Why do you think that the first thing you learn about in a Biology course is Chemistry? How does learning chemistry apply to Biology?

In genetics, the color of guinea pigs is determined by the presence or absence of a dominant allele for black coat color. The F2 ratio of 3:1 (black:white) suggests that black coat color is dominant. In the exceptional cases, the black guinea pigs were homozygous recessive and the white guinea pigs were heterozygous.

In genetics, the color of guinea pigs is determined by the presence or absence of a dominant allele for black coat color. Since all the F1 offspring were black in 94 of the crosses, this suggests that black coat color is dominant. The F2 ratio of 3:1 (black:white) indicates that the black guinea pigs were heterozygous, and the white guinea pigs were homozygous recessive.

In the 6 exceptional cases where half of the F1 animals were black and the other half were white, it suggests that the black guinea pigs were homozygous recessive and the white guinea pigs were heterozygous. When these guinea pigs are crossed, the expected result would be a ratio of 1:1 (black:white) in the F2 generation.

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Answer: Option C, D, E and A

Explanation:

Sympatric speciation is the evolution or isolation of new species from the original population of species occupying the same geographic area. It is also due to sexual selection of mates leading to reproductive barriers. A plant with extra set of homologous chromosomes is an example. Sympatric speciation is due to isolation of new species from the population of species who arise from common anscetor.

Answer:

C) Because apoptosis genes kill cells, natural selection is seldom involved in apoptosis pathways.

Explanation:

Apoptosis is a natural process taking in the cell which causes the physiological and biochemical changes in the cell which could cause the death of the cell.

The apoptosis is controlled at the genetic level therefore apoptosis is also known as the programmed cell death.

The studies have shown that the process is involved in the development of the finger and the toes in humans and the sequences controlling the process has been conserved during the evolution in a different organism.

This shows that humans and nematode have the same conserved sequence of apoptosis but the natural selection does not control the apoptosis.

Thus, the selected option is the correct answer.

Apoptosis is involved in human development and shared across species like humans and nematodes. It's incorrect to state that natural selection plays no role in apoptosis pathways, as these pathways have evolved to benefit the overall health of the organism.

The question asks which statement about apoptosis is false. Let's go through each item:

a) Apoptosis does play a key role in the development of human fingers and toes. During embryological development, apoptosis is meant to eliminate the web-like tissues between individual fingers and toes to form fully separated digits.
b) Indeed, despite the vast evolutionary divergence, humans and nematodes share similar apoptosis pathways.
The false statement here is c) Even though apoptosis does involve killing off cells, natural selection plays a role here as well. It's involved in apoptosis pathways as these pathways have evolved to maintain the health of the organism by removing damaged or unneeded cells and preventing dangerous ones from proliferating.

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Answer:

The earth is extremely old,and earth processes acted gradually to twist or bend rocks.

Explanation:

The transformation takes place through many years as the molten rock was pushed upward and sediment form into the rock that was lifted above sea level by the force generated by uprising rocks.The changes are very little and gradually takes place which gives rise to the formation of rocks.

Answer: Assuming the population is at Hardy Weinberg's equilibrium, the frequency of the homozygous recessive individuals is 0.107

Explanation: Hardy-Weinberg law provides an equation to relate genotype frequencies and allele frequencies in a randomly mating population. p² + 2pq + q² = 1 where p² = homozygous dominant, q² = homozygous recessive and 2pq = heterozygous.

Also, for two alleles, p and q, p + q = 1

Using these equations and the information provided in the question given, the total number of squirrels is 600.

Grey squirrels frequency, p² = 536/600 = 0.893

Red squirrels frequency, q² = 64/600 = 0.107

Therefore, the frequency of recessive individuals (Red) = 0.107

The frequency of the homozygous recessive individuals in the squirrel population, assuming Hardy-Weinberg equilibrium, is 10.67%, calculated by dividing the number of red squirrels by the total population.

To determine the frequency of the homozygous recessive individuals in a squirrel population, we can use the Hardy-Weinberg equation. In this case, the recessive phenotype is represented by the red squirrel coloration. We are given that there are 64 red squirrels, which are the homozygous recessive individuals (rr), out of a total of 600 squirrels. The frequency is the number of homozygous recessive individuals divided by the total population.

Since the population is at Hardy-Weinberg equilibrium, we know that the sum of the genotype frequencies (p² + 2pq + q²) equals 1. In this context, 'p' represents the dominant allele frequency, and 'q' represents the recessive allele frequency. The frequency of the homozygous recessive individuals (q²) equals the number of red squirrels (recessive phenotype) divided by the total population.

Therefore, the frequency of the homozygous recessive squirrels is:

q² = 64 red squirrels / 600 total squirrels = 0.1067 (or 10.67%)

This value represents the frequency of the homozygous recessive individuals in the squirrel population.

Answer:

Explanation:

"Bacteria and other microbes can be used to "clean up" an oil spill by breaking down oil into carbon dioxide and water. Two samples isolated from the Deepwater Horizon leak in the Gulf of Mexico were labeled A and B. The DNA of each was isolated and the percent thymine measured in each sample. Sample A contains 19.3 % thymine and sample B contains 29.7 % thymine. Assume the organisms contain normal double‑stranded DNA and predict the composition of the other bases."

DNA is made up of four bases: Adenine (A), Thymine (T), Cytosine (C) and Guanine (G).

If Sample A contains 19.3% thymine, therefore it contains the same percentage of adenine. Added together 19.3+19.3=38.6%.

Since A-T is 38.6%, therefore C-G is 100 - 38.6% = 61.4%

Dividing 61.4% in half, C and G are 30.7% each.

In summary, for Sample A:

Adenine = 19.3%

Thymine = 19.3%

Cytosine = 30.7%

Guanine = 30.7%

For Sample B, since thymine is 29.7%, adenine is also 29.7%

So A-T is 29.7 +29.7 = 59.4%

C-G = 100 - 59.4 = 40.6%

Dividing 40.6% in half, C and G are 20.3% each

In summary, for Sample B:

Adenine = 29.7%

Thymine = 29.7%

Cytosine = 20.3%

Guanine = 20.3%

Answer:

A. same genotype as the father (AaBbCcDdee) = (½ x ½ x ½ x ½ x ½) = 0.03

B. phenotypically resemble the father (A_B_C_D_ee) = (½ x ½ x ¾ x 1 x ½) = 0.09

C. same genotype as the mother (aabbCcDDEe) = (½ x ½ x ½ x ½ x ½) = 0.03

D. phenotypically resemble the mother (aabbC_D_E_) = (½ x ½ x ¾ x 1 x ½) = 0.09

E. phenotypically resemble neither parent (simply subtract the probability of phenotypically resembling each parent from 1) = 1 – (0.09375 + 0.09375) = 0.81

Explanation:

The computation of the total probability would simply be the product of individual chances for each particular gene.

This can be better gotten with the punnette sqaure

The study of genes and inheritance is called genetics.

The correct answer is 0.81

The answer is as follows:-

Phenotypically resemble neither parent (simply subtract the probability of phenotypically resembling each parent from 1)[tex] = 1 - (0.09375 + 0.09375) = 0.81[/tex]

Hence, the correct answer is 0.81

For more information about the genes, refer to the link:-

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Answer:

WwNn

Explanation:

A fly that is homozygous dominant for both traits (WWNN) is crossed with a fly that is homozygous recessive for both traits (wwnn).

All gametes produced by the WWNN individual will be WN, and all gametes produced by the homozygous recessive individual will be wn.

When the gametes from the two parents combine, the zygote will be heterozygous for both genes, with the genotype WwNn.

The offspring from a homozygous dominant fruit fly (WWNN) and a homozygous recessive fly (wwnn) will all have the heterozygous genotype WwNn, displaying long wings and brown pigments.

In classical Mendelian genetics, long wings (W) are dominant over short wings (w), and brown pigments (N) are dominant over yellow pigments (n). When a homozygous dominant fruit fly for both traits (WWNN) is crossed with a homozygous recessive fly (wwnn), the resulting genotypes for all the offspring will be heterozygous for both traits (WwNn), expressing the dominant phenotypes—long wings and brown pigments.

The scenario is straightforward as both traits are being considered separately, and the inheritance pattern follows simple Mendelian inheritance, not taking into account potential linkage or sex linkage as in X-linked crosses. Therefore, all offspring produced from this cross will have the same genotype, with one dominant and one recessive allele for each gene, showcasing the principle of uniformity in the F1 generation.

Answer: 0.18

Explanation:

For the alleles, the percentage distribution of each is 'A' (90% = 0.9)

While 'a' (10% = 0.1)

Hence, 0.9 and 0.1 are the respective frequencies of each allele

Now, apply Hardy-Weinberg Equilibrium equation, where heterozygotes are represented by the 2pq term.

Therefore, the number of heterozygous individuals (Aa) is equal to 2pq which equals

2 × 0.9 × 0.1 = 0.18

Thus, the frequency of heterozygote is 0.18, while the percentage distribution in the population is 18%

Answer: It is greater when sucrose concentration went from 2.5 to 7.5g/l.

Explanation: The rate of reaction of an enzyme is known to be affected by the rate of concentration of its substrate, which in this case is the sucrose Solution.

If the rate of increase of concentration is high,the activities of the enzyme SUCRASE will increase accordingly, in order to breakdown the substrate.

The rate of increase of Sucrose from 2.5 to 7.5g/l is higher(300%) than the rate of Increase of Sucrose from 22.5 to 27.5g/l (1.22%). It is expected under circumstances that the action of SUCRASE will increase at a rate higher in the first Solution than in the second Solution.

The rate of increase in sucrase activity depends on the concentration of sucrose and whether or not the enzyme is saturated. The increase could be greater at lower concentrations (2.5 to 7.5 g/l) if sucrase is not yet saturated. The increase might be less at higher concentrations (22.5 to 27.5 g/l) if sucrase is near or at saturation point.

The increase in sucrase activity is generally considered to be a response to the concentration of substrate present, in this case, sucrose. The increase in activity happens because more substrate (sucrose) is available for the enzyme (sucrase) to act upon. However, there is a limit to this increase. Once the enzyme is saturated with substrate, further increases in substrate concentration do not increase the enzyme's activity. This is known as the saturation point.

To determine whether sucrase activity increased more when sucrose concentration increased from 2.5 to 7.5 g/l or from 22.5 to 27.5 g/l, we would need specific data on the rate of sucrase activity at these different concentrations. It's possible that the increase from 2.5 to 7.5 g/l was greater if this is in the ascending portion of the enzyme activity curve and the sucrase was not yet saturated with sucrose. Conversely, the increase from 22.5 to 27.5 could be lesser if the sucrase is near or at saturation point.

Question reformatted

In tomato plants, the production of red fruit color is under the control of an allele R. Yellow tomatoes are rr. The dominant phenotype for fruit shape is under the control of an allele T, which produces two lobes. Multilobed fruit, the recessive phenotype, have the genotype tt. Two different crosses are made between parental plants of unknown genotype and phenotype. Use the progeny phenotype ratios to determine the genotypes and phenotypes of each parent.

Cross 1 Progeny: 38 two-lobed red, 18 two-lobed yellow, 38 multilobed red, 18 multilobed yellow.

Cross 2 Progeny: 14 two-lobed, red 14 two-lobed, yellow 14 multilobed, red 14 multilobed, yellow

For cross 1, determine two appropriate genotypes for both parents.

a) Rrtt × Rrtt

b) Rrtt × rrTt

c) RrTt×Rrtt

d) Rrtt × RrTt OR rrTt × RrTt

For cross 1, determine two appropriate phenotypes for both parents.

a) yellow fruit, two lobes AND red fruit, multiple lobes

b) red fruit, two lobes AND red fruit, two lobes

c) red fruit, two lobes AND red fruit, multiple lobes

d) red fruit, two lobes AND yellow fruit, two lobes

For cross 2, determine two appropriate genotypes for both parents.

a) RrTt × Rrtt

b) RrTt × rrtt

c) RrTt × rrTt

d) RrTt × rrtt OR Rrtt × rrTt

Answer:

Part A - C

Part B - C

Part C - D

Explanation:

The genotype for two lobed red must be either RRTT, RrTT, RRTt or RrTt, as these are both dominant traits

The genotype for two lobed yellow must be rrTt or rrTT, as yellow is a recessive trait but two lobed is dominant.

The genotype for multilobed red must be Rrtt or RRtt, as multilobed is recessive but red is dominant

The genotype for multilobed yellow must be rrtt, as both these traits are recessive.

The answer cannot be a) as we have multilobed fruit which would require a T allele.

For a dihybrid cross between two heterozygous individuals (i.e. both RrTt) gives the expected ratio of 9:3:3:1, a classic Mendelian ratio for this type of cross. That is not the case for this cross, as there are double the amount of multilobed yellow than we would expected, and half the amount of two lobed red. This suggests one of the parents is homozygous for the t allele, but that they are both heterozygous for the R/r allele. Therefore, we can first check the ratios of RrTt x Rrtt (answer c) by a punnett square (see attached).

The possible gametes are RT, rT, Rt, rt for one parent (RrTt), and RT, Rt and rt for the other parent (Rrtt). This gives us the correct progeny.

If you ever struggle with these questions, just draw out all the possible punnett squares!

We have determined that the parental genotypes are RrTt x Rrtt. This means the parental phenotypes are Red two-lobed and red multi-lobed, corresponding to answer C.

The second cross has a ratio of 1:1:1:1. This gives the clue that the gametes are all present in equal numbers (i.e. each parent is heterozygous for one trait and homozygous for the other), OR that one parent is heterozygous for both and one is homozygous to both. This points to option d.

Answer:

Option C, a testable prediction leading to design of an experiment

Explanation:

The statement here predicts the outcomes of raising tadpoples is water with atrazine levels of 0.1 ppb as compared to the ones raised in pure water. Hence, it cannot be question. Now since this prediction can be tested, an experiment can be designed where a certain number of tadpoles can be raised in pure water and the same number of tadpoles can be raised in water with 0.1ppb of atrazine level. The difference in two populations can be then compared to either support the prediction or contradict it.

Hence, option C is correct

Answer:

Because of their property of preciseness and rhythmic repetitions like a normal clock sequence...

Explanation:

Evolutionary biologist use these neutral polymorphisms as molecular clock because of their gradual accumulation of  mutations that remain on changing gradually in a uniform sequence overlarge number of generations. These changing sequences are highly precise over and over, that's why these are also termed as neutral mutations.

Answer:

There will be null activity for a gene within in the deletion, this means if a genotype has similar phenotype. This suggests that mutant allele and the allele with deletion has the same level of activity.

Explanation:

The limitation of such interpretation is that the phenotypes have the enzymatic activity according to a threshold level. This means that the mutant phenotype is visible till the enzymatic activity is lower than a threshold. If the level rises above the threshold the wild type phenotype is observed.

Answer:all of

Explanation:

Answer:

mass(g) of colchicine = 3.99 × 10⁻⁵g

Explanation:

Given that;

the number of moles colchicine =  100 nM = 100 × 10⁻⁹M

Density of chlamydomonas culture = 5 × 10⁷ cells/mL

∴ In 1mL of chlamydomonas culture there are 5 × 10⁷ cells present.

To decide the number of one cell, we need to use [tex]\frac{1}{5*10^7}[/tex] mL of the cellulose

However, 1 × 10⁸ cells will be present in [tex]1*10^8*\frac{1}{5*10^7}[/tex] mL , which in turn give us;

[tex]\frac{10^8}{5*10^7}[/tex]mL

Afterwards, to get the required 1 × 10⁸ cells, 2mL(molarity, since molarity=  no of moles/litre) has to be taken

If colchicine has to be treated, we need to determine the mass of colchicine that is required in the process as well;

since,  the number of moles colchicine =  100 nM = 100 × 10⁻⁹M

And, the given molecular weight = 399.44;

we can determine the mass of colchicine as;

∴ [tex]numberofmoles = \frac{mass(g)}{molar mass(g)}[/tex]

substituting the parameters given, we have:

[tex]100*10^{-9}= \frac{mass(g)}{399.4}[/tex]

mass(g) = 399.4 × 100 × 10⁻⁹

            = 3.99 × 10⁻⁵g

Hence, the mass of colchicine that is required in the process to make 100 nM dissolve in the in 2mL of the culture in one Litre of water  is 3.99 × 10⁻⁵g.

Answer:

43.6 hours, which is less than two days.

Explanation:

To calculate Growth for an exponentially growing populations

Nt = N▪ * 2^n

Where,

N▪= cell number at initial time

Nt = cell number at later time

n = number of generations

Assuming exponential phase and limitless nutrients

How long until E.coli conquers the earth?

Given,

1 doubling = 20 min = 0.33hr

N▪= Mass = 9.5 x 10^-13 g/bacterium

Nt= Mass = 5.9 x 10^27 g/Earth

Nt = N▪ * 2^n

5.9 x 10^27 = 9.5 x 10^-13 * 2^N

nlog(5.9 x 10^27) = log(9.5 x 10^-13) + nlog(2)

27.7 = -12.0 + n(0.3)

27.7 + 12.0 = n(0.3)

39.7 = n(0.3)

132 = n

Therefore,

132 generations * 0.33 hour/generation = 43.6 hours

43.6 hours is less than two days.

Our answer for a single cell, it will take less than two days for the mass of an E. coli culture to equal that of the Earth on assuming exponential phase and limitless nutrients.

a). 2 genes are involved in production of colour in F2 generation.

b). 2 possible alleles for each phenotype

c).Rr (medium red), rr light red

d) F1 generation  1:2:1 ( 1 light red stain, 2 medium red) and F2 generation 1:1

1 medium red and 1 light red.

Explanation:

F1 generation:

dark red strain X white strain

RR X rr

  r    r

R rR  Rr

R  rR Rr

genotype ratio 4:1, all medium red

Cross between F1 generation of true breeding plants

  R  r

R RR Rr

r  Rr  rr     1 light red stain, 2 medium red, 1 white genotype ratio 1:2:1

In F2 Generation, F1 plants interbred ie true breeding

F2 generation:

    R r

r   Rr  rr

r  Rr   rr  

It produces 1 medium red and 1 light red plant. genotype ratio 1:1

Here alleles Rr is for medium red and rr is for light red

Additive allele depends on the allele concentration for a phenotype trait to appear.

Final answer:

The ratio of phenotypes in the F2 generation indicates that two genes are involved in the production of color in wheat, with a total of four additive alleles influencing the phenotype. True-breeding medium red (RaRa) crossed with white (rraa) would yield medium-red F1 offspring. F2 would show a ratio of 3 medium-red to 1 white.

Explanation:

The phenotypic ratio of the F2 generation, which is given as 1 dark-red : 4 medium-dark-red : 6 medium-red : 4 light-red : 1 white, suggests the involvement of two genes in color production due to the pattern not matching a simple 3:1 monohybrid cross. It fits a modified dihybrid cross ratio of 1:4:6:4:1, which is reminiscent of a 9:3:3:1 ratio with intermediate forms due to incomplete dominance.

To determine the number of additive alleles needed for each phenotype:

Dark-red: All four additive alleles are present (2 from each gene).

Medium-dark-red: Three additive alleles are present.

Medium-red: Two additive alleles are present.

Light-red: One additive allele is present.

White: No additive alleles are present.

We can assign the symbols R for the allele contributing to red color and r for its absence from one gene, and A for the allele contributing to red color and a for its absence from the second gene.

Thus, possible genotypes for medium red are RaRa or rrAA. For light red, the genotype could be Ra with the other gene being rr or aa, such as Rraa or Aarr.

For a cross between a true-breeding medium red plant (RaRa) and a white plant (rraa), the F1 would all be Rara (medium-red), and the F2 generation would show a 3:1 phenotypic ratio of medium-red (R-) to white (-rr) since the white parental plant can only contribute recessive alleles.

Answer: Closer to Acanthostega

Explanation: Phylogenetic tree is a diagram that represents evolution patterns of species. The ramification of the tree indicates how the species evoluted from a series of commom ancestor. Two species are closely related if the ancestor is closer to them. By it, it means that two species are closer if their characteristics are similar to each other. As the question stated, Ichthyostega had digits and reduced tail. Among the three others, Acanthostega has arms with digits, gills, tails and no fin.

So, in the phylogenetic tree, the branch that has acanthostega would be closer to the one with Ichthyostega, as this one is likely the descendent of the previous one.

Answer:

Tulerpeton

Explanation:

Answer:

a. 6.26%

b. 3.65%

Explanation:

A food chain is a series of event showing how organism feed on one another, it also shows the transfer of energy in an ecosystem.

From the question; we can have a food chain showing the transfer of energy right from the producer to the tertiary consumer

(Producer)                  Primary                 Sedcondary           Tertiary

                                   consumer (1°)       consumer (2°)        consumer (3°)

Phytoplankton    ⇒    Copepods      ⇒  small fish         ⇒    Marine birds

120,000 kJ/m2         7,514 kJ/m2            383 kJ/m2              14 kJ/m2

W                                    X                              Y                          Z

Efficiency of energy transfer is given as: [tex]\frac{EnergyAvaliableAfterTheTransfer}{EnergyAvailableBeforeTheTransfer}*100%[/tex]

The first question (a) says,  Calculate the efficiency of energy transfer to the copepods.

i.e efficiency of energy transfer from phytoplankton to copepods =[tex]\frac{X}{W}*100%[/tex]

=[tex]\frac{7514}{120000}*100[/tex]

= 6.25%

b) Calculate the efficiency of energy transfer from the secondary consumer to marine birds.

i.e efficiency of energy transfer from small fish to marine birds = [tex]\frac{Z}{Y} *100[/tex]

=[tex]\frac{14}{383} *100[/tex]

=3.65%

Answer:

Assuming the dominant allele is represented as A, and the genotype as AA and the recessive as a, and genotype as aa; then the F1 will produce the offispring. Aa, Aa, Aa,Aa in ratio: 1:1:1:1:1

This shows the dominance of allele A.

However if two of  the F1 generation are crossed then:

The  F2 is AA,Aa,Aa,aa the phenotypic ratio is

                        3   ;     1

While the Genotypic ratio is ratio;

  1AA : 2Aa Aa; 1aa(1 ;2;1)

Explanation:

In a typical monohybrid cross in which one allele is dominant to the other, the expected phenotypic ratio when F1 generation is crossed is 3:1. This means that 3 out of 4 offsprings will express the dominant phenotype, while one will express the recessive phenotype.

In a typical monohybrid cross where one allele is completely dominant to the other, the F1 generation would be heterozygous - meaning they carry one each of the dominant and recessive alleles. When these are crossed, the Mendelian genetics laws predict that there will be a phenotypic ratio of 3:1. This means that 3 out of 4 offsprings are expected to express the dominant phenotype, while 1 out of 4 is expected to express the recessive phenotype. For instance, in a cross between two purple-flowered pea plants (Pp) where purple color (P) is dominant over white (p), the expected phenotypic ratio in the F2 generation would be 3 purple-flowered to 1 white-flowered.

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Final answer:

Given a family history of a recessive condition, the probabilities are approximated as 1/2 for each parent (III-3 and III-4) being carriers (Rr) and 1/4 for their child (IV-1) being affected (rr), based on genetic inheritance patterns.

Explanation:

The question revolves around determining the probability of carriers and affected individuals in a family with a recessive condition, specifically focusing on individuals III-3, III-4, and their potential child, IV-1. Given the nature of recessive conditions and the information provided, calculating these probabilities involves understanding genetic inheritance patterns.

The probability that III-3 is a carrier (Rr) depends on the exact family history not detailed here, but following general rules of recessive inheritance, if a direct parent is affected or a carrier, the probability could range up to 1/2.

Similarly, for III-4, without explicit information about the parents, a likely assumption would also be 1/2, as they have a family history of the condition, suggesting a chance of being a carrier.

The probability that their child (IV-1) will be affected (rr) if both parents are carriers (Rr) follows the Punnett square rule, resulting in a 1/4 chance. This is because each parent has a 50% chance of passing on the recessive allele. When both do, the child will express the recessive condition.

Therefore, the potential probabilities are 1/2 for each parent being a carrier and 1/4 for their child being affected by the recessive condition.

Answer:

Explanation:

The presence of an oceanic trench, a chain of volcanic mountains along the edge of a continent and deep seated earthquakes is characteristics of ocean-continent plate convergence

Hope this help

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Have a nice day!!!!!

The presence of an oceanic trench, a chain of volcanic mountains along the edge of a continent and deep-seated earthquakes is characteristic of a convergent plate boundary, which typically forms when an oceanic plate is subducted under a continental plate.

The presence of an oceanic trench, a chain of volcanic mountains along the edge of a continent, and deep-seated earthquakes, are indications of a convergent plate boundary. This typically happens when an oceanic plate is subducted or pushed beneath a continental plate, forming a deep oceanic trench. The intense heat and pressure cause the subducted plate to partially melt, and this molten material can rise to form a chain of volcanic mountains along the edge of the continent, often known as a volcanic arc. The process of subduction also leads to seismic activity or earthquakes that are deep-seated.

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Answer:

Water shed is created when an area or land separates the water flowing to different rivers, basin and seas.

Explanation:

Water shed are of different types-

The number of watersheds in a model depends on its design. Watersheds are areas of land that drain into common outlets, such as a reservoir or bay. To determine how many watersheds are in a model, identify the common outlets and each individual area that drains into one of these outlets is a separate watershed.

The number of watersheds in your model depends on how your model is designed. A watershed is an area of land that drains all the streams and rainfall into a common outlet such as the outflow of a reservoir, mouth of a bay, or any point along a stream channel.

To determine the number of watersheds in your model, you need to identify the points at which water flows together, and designate those as common outlets. Each individual area that drains into one of these common outlets comprises a separate watershed. For example, if your model includes three rivers which all converge at a single point, you would have three watersheds.

This definition also implies that all of the land on earth can be divided into watersheds. The boundaries of a watershed are determined by the topography (i.e., the hills and valleys) that guides the direction of water flow.

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Answer:

True.

Explanation:

Answer: a. True

Explanation: Through evolutionary time, animals have developed more-complex body plans, including  true tissues, non-segmentation and bilateral symmetry which is possible as a result of evolutionary innovation--the introduction and progression of novel traits compared to what exists before leading to a more advanced or complex form. Non-segmentation indeed allows for evolutionary innovation in body form.

Answer:

False

Explanation:

Answer:

According to biologists the coelacanth is NOT the closest living relative to amphibians,

Explanation:

In this modern era of molecular biology there is evidence that the coelacanth and  tetrapods are not closely related. While in the other hand, evidences indicated a close relationship between lungfishes and tetrapods. The molecular analysis was based on mitochondrial DNA sequences. I have attached picture of evolutionary relationship.

FIG. 1. Alternative hypotheses of sister group relationships between sarcopterygii and tetrapods.  

(A) Lungfish as the sister group of

tetrapods.  

(B) Coelacanth as the closest living relative of tetrapods.

(C) Coelacanth and lungfish equally closely related as sister groups of  tetrapods.

Reference: Zardoya, R., & Meyer, A. (1996). Evolutionary relationships of the coelacanth, lungfishes, and tetrapods based on the 28S ribosomal RNA gene. Proceedings of the National Academy of Sciences, 93(11), 5449-5454.

Answer:

No, it doesn't support

Explanation:

The DNA based alternative hypothesis shows that the lung fish is actually the closest relative to amphibians rather than the coelacanth.

Answer:

Plants and animals both are living organisms that are directly dependent on the air. The air is comprised of various gases with definite proportions such as  78% of nitrogen, 21% of O₂, 0.9% of argon, 0.03% of CO₂, and other minor gases.

Oxygen (O₂) is the main gas that is essential for life to exist. Without this gas, no organisms can survive.

The organisms inhale oxygen (O₂) gas and exhale carbon dioxide (CO₂) gas.

These gases play an important role in the lives of plants and animals as it helps in carrying out the process of respiration. This process is essential in order to generate energy in the body of the organisms where the food particles (commonly known as sugar) are disintegrated by the cells. This is how an organism respire by inhaling O₂ gas and exhaling CO₂ gas.

Plants and animals die without air because they need to interchange oxygen and carbon dioxide with their surrounding environment to carry out cellular respiration (plants and animals) and photosynthesis (only plants).

In conclusion, plants and animals die without air because they need to interchange oxygen and carbon dioxide with their surrounding environment to carry out cellular respiration (plants and animals) and photosynthesis (only plants).

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Answer:

The amino acids in such a helix would be hydrophobic in nature.

An α helix is particularly suited to cross a membrane because all of the carbonyl oxygen atoms and the hydrogen atoms of amide of the peptide backbone take part in intrachain hydrogen bonds, thus stabilizing these polar atoms in a hydrophobic environment.

Explanation :

Many transmembrane protiens use several alpha-helices wrapped up together.

It is usually seen as the helical structure can internally satisfy all the hydrogen-bonds , it doesn't leave any polar groups that are exposed to membrane if the sidechains are hydrophobic.

Sometimes, 2 to 3 alpha helices will wrap around each other , forming coiled coil. In an aphipathic alpha helix , the hydrophobic R groups on one side of each helix interact with each other while the hydrophilic R groups on the other side of each helix will interact with water.

Answer:

→alpha-helices

→Non-polar Amino acids

→because they and non -polar and hydrophobic.

Explanation:

Membrane proteins can be intrinsic (integral ) that is embedded in the membrane bilayer or extrinsic(peripheral) attached to the outer membrane layer.

These  integral protein transcend  the entire phospholipid bilayer, with the alpha- helices. The latter have hydrophobic side chains of non-polar amino acids. They are held to  the cell membrane  with  these  side chains  which forms hydrophobic interactions   with fatty acyl group of the  phosphoslipid  bilayer, and  sometimes  ionic bond with the polar head of phospholipid.

These alpha helix are non-polar(uncharged) and hydrophobic.A characteristic  feature that make them  to   interact and fixed into  the integral   phospholipid hydrophobic medium.