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Y_Kistochka [10]
4 years ago
7

A lab technician provides you with a bacterial pellet labeled "G3". When you observe the pellet under UV light, there is little

to no fluorescence. Other than contamination, provide two scientific explanations for what might have happened to cause this outcome.
Biology
1 answer:
umka21 [38]4 years ago
4 0

Answer: The two possible explanation for the outcome are as follows:

Explanation:

1. The bacterial DNA in the pellet was not completely isolated properly, therefore, no fluorescence was observed in the pellet under the exposure of UV radiation.

2. After 3 hours of induction with the IPTG and labeling with the G3. The IPGT might not be able to bound to the bacterial DNA that could show response of fluorescence in the ultraviolet light.

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With a population in Hardy-Weinberg equilibrium, the A allele has a frequency of 0.60, and the frequency of the a allele is 0.40
valkas [14]

Following the Hardy-Weinberg equilibrium theory, the frequency of the heter0zyg0us genotype is 2pq. In the exposed example, 2pq = 0.48.

<h3>Hardy-Winberg equilibrium</h3>

The Hardy-Weinberg equilibrium theory states that the allelic frequencies in a locus are represented as p and q.

Assuming a diallelic gene,

→  The allelic frequencies are

  • p is the frequency of the dominant allele,
  • q is the frequency of the recessive allele.

→  The genotypic frequencies after one generation are

  • p² (H0m0zyg0us dominant genotypic frequency),
  • 2pq (Heter0zyg0us genotypic frequency),
  • q² (H0m0zyg0us recessive genotypic frequency).

If a population is in H-W equilibrium, it gets the same allelic and genotypic frequencies generation after generation.

The addition of the allelic frequencies equals 1  ⇒ p + q = 1.

The sum of genotypic frequencies equals 1 ⇒ p² + 2pq + q² = 1

If the allele A has a frequency of 0.6, and the allele B has a frequency of 0.4, then the frequency of the heter0zyg0us genotype is

2pq = 2 x 0.6 x 0.4 =<u> 0.48</u>

You can learn more about the Hardy-Weinberg equilibrium at

brainly.com/question/3406634

7 0
2 years ago
which statement explains why tilted layers of rack appear through the Tapeats Sandstone? A. The ground was uneven at the time th
klemol [59]
C. and the sandstone formed around the rocks which are most likely metamorphic
6 0
3 years ago
Read 2 more answers
What is mtDNA testing?
Yuliya22 [10]
Traces a person's matrimonial or motherline
4 0
3 years ago
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The metabolically active and growing phase of endospore forming bacteria is called the?
Paraphin [41]

The metabolically active and growing phase of endospore-forming bacteria is called the <u>vegetative cell.</u>

In bacteria, vegetative cells are ordinary living cells that are produced as a result of asexual reproduction.

Endospores consist of cytoplasm and DNA genome in bacteria and their formation is beneficial in many ways such as providing resistance against heat, adverse external conditions, and harmful radiation.

To put it simply, the vegetative cell is a restive stage of endospore bacteria that provides protection against harsh environments during the life cycle of bacteria by forming a protective layer.

If you need to learn more about endospore-forming bacteria click here:

brainly.com/question/16031149

#SPJ4

7 0
2 years ago
You arrive at an mci. you see a non-ambulatory child who has been evaluated using the jumpstart algorithm with a yellow tag. wha
eimsori [14]

Answer:

it signifies delayed  response to the victim,  that there is no sign of any life threatening injury , but it is needed to check for any noticeable signs of external  injuries.

Explanation:

Jump-start algorithm are generally used to sort children into categories based on the degree or extent of the injuries in multiple -causality situations.

it is applicable to children from infancy to age 8yrs,

It is divided into 4-categories called Triage categories,with certain 4- colors where each  indicates the extent of  severity of the child conditions  represented with tags.

Thus Red- tag indicated need for  urgent medical attention,because of life threatening injuries.

Yellow indicated although medical attention is needed, the injury is not life threatening. Intervention may be delayed for few hours.

Green-indicated  minor injury, that  is not so serious,medical intervention may be delayed for some days

Black indicated diseased or the injury is so severe that the life of the victim can not be saved with available resources.

4 0
4 years ago
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