The resistant bacterial cells illustrate what characteristic of life? A) All living things respond to stimuli. B) All living things maintain homeostasis. C) All living things adapt to changes in their environment. D) All living things contain genetic material and reproduce.

Answers

Answer 1
Answer: Your answer would be C. All living things adapt to change in their environment.
 
For instance, if people continuously take antibiotics, then the bacterial cells will slowly start to learn and evolve in a way where they resist these antibiotics. This is actually an issue and if the current bacterial cells resist our current antibiotics, then scientists will need to create a new anti-resistant antibiotic. 

I hope your answer was satisfyingly answered. :)
Answer 2
Answer: The answer is c, All living things adapt to changes in their environment. The nonresistant bacteria dies off because it hasn't adapted, while the resistant bacteria is still ok. It's kind of like the super-flu, but bacterial, not viral.

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A woman has a rare eyelid abnormality called ptosis, which makes it impossible for her to open her eyes all the way. The condition is caused by a dominant allele. The woman's father had ptosis but her mother was normal. Her father's mother also had normal eyelids.a. What are the genotypes of the woman, her father, and her mother? b. What proportion of the woman's children will have ptosis if she marries a man with normal eyelids?

Structure moves up against epiglottis when food is swallowed to prevent passage of food into it.a. The larynx
b. The nose
c. The pharynx

Answers

Answer:

The correct answer is A. The larynx moves up against epiglottis when food is swallowed to prevent passage of food into it.

Explanation:

The epiglottis is a moist, cartilaginous structure that is part of the cartilaginous skeleton of the larynx. It also marks the boundary between the oropharynx and the laryngopharynx. The epiglottis obstructs the passage of the bolus at the time of swallowing preventing it from going to the respiratory system.

Larynx closure occurs when the vestibular and vocal folds approach the midline during swallowing. Occasionally, when you eat very fast, solid foods or liquids can enter the larynx.

Strepsirhines have a special lower incisor called a Group of answer choices tooth comb. diastema. two-ridge tooth. bilophodont.

Answers

Answer: tooth comb

Explanation: Strepsirhines  are any member of the clade (a group of animals or other organisms derived from a common ancestor species) Strepsirrhini, one of the two suborders (a taxonomic category below order and above infraorder) of primates. They have a special lower incisor called a tooth comb which consists of long, flat teeth with microscopic grooves, and used for grooming the fur.

Final answer:

Strepsirhines, a suborder of primates that includes lemurs and lorises, have a distinct structure called a toothcomb, made up of closely packed lower incisors that stick out from the mouth. Apart from other unique features like a grooming claw, this toothcomb sets strepsirhines apart from other primates.

Explanation:

Strepsirhines, a suborder of primates that includes lemurs and lorises, have unique anatomical features that set them apart from other primates. They have incisors in their lower jaw that are packed closely together and stick out from the mouth in a structure called a toothcomb. This toothcomb, in combination with a clawlike second toe known as a grooming claw, is a hallmark of strepsirhines. The toothcomb is used for grooming - removing debris and parasites from their fur - and is also thought to assist in feeding.

Unlike the majority of mammals, which are diphyodonts and have two sets of teeth in their lifetime, strepsirhines' teeth do not necessarily get replaced. Strepsirhines are found primarily in the 'Old World' - parts of Africa, Asia, and Europe - though a number of species are indigenous to islands such as Madagascar.

Learn more about Strepsirhines' Toothcomb here:

brainly.com/question/36535533

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) How many cells can be grown in a 5 mL culture using minimal medium before the medium exhausts the carbon?

Answers

Answer:

5 * 10^(10)

Explanation:

The question is not complete. Remaining part of the question is as follows - Minimal growth medium for bacteria such as E. coli includes various salts with  characteristic concentrations in the mM range and a carbon source. The carbon source is  typically glucose and it is used at 0.5% (a concentration of 0.5 g/100 mL). For nitrogen,  minimal medium contains ammonium chloride (NH4Cl) with a concentration of 0.1g/100 mL

How many cells can be grown in a 5 mL culture using minimal medium before the medium exhausts the carbon?

Solution -

We will first find the mass concentration of 0.5 g/100 mL of solution.

(0.5)/(100) gram per ml of glucose

The chemical formula of glucose is C_6H_(12)O_6

The molecular weight of glucose molecule is 180 grams per mole

Now, we will find the number of moles of glucose in a 5 ml medium -

((0.5)/(100) * 5)/(180) \n1.39 * 10^(-4) mole

The number of carbon atom in each glucose molecule is equal to six, thus, number of minimal carbon mole is equal to

1.39 * 10^(-4) * 6\n= 8.34* 10^(-4)mole

Number of carbon atoms is equal to

8.34* 10^(-4) * 6.023 * 10^(23)\n= 5 * 10^(20)\n Carbons

One bacteria has 10^(10) carbon molecule.Thus, 5 ml medium will have 5 * 10^(10) bacteria

Can the genotype for a gray-bodied fly be determined? Why or why not? Describe all of the possible genotypes for a fly with that phenotype

Answers

Answer:

The genotype of a fly can be determined gray bodied is dominant over black body flies. The possible genotype of a fly is either YY or Yy.

Explanation:

The genotype of a gray bodied fly can be determined with the help of punnett square. The gray bodied fly is crossed with black bodied fly, the original genotype of the fly can be determined. This type of cross is known as test cross.

If the cross results in the formation of all gray bodied progeny, the fly is homozygous dominant with the genotype YY. The cross results in  the mixture of progeny with gray and black body , then the genotype of a fly is heterozygous dominant Yy.

Thus, the genotype of a fly can be determined by the test cross. The fly may have genotype either YY or Yy.

What is worm "poop" called?​

Answers

Answer:

Vermicast.

Explanation:

Vermicast are worm castings, worm manure, or worm faeces.

Answer:

Vermicast.

Explanation:

lol I copied his answer to get points LOL

In humans there is a dominant allele (A) for the absence of moles; while the recessive allele (a) results in the presence of moles. A couple who are both heterozygous for this locus (Aa) plan to have seven children. a. Expand the binomial (p + q)7.
b. What is the probability that:
(i) The first child that is born will not have moles.
(ii) All of the children will have moles.
(iii) The first two children will have no moles and the last five will have moles.
(iv) Of the 7 children, 4 will have no moles and 3 will have moles
c. Assume this couple now have two children, one with moles and one without moles. What is the probability that the child born without moles is a carrier of the a-allele (ie heterozygous)?

Answers

Answer:

a. (p + q)^7 = p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7

b. i. 0.75

   ii. 0.000061

   iii. 0.012

   iv. 0.17

c. 0.67

Explanation:

a. The expansion of the binomial (p + q)7 would be such that:

(p + q)^7 = p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7

b. Both couples are heterozygous:

             Aa    x    Aa

         AA   Aa   Aa   aa

Since A is dominant over a,

probability of having mole (aa) = 1/4

probability of not having moles = 3/4

Therefore, the probability of the first child not having moles = 3/4 or 0.75

ii. Let the probability of not having mole = p and the probability of having mole = q. From the binomial expansion:

(p + q)^7 = p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7

Probability that all of the children will have moles = p^0q^7

since p = 3/4 and q = 1/4

p^0q^7 = (3/4)^0(1/4)^7 = 0.000061

iii.Probability that the first two children will have no moles and the last five will have moles = 21p^2q5

                       = 21(3/4)^2(1/4)^5

                         = 0.012

iv. Probability that 4 will have no moles and 3 will have moles out of the 7 children = 35p^4q^3

               = 35(3/4)^4(1/4)^3

                      = 0.17

c. Probability that the child born without moles is a carrier of the a-allele  = probability of heterozygous.

From the cross in (b), the genotypes of those born without moles are AA and 2Aa. Therefore, the probability of not having moles and be Aa is:

      = 2/3 or 0.67

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