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jarptica [38.1K]
3 years ago
14

Teratogens may cause anatomical abnormalities in an embryo, but do not cause anatomical abnormalities in a pregnant woman becaus

e ______.
Biology
1 answer:
9966 [12]3 years ago
3 0
Teratogens affect the fetus and not the mother because teratogens are "environmental exposures" done by the mother. It is an agent that can disturb the development of an embryo of fetus such as drugs or ingestion of harmful substances such as smoking and drinking. What happens is that the toxins are being ingested by the fetus through the umbilical cord where they are being supplied by oxygen and nutrition and instead of providing nutrients harmful substances are being ingested by a growing fetus.
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You're answer would be B) Terrae.

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How high can creatinine and bun levels go before death?
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Creatinine is a by-product of muscular metabolism. In the natural and normal scheme of things, this substance or waste product can be eliminated from the body. A high-serum creatinine level may cause kidney damage. In relation to the above question as to how high can creatinine levels go before death, it must be noted that kidneys have strong compensatory ability and by that as long as its still 50 percent functional, creatinine level won't be that high. Which leads us to a conclusion that, the lesser the kidney function level is, the higher the creatinine level.

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The scientific method is
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Imagine that a DNA sequence of 5'-A-C-G-T-A-C-G-T-3'
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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

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