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3 years ago

9

in a experiment 4 liters of water was observed to evaporate in 20 hours. What rate of evaporation was observed in this experimen

t?

2 answers:

Natalka [10]3 years ago

8 0

0.25 liters per hour

kolbaska11 [484]3 years ago

5 0

1 liter evaporates in 5 hours. just set up a proportion (:

hope i helped!

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3. Which cells are responsible for creating keratin? *<br> Melanocytes<br> Keratinocytes

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Answer:keratinocytes

Explanation:

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3 years ago

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Phenylketonuria (PKU) is a genetic disease caused by the inability to break down the amino acid phenylalanine. If untreated it l

tiny-mole [99]

Answer:

Therefore, 98% of the US population have no alleles for PKU

Explanation:

The Hardy-Weinberg equilibrium states that the amount of genetic variation in a population will remain constant from one generation to the next in the absence of disturbing factors.

The Hardy-Weinberg equilibrium is expressed quantitatively using a mathematical equation known as the Hardy-Weinberg equation. the equation is given below:

p² + 2pq + q² = 1

also, p + q = 1

Given a pair of alleles, S and s with A dominant and a recessive

where p is the frequency of the dominant allele in the population,

q is the frequency of the recessive allele in the population,

p² represents the frequency of the (SS) dominant genotype,

q² represents the frequency of the (ss) recessive genotype,

2pq represents the frequency of the heterozygous genotype

From the given question,

q² = 1/10000 = 0.0001

q = 0.01

from p + q = 1

p = 1 - 0.01 = 0.99

p² = 0.98

Therefore, 98% of the US population have no alleles for PKU

4 0

3 years ago

Drosophila eye color is an X-linked trait. Red eye color is dominant, and white eye color is recessive. Which Punnett square sho

Anestetic [448]

The genotype of the mother must be heterozygous.

<h3>Punnet' square</h3>

In order to obtain a probability of 25% red-eyed males from the offspring, the mother must be heterozygous for the trait. The father can be of any genotype.

Thus, the likely genotype combination that will result in 25% red-eyed males are:

$X^RX^r$ x $X^RY$

$X^RX^r$ and $X^rY$

When you cross the first one above, the genotype of the offspring would be $X^RX^R, X^RY, X^RX^r, X^rY$ .

The second cross will give offspring with the genotypes $X^RX^r, X^RY, X^rX^r, X^rY$

Both crosses give a 25% chance of producing re-eye males.

More on Punnet's square can be found here: brainly.com/question/3154616

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2 years ago

Consider a locus with two alleles - B and b. B is dominant, while b is recessive. There is no mutation. B has a selective advant

salantis [7]

The given question is incorrect. The correct question is as follows:

Consider a locus with two alleles - B and b. B is dominant, while b is recessive. There is no mutation. B has a selective advantage relative to b, so that the fitnesses of the three genotypes are BB = 1, Bb = 1, and bb = 1-s. In this case, s = 0.50, so that bb homozygotes have 50% fitness of heterozygotes and BB homozygotes. If the population has the following genotypic counts prior to selection of BB = 500, Bb = 250, and bb = 250, what is the frequency of B after one generation with selection? Please give your answer to two decimal places.

Answer:

0.63

Explanation:

According to hardy Weinberg equilibrium the frequencies of the population remains stable from one generation to the next generation unless no selection or mutation is experienced by the population.

The frequency of B after one generation with selection can be calculated as follows:

Given, BB = 500, Bb = 250, and bb = 250.

Frequency (B) = 2 × BB + Bb / 2 × total number of individual.

Frequency (B) = 2 × 500 + 250 / 2 × 1000.

Frequency (B) = 1250 / 2000 = 0.625 = 0.63.

Thus, the frequency of B in population is 0.63.

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3 years ago

Brian is a 15-month-old male whose screening hemoglobin is 10.4 g/dl. treatment for his anemia should be:

schepotkina [342]

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3 years ago