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rodikova [14]
3 years ago
6

How to prevent yeast infection while on antibiotics?

Biology
1 answer:
Yuliya22 [10]3 years ago
8 0

Answer: some things you can do

Explanation: talk to your doctor, use yogurt, replenish your good bacteria, and use an over-the-counter antifungal.

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Students are studying the rate of yeast fermentation under different conditions. Glucose is a monosaccharide. Sucrose is a
Akimi4 [234]

Answer:

The correct option is B. Fermentation would occur at a similar rate to that of glucose.

Explanation:

Fermentation can be described as a process in which alcohol is released by using sugars. In fruit juices, fructose is present mainly to get the sweetness. It is used as a substitute for glucose because fructose is more sweeter than glucose. Both fructose and glucose are monosaccharides with the formula C6H12O6. The only difference is in the arrangement of molecules in the atoms of these compounds. Hence, the rate of fermentation would be same for these two sugars.

8 0
3 years ago
What does muscle tissue do?
Oliga [24]
Muscle tissue is the outer layer that protects your muscles and helps them to move without falling off.
4 0
2 years ago
A group of cells is assayed for DNA content immediately following mitosis and is found to have an average of 4 picograms of DNA
damaskus [11]

Answer:

b: 8;8

Explanation:

Mitotic or meiotic cell division constitute the m phase of the cell cycle. At the end of the m phase, the new cells enter the interphase stage of the cell cycle. The interphase is further sub-divided into;

  • <em>the G_0 phase,</em>
  • <em>the G_1 phase,</em>
  • <em>the S phase; and</em>
  • <em>the G_2 phase</em>.

The G_0 phase is essentially a resting phase. Cells that do not need to divide except when necessary move into this phase after exiting the m phase.

Actively dividing cells enter the G_1 phase after exiting the m phase. Cell development and growth takes place. From there, the cells enter the S phase where DNA replication/synthesis takes place. The cells then enter the G_2 phase where proteins are synthesized in preparation for division or m phase.

At the S phase, the amount of DNA a cell carries is doubled but the chromosome number remains the same. For example, if a cell enters the S phase with 2 g of DNA containing 10 chromosomes, at the end of S phase, the amount of DNA would have come 4 g while the number of chromosomes will remain 10.

Hence, if the average amount of DNA in the assayed cells immediately after mitosis is 4 picograms, the amount would be 8 picograms at the end of S phase and will still remains 8 picograms at the end of G_2 phase.

The correct option is b.

8 0
2 years ago
What possible offspring genotypes and probabilities for each can be generated by an Aa parent through selfing?
GenaCL600 [577]

.75. 75% A because is is a dominant gene. a is 25% because it is recessive, and is mostly overcome by a dominant gene.

6 0
2 years ago
A population of 100 tigers has 30 with horizontal stripes. Using pedigrees, you ascertain that horizontal striping is caused by
bixtya [17]

Assuming that this population is in Hardy-Weinberg equilibrium, we can sayt that  <em>the</em><em> probability</em><em> that, given enough time, all tigers in the population will have horizontal stripes is</em><em> 30% = 0.3.</em>

--------------------------

<u>Available data:</u>

  • Population size, N = 100 tigers
  • Number of Tigers with horizontal stripes = 30
  • Horizontal stripes is caused by an autosomal recessive mutation.

We need to know the probability that all tigers in the population will have horizontal stripes, which is the recessive phenotype.

We assume that the population is in Hardy-Weinberg equilibrium, so there should be no evolution.

Since the population is in H-W equilibrium, the allelic, genotypic and phenotypic frequencies will remain the same generation after generation.

Now, we will calculate the allelic and genotypic frequencies of horizontal striped individuals in the population.

There are 100 individuals, and only 30 have horizontal stripes.

So, the phenotypic frequency, F(HS) is 30/100 = 0.3 = 30%

Since this is a recessive phenotype, this value equals the genotypic frequency, F(hh) = 30%

Finally, we can get the allelic frequency by taking the square root of this value.

F(hh) = q² = 0.3

f(h) = q = √0.3 = 0.55 = 55%

According to these calcs, the probability for the fixation of the recessive allele is 55%, and the probability that all individuals express the recessive horizontal stripes phenotype is 30%.

--------------------------

You can learn more about Hardy-Weinberg equilibrium at

brainly.com/question/12724120?referrer=searchResults

brainly.com/question/9870841?referrer=searchResults

brainly.com/question/419732?referrer=searchResults

brainly.com/question/13603001?referrer=searchResults

brainly.com/question/5028378?referrer=searchResults

6 0
2 years ago
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