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3 years ago

How liver performs homeostatic functions?

Biology

1 answer:

mestny [16]3 years ago

5 0

Answer:

See below

Explanation:

Answer in my own words:

The liver helps with preserving homeostasis by controlling the production of glucose

Key words:

1: Homeostasis

2: Glucose

3: Liver

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Hope this helps.

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Cellular differentiation and morphogenesis in plants depends primarily on _____.

rosijanka [135]

Regulation Of Gene Expression.

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3 years ago

During the light reactions part of photosynthesis, chlorophyll in the chloroplast captures energy from the sun. What is this lig

vladimir1956 [14]

In light-dependent reactions, the energy from sunlight is absorbed by chlorophyll and converted into chemical energy in the form of electron carrier molecules like ATP and NADPH. Light energy is harnessed in Photosystems I and II, both of which are present in the thylakoid membranes of chloroplasts.

I hope this helped

7 0

3 years ago

What is the difference between bracketed key and indented key?

____ [38]

Answer:

Bracketed and indented keys are two dichotomous/or branching keys with an ordered, numbered couplet, and spatially differentiated respectively.

Explanation:

Dichotomous keys are used in the identification and classification of taxa.

Bracketed keys, or parallel keys, are easy to understand- for efficiency, these keep the first two entries of every couplet or pair being compared. It also makes use of a numbering system in the couplet choices for easy tracking.

Indented keys maintain an equal distance from the margin on the left of the pages. Successive couplets are indented til the taxon is identified.

5 0

3 years ago

White matter is found in all of the following locations EXCEPT the ________. a) cerebral cortex b) corpus callosum c) corticospi

Eddi Din [679]

Answer:

cerebral cortex

Explanation:

The tissue through which the messages are relayed between different areas of the grey matter is termed the white matter.
The nerve fibers present in the white matter are surrounded by myelin due to which they appear white and hence, the name.
White matter is present in deeper parts of the brain and on the superficial parts of the spinal cord in the central nervous system.
White matter under normal conditions is absent from the cerebral cortex.

3 0

3 years ago

Cystic fibrosis (CF) is one of most common recessive disorders among Caucasians it affects 1 in 1,700 newborns. What is the expe

Phantasy [73]

Answer: The expected frequency of carriers is P(Aa)=0.046.

The proportion of childs with CF is P(aa)=0.024.

25% of having a child with CF (aa).

Explanation:

Hardy-Weinberg's principle states that in a large enough population, in which mating occurs randomly and which is not subject to mutation, selection or migration, gene and genotype frequencies remain constant from one generation to the next one, once a state of equilibrium has been reached which in autosomal loci is reached after one generation. So, a population is said to be in balance when the alleles in polymorphic systems maintain their frequency in the population over generations.

Given the gene allele frequencies in the gene pool of a population, it is possible to calculate the expected frequencies of the progeny's genotypes and phenotypes. If P = percentage of the allele A (dominant) and q = percentage of the allele a (recessive), the checkerboard method can be used to produce all possible random combinations of these gametes.

Note that p + q = 1, that is, the percentages of gametes A and a must equal 100% to include all gametes in the gene pool.

The genotypic frequencies added together should also equal 1 or 100%, and all the equations can be summarized as follows:

$p+q=1\\(p+q)^{2} = p^{2} +2pq+q^{2} = 1\\P(AA)=p^{2} \\P(aa)=q^{2} \\P(Aa)=2pq1$

So, there are 1700 individuals and only one is affected. Since it is a recessive disorder, the genotype of that individual must be aa. So the genotypic frequency of aa is 1/1700=0.000588.

Then, $P(aa)=q^{2}=0.000588$ . And with that we can calculate the value of q,

$P(a)=q=\sqrt{0.000588}=0.024$

And since we know that p+q=1, we can find out the value of p.

$p+0.024=1\\1-0.024=p\\p=0.976$

Next, we find out the genotypic frequency of the genotype AA:

$P(A)=p=0.976\\P(AA)=p^{2} = 0.976^{2}=0.95$

Now, we can find out the genotypic frequency of the genotype Aa:

$P(Aa)=2pq=2 x 0.976 x 0.024 = 0.046$

Notice than:

$p^{2} + 2pq + q^{2} = 1\\x^{2} 0.976^{2} + 2 x 0.976 x 0.024 + 0.024^{2} = 1$

Then, the expected frequency of carriers is P(Aa)=0.046

The proportion of childs with CF is P(aa)=0.024

If two parents are carriers, then their genotypes are Aa.

Gametes produced by them can only have one allele of the gene. So they can either produce A gametes, or a gametes.

In the punnett square, we can see that there genotypic ratio is 2:1:1 and the phenotypic ratio is 3:1. So, there is a probability of 25% of having an unaffected child, with both normal alleles (AA); 50% of having a carrier child (Aa) and 25% (0.25) of having a child with CF (aa).

5 0

3 years ago