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4 years ago

13

When babies engage in physical activity or use language, some synaptic connections will be strengthened, while the unused ones a

re replaced by other pathways or disappear. a neuroscientist would identify this process as:?

1 answer:

RUDIKE [14]4 years ago

8 0

A neuroscientist would identify this certain process as "pruning", which means to trim or remove.

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How would I make a food chain out of these options? Which would come 1-2-3 so on?..

sammy [17]

Its a transition of energy to one animal to another. dont put a animal ate this animal and got the energy. you would put sun goes into plant, then plant goes into grasshopper, then grasshopper goes into spider, spider goes into mouse, mouse goes into wiesel, wiesel goes into hawk, hawk goes into owl

6 0

3 years ago

Scientists have also used fossil evidence to determine _____ on Earth.

adoni [48]

Answer: b

Explanation:

Plz give brainlest

3 0

3 years ago

Read 2 more answers

In ________ eukaryotes, photosynthesis takes place in the ________.

inn [45]

Answer:

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chloroplast

chloroplastIn all phototrophic eukaryotes, photosynthesis takes place inside a chloroplast, an organelle that arose in eukaryotes by endosymbiosis of a photosynthetic bacterium (see Unique Characteristics of Eukaryotic Cells). These chloroplasts are enclosed by a double membrane with inner and outer layers.

5 0

3 years ago

What weight of Tris-HCL (MW=121) is required to prepare 2L of a 15% w/v solution?

podryga [215]

Explanation:

Percentage w/v means that weight of solute present in 100 ml of a solution. Hence, 15% w/v means that 15 gram of solute per 100 ml of solution.

Hence, for 2 L (or 2000 mL) preparation the amount of mass will be calculated as follows.

$\frac{15 g}{100 ml} \times 2000 ml$

= 300 g

Thus, we can conclude that weight of Tris-HCL required is 300 g.

7 0

4 years ago

An autosomal recessive disease has an incidence of 1/10,000. What is the approximate frequency of heterozygote carriers for this

goblinko [34]

Answer:

Option B

Explanation:

Given

Number of incidences or frequency of an autosomal recessive disease $= \frac{1}{10000}$

As per Hardy Weinberg's equation, frequency of a recessive genotype is $q^{2}$

$q = \sqrt{\frac{1}{10000} }\\ q = \frac{1}{100}$

As per first equation of Hardy Weinberg's -

$p+q=1$

so ,

$p = 1-q\\p = 1-\frac{1}{100}\\p = \frac{99}{100}$

$p^2 = (\frac{99}{100})^2\\p^2 =\frac{9801}{1000}$

As per second equation of Hardy Weinberg's -

$p^{2} + q^{2} + 2pq=1$

Substituting the given values in above equation, we get -

$\frac{9801}{10000} + \frac{1}{100} + 2pq=1\\2pq = 1-(\frac{9801}{10000} + \frac{1}{100})\\2pq = 0.99%\\OR[tex]2pq=1$ %

$2pq = \frac{1}{100}$

Hence, option B is correct.

8 0

3 years ago