) can initiate transcription on its own.
More differences between prokaryotic and eukaryotic transcription:
• In prokaryotes transcription occurs in the cytoplasm (unlike in eukaryotes in nucleus) and it is simultaneous with translation,
• In prokaryotes there is only one type of RNA polymerase (in eukaryotes there are three types of them).
• There is no sigma subunit in eukaryotes, the initiation of transcription begins thanks to initiation factors.
• Promoter region in prokaryotes contains pribnow box, while in eukaryotes it contains TATA and CAT box.
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
