The P and Q genes of a plant are 20 m.u. apart. A PP qq and a pp QQ individual cross, and the F1 cross with each other to produce the F2. The P and p, and Q and q alleles are completely dominant and recessive, and there are no interactions between genes P and Q. What phenotypic ratio would you expect in the F2

Explanation:

General way of circulation through the cat:  

Final answer:

The circulation pattern in the thoracic cavity involves the aortic arch, which leads to various arteries and returns blood to the right side of the heart via the superior and inferior vena cava. Both humans and cats have a diaphragm that aids respiration crucial for systemic circulation. Differences between species may include the branching pattern and vessel size from the aortic arch.

Explanation:

The question asks us to describe the differences in the circulation pattern in the thoracic cavity for the right side versus the left side after the vessels leave the aortic arch in humans compared to cats. Humans, as well as cats, have a four-chambered heart with the systemic and pulmonary circuits. However, the branching pattern of the aorta and the arrangement of the major arteries can vary between species. In humans, the aorta arcs to the left and gives rise to the brachiocephalic artery, the left common carotid artery, and the left subclavian artery. After the blood flows through the systemic circuit, it returns to the right side of the heart via the superior vena cava and inferior vena cava. The diaphragm, present in all mammals, plays a crucial role in separating the thoracic and abdominal cavities and assisting with respiration which is pivotal for the systemic circulation. In contrast, cats might display variations in the branching pattern from the aortic arch and the relative size of the blood vessels, which may affect the blood flow dynamics on either side of the heart after leaving the aortic arch. For both humans and cats, the heart pumps deoxygenated blood from the right side through the pulmonary circuit to the lungs and oxygenated blood from the left side through the systemic circuit to the body.

Answer: Option A ,Band C

Explanation:

Red blood cells are called erythrocytes. They are produced from the bone marrow and deposited into the blood stream when they are matured. The red blood cells carry oxygen through out the body.

Lymphocytes are white blood cells that are produced from the bone marrow and help in body defence, they are found in lymph tissues. They are immune cells.

Platelets are tiny cells that contribute to blood clots

Answer:

loops

Explanation:

An enhancer is a region in the DNA containing about 50-1500 base pairs that can be bound to proteins to increase the likelihood of transcription of a certain gene.

In the cell, the enhancer functions are limited by the formation of loops that hold specific genes and enhancers in close proximity.

By doing so there will be structural distortion of the DNA which is important for limiting enhancer functions.

Enhancer sequences in the cell facilitate gene regulation by forming chromatin loops within Topologically Associating Domains, ensuring proximity between enhancers and promoters.

In the cell, enhancer sequences function to regulate gene expression by facilitating the proximity of enhancers and promoters despite their linear distance on the DNA strand. This is achieved through the formation of chromatin loops which hold specific genes and their regulatory elements, like enhancers and silencers, in close spatial proximity within a three-dimensional space. These loops are formed within specific regions known as Topologically Associating Domains (TADs), which ensure that the effects of enhancers and silencers are limited to a subset of genes within the same domain, preventing unwanted interference with other genes located in different TADs.

Answer:

B. 3. Right-handed coil: 1.left-handed coil

Explanation:

Phenotype is what you see - the visible or observable expression of the results of genes, combined with the environmental influence on an organism's appearance or behavior.

When Nn is crossed with Nn, they will produce offspring with NN, Nn, Nn and nn genotype.

N - Dominant allele

n - recessive allele

The phenotypic ratio of this offspring is 3 right-handed coil and 1 left-handed coil.

Answer:

B. 3 right-handed coil: 1 left-handed coil

Explanation:

Female genotype : Nn

Male genotype: Nn

N: dominant allele (right handed coil)

n: recessive allele (left handed coil)

When we cross the Male (Nn) x Female (Nn) the phenotypic ratio will be right handed coil 3 : left handed coil 1.

  Solution:

                               Nn     x       Nn

Four possible genotypes = NN, Nn, Nn, and nn

So, NN = right handed coil

Nn (2) = right handed coil

nn= left handed coil

Answer:

here, marbled is dominant (MM) ; spotted is co-dominant (Ss)

in F1 generation → MS (marbled spotted coat)

    2. dotted and clear

here, dotted is co-dominant (Dd) ; clear is recessive (dd)

in F1 generation → Dd (clear background with dotted coat)

if     MS × Dd , then, it will give,

25% spotted and dotted ( as both are co-dominant {Ss=Dd}) : 50% marbled (as marbled is dominant{ MM}) : 25% spotted ( as spotted {Dd} is co-dominant and clear {dd} is recessive)

a. 25% spotted dotted: 50% marbled: 25% spotted

MS (marbled)- as marbled is the most dominant among all, it won't let spotted to be expressed.

and Dd - as clear is recessive only dotted will be expressed.

b.marbled and dotted.

Final answer:

The expected phenotypes from the mating of two F1 lentil plants from different crosses are 1:1 marbled to spotted/dotted, as marbled is dominant while spotted and dotted show codominance, and clear is recessive. The F1 offspring from the first cross will have a marbled phenotype, and from the second cross, a dotted phenotype due to codominance.

Explanation:

We are addressing complex inheritance patterns in lentils involving multiple alleles with a dominance series. To predict the offspring's phenotypes, we first need to understand Mendel's laws and how dominance affects the genotype-phenotype relationship. We also employ Punnett squares to visualize genetic crosses.

a. Expected Phenotypes in Offspring from F1 Mating

The F1 plants from the first cross (marbled × spotted) will all exhibit the marbled phenotype, as marbled is dominant over spotted. The F1 plants from the second cross (dotted × clear) will all show the dotted phenotype due to codominance between dotted and spotted alleles, and dominance over clear.

When these two F1 plants are crossed (marbled/dotted × dotted), the expected offspring phenotypes are as follows, assuming each allele has equal chance of passing on:

Marbled: Represents the dominant allele and is expected to show up in half the offspring if the marbled allele is present.

Spotted/Dotted (Codominance): Should appear in the other half of the offspring, as these are the alleles present in both F1 parents. The spotted and dotted patterns are codominant and will be expressed equally if one of each is inherited.

Clear: Will not be seen in the offspring, as it is recessive to all other alleles.

Thus, the phenotypic ratio of the offspring from the cross between the two F1 plants is expected to be 1:1 marbled to spotted/dotted with no clear phenotypes.

b. Expected F1 Phenotypes from Original Parental Crosses

The F1 phenotype resulting from a homozygous marbled crossed with homozygous spotted will be all marbled, following Mendelian inheritance patterns. In the cross of homozygous dotted with homozygous clear, all offspring will display a dotted pattern, indicative of codominance interaction over the clear allele.

Answer:

x = oxygen

Explanation:

Food and water is essential to life at cellular level because this water is responsible for making food with the help of CO2 during photosynthesis in plant and this food provides energy to the cell and support the life of plant cell. In an animal cell, H2O helps in releasing waste product from cell and provide a medium for reactions to occur.

Apart from food and water oxygen is also required to support life because oxygen is required by cells to oxidize food and release energy from food so this energy is used to perform metabolic function of cells that support cell life.

Final answer:

Macromolecules and water provide essential nutrients vital for cellular functions. Biological macromolecules are carbohydrates, lipids, proteins, and nucleic acids, synthesized through dehydration synthesis. The additional factor 'x' needed for life is oxygen, essential for cellular respiration and ATP production.

Explanation:

Answer:

Penguins in general are distributed close to each other to be able to withstand the extreme cold, characteristic of the ecosystems they inhabit. They also group together to incubate their eggs.

They perform a series of coordinated movements that help them fight low temperatures, being careful to drop the egg.

Penguins are most likely to use a clumped distribution. This allows them to balance the need for warmth from proximity to others while minimizing the risk of nest disturbances. This pattern is common amongst many species in the animal kingdom.

The behavior described in the question for penguins refers to clumped distribution. This pattern occurs when individuals in a population find it advantageous to stay close to one another, often for reasons of protection, access to resources, or reproduction. However, as described, being too close can also have its drawbacks, such as the risk of having a nest stepped on by a neighbor. This balance between proximity for warmth and space for safety is indeed what defines a clumped distribution.

In the animal kingdom, many species use clumped distribution as a defense mechanism against predators. For penguins in particular, communal nesting not only provides safety in numbers but also helps conserve heat in the harsh, cold climate where they live.

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Answer: Option D) osteocalcin; regulates glucose metabolism is not correctly matched

Explanation:

The function of osteocalcin, secreted by cells of the bone (osteoblasts) is to regulate body metabolism by increasing the level of calcium in the bone.

Insulin and Glucagon secreted by the pancreas, on the other hand, regulate glucose metabolism by lowering and increasing the level of blood sugar respectively.

Thus, osteocalcin; regulates glucose metabolism is not a correct match

Hormone is NOT matched with its appropriate function c. calcitonin; increases blood calcium levels Therefore , c. calcitonin; increases blood calcium levels is correct .

Calcitonin is a hormone produced by the thyroid gland. Its main function is to regulate calcium levels in the blood by decreasing calcium levels. When blood calcium levels rise too high, calcitonin is released to lower the levels by promoting calcium deposition in bones and inhibiting calcium absorption in the intestines.

a. Growth hormone; lengthens bones - Growth hormone, also known as somatotropin, is produced by the pituitary gland and plays a crucial role in stimulating growth, cell regeneration, and cell reproduction.

It indeed promotes the lengthening and thickening of bones, among other growth-related functions.

b. Leptin; regulates bone density - Leptin is a hormone primarily produced by adipose cells. Its main function is to regulate energy balance and inhibit hunger.

While it primarily regulates appetite and body weight, there's some evidence suggesting it may indirectly influence bone metabolism.

d. Osteocalcin; regulates glucose metabolism - Osteocalcin is a hormone produced by osteoblasts, cells that create bone.

Recent research has indicated that osteocalcin might play a role in glucose metabolism, affecting insulin secretion and sensitivity.

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Answer:

A compound is made up of atoms of different elements combined together in a chemical reaction with fixed ratio and in a defined manner via chemical bonding.

Elements on the other hand are pure chemical substance made of same type of atom.

The properties of the compound is related to the ratio of elements in a way that the elements individual properties will manifest itself in the compound giving it a hybrid property.

Answer: the child's phenotype will be normal.

The child will acquire that disease.

Explanation:

The mitochondrial DNA is a small circular chromosome that exists in the mitochondria. The mitochondria is the power house of the cell and all its 37 DNA are essential for a normal function of the mitochondria. the mtdna is passed directly from the mother (through the egg cell) to the child either a male or female but the son cannot pass his mtdna to his child. Thus, if a man with this disease marries a woman not affected, the child's phenotype will be normal. But if the wife has the disease, this willl be passed from her to her child me it a male or female.

Answer:

Explanation:

Answer:

A neuron or a nerve cell is a unique cell, which performs an essential function of transmitting the nerve impulse in the form of action potential from one nerve cell to another. The point of communication between two neurons is known as a synaptic junction from where the transmission of a signal between presynaptic axon end and the postsynaptic dendrite occurs.  

The synapse present in the dendrite receives a signal in the form of neurotransmitter from the presynaptic axon. This results in the formation of an action potential that gets transmitted towards the cell body of a neuron. From the cell body, the conduction of impulse takes place via a long tubule composition known as an axon, which constitutes the nodes of Ranvier and reaches the nerve endings or the axon terminals.  

In the terminals, the change in potential results in the opening of the ions channels that discharge neurotransmitters in the form of acetylcholine, which further combines with the ligand-receptor situated on the next dendrite and thus repeats the process. Thus, the action potential travels from the dendrite to the cell body, and from there it travels to the terminals of the axon and eventually towards the ion channels and the ligand-receptor.  

Answer:

A. by reacting animal fat or vegetable oil with water.

D. by treating animal fat or vegetable oil with hydrogen

Explanation:

Water is insoluble in animal fat or vegetable oil so it makes a layer on the surface of oil. When animal fat or vegetable oil react with hydrogen, it produces solid oil called ghee not biofuels. Biofuel is produced when animal fat or vegetable oil react with alcohol.

Answer: Option A and D.

Explanation:

Biofuel is fuel that produced from either plants, algae or animals. There are four main types;

Ethanol that is made from corn.

Biodiesel made from animal fats and vegetable oil.

Green diesel made from algae and plant sources.

Biogas e.g methane made from animal manure and digested organic material.

Ethanol a type of biodiesel is made from fermentation of sugar components in starch crops.

Biodiesel is made from reaction between animals fats or vegetable oil with alcohol.

The question is incomplete. I have attached the complete question  

Answer:

Ears and starch are linked, glossy isn't linked to either

Explanation:

The deviation from expected frequencies indicates that ears and starch genes are linked  

Answer:

B) Replication of their genomes does not involve the proofreading steps of DNA replication.

Explanation:j

1. This is because RNA polymerase do not proofread proteins or polymerase components to check for errors and repair.

2.Because RNA viruses usually encoded their replication machinery, therefore they  have the ability to  replication at high rate   which  increases chances  for  mutation, to any length and  extent to  suits their needs.

Therefore,since there is no check for the number of replication  they can go through and  no need for correction of  replication errors, they have rapid chances for  mutation  and therefore virulence. Different  offspring, with distinct genetic composition  from wrong bases insertion and translation and therefore of  virulence   totally different from  the parents will be produced. Thus mutagensis, and virulence continues to increase.

RNA viruses have higher mutation rates primarily because their replication lacks DNA's proofreading mechanisms, making RNA-dependent RNA polymerases more error-prone, facilitating genetic variability and adaptation. So option B is correct choice.

RNA viruses tend to have higher mutation rates primarily because their replication process lacks the proofreading mechanisms present in DNA replication.

During DNA replication, enzymes like DNA polymerase have built-in proofreading capabilities, which help to correct errors in base pairing, resulting in a relatively low mutation rate.
In contrast, RNA viruses typically lack these proofreading mechanisms during genome replication. RNA-dependent RNA polymerases, the enzymes involved in replicating RNA genomes, are more error-prone, leading to a higher likelihood of mutations.

This elevated mutation rate contributes to the genetic variability seen in RNA viruses, allowing them to adapt more rapidly to changing environments, evade host immune responses, and potentially develop resistance to antiviral treatments.
While factors such as the instability of RNA nucleotides (option A) and the speed of replication (option C) can contribute to higher mutation rates, the absence of proofreading during replication (option B) is the primary reason for the observed higher mutation rates in RNA viruses.

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To determine where each species belongs on the phylogenetic tree, examine their shared traits and make inferences about their evolutionary history. Five additional traits can be added to the phylogeny, such as anatomical structures, biochemical reactions, reproductive strategies, behaviors, and genetic sequences.

When examining specimens in the lab, you can determine where each species belongs on the phylogenetic tree based on their shared traits. By comparing the traits of different species, you can identify common characteristics and make inferences about their evolutionary history.

For example, if two species share a trait that is not found in any other species, they are likely more closely related and would appear closer on the phylogenetic tree.

To add five additional traits to the phylogeny, you can consider various characteristics such as anatomical structures, biochemical reactions, reproductive strategies, behaviors, and genetic sequences. By including these traits, you can further refine the phylogenetic tree and understand the relationships between different species more comprehensively.

Answer:Red is 91%, white is 9%

Explanation:3000-270 =2730 red in percentage is 2730/3000 *100/1 =91%

White is 270/3000*100/1= 9%.

Allele frequency or gene frequency is the relative frequency of an allele expressed as a fraction or percentage.

Answer:

0.08

Explanation:

According to Hardy-Weinberg equilibrium, in absence of an evolutionary force allele frequencies in a population remain constant. In case of polyploid organisms, the formula for Hardy-Weinberg equilibrium is :

(p+q)^c = 1  where,

p = frequency of dominant allele

q = frequency of recessive allele

c = ploidy number

Here, the ploidy number is 4 since there are four chromosomes at a locus instead of the usual two.

f(C1)(C1) = 0.28

f(C2)(C2) = 0.72

Black rabbits = C1C1C1C1

Frequency of black rabbits= f(C1)(C1)*f(C1)(C1)

= 0.28 * 0.28

= 0.0784

= 0.08

The given question is incomplete. The complete question is as follows.

Carbon absorbs energy at a wavelength of 150 nm. The total amount of energy emitted by a carbon sample is [tex]1.93 \times 10^{5} J[/tex]. Calculate the number of carbon atoms present in the sample, assuming that each atom emits one photon.

Explanation:

It is given that the energy at which C-atom absorbs energy is 150 nm. So, energy emitted by the carbon atom will have same wavelength at which C-atom absorbs the energy.

As we know that relation between energy and wavelength is as follows.

                  E = [tex]\frac{hc}{\lambda}[/tex]

where,  h = Planck's constant = [tex]6.624 \times 10^{-34} J sec[/tex]

             c = speed of light = [tex]3 \times 10^{8} m/s[/tex]

             [tex]\lambda[/tex] = 150 nm = [tex]150 \times 10^{-9}[/tex]

Therefore, energy of one carbon atom is calculated as follows.

                E = [tex]\frac{hc}{\lambda}[/tex]

                  = [tex]\frac{6.624 \times 10^{-34} Js \times 3 \times 10^{8} m/s}{150 \times 10^{-9}}[/tex]

                 = [tex]1.324 \times 10^{-18} J[/tex]

As the total energy emitted by the carbon sample is  [tex]1.93 \times 10^{5} J[/tex]. Let us assume that the number of C-atoms in the sample be x and it is calculated as follows.

           [tex]E_{total} = n \times E_{1C-atom}[/tex]

                     n = [tex]\frac{E_{total}}{E_{1C-atom}}[/tex]

                        = [tex]\frac{1.93 \times 10^{5}}{1.324 \times 10^{-18} J}[/tex]

                        = [tex]1.45 \times 10^{23}[/tex]

Thus, we can conclude that number of carbon atoms present in the sample, are [tex]1.45 \times 10^{23}[/tex].

Answer: Option E) Facilitated diffusion

Explanation:

Facilitated diffusion engage selective transport proteins to move non-lipid soluble solutes like glucose across the lipid bilayer along concentration gradient (from high to low concentration) without the supply of energy (ATP).

Thus, transport proteins embedded in the lipid bilayer help move large quantities of glucose rapidly in intestinal cells by facilitated diffusion.

Answer:

it doesnt make sense can you try to put it some other way

Explanation:

Answer:

Specialist of High Average Level

Explanation:

While studying all these factors affecting the environment at the same time. One should have a High Average intellect of knowledge related to Aquatic, Marine, Soil, Microbial, Forest, Zoology, Botany, Demography as well as enough knowledge related to Bio-Physics. Only then, he'll be able to tell about the consequences related to the situation.

A scientist studying the effects of beaver dams on river ecosystems would be an ecosystem ecologist, a role that involves examining the intricate interactions between organisms and their environment, including the impact on water flow, soil nutrients, and species dynamics.

If you were a scientist studying the impact of beaver dams on river water flow, soil nutrient content, and the dynamics of other populations such as fish and amphibians, you would be operating at the ecosystem level of biology within the field of ecology. Ecosystem ecologists examine the relationships between living organisms, including humans, and their physical environment. They focus on understanding how ecosystem components, such as the water flow and soil nutrients affected by beaver dams, interact and affect the distribution and abundance of organisms in a particular habitat.

Beaver dams can have profound effects on river ecosystems. By altering water flow and sediment distribution, these ecosystem engineers can change the dynamics of aquatic and terrestrial interfaces, influence nutrient cycles, and affect the biodiversity and health of wetlands. Changes to these ecosystems can notably impact populations of fish and amphibians, which depend on specific flow regimes and access to aquatic and terrestrial habitats for their life cycles.

Ecologists interested in conservation biology might also draw on habitat modification studies, as damming can drastically influence ecosystem dynamics. Key informed decisions for biodiversity conservation and restoration often rely on understanding such complex interactions. By investigating and modeling these impacts, ecosystem ecologists contribute valuable insights into sustainable management practices that can mitigate the effects of human intervention on natural habitats.

Answer:

what is the question you basicly just gave us a paragraph to comment on

Explanation:

The security level, represented by the bit length, impacts the algorithm performance. RSA and ECC illustrate this point, providing the same security level, yet ECC is faster due to its efficiency and the use of fewer bits for the same security.

The level of security indeed has a direct impact on the performance of an algorithm, and this can be clearly seen in your given example with RSA and ECC encryption algorithms. RSA-1024 takes 15.7ms for signature generation, whereas ECC-160 only takes 1.3ms. This speed discrepancy can be attributed to the fact that ECC (Elliptic Curve Cryptography) is much more efficient than RSA. Although ECC delivers the same level of security with fewer bits, thus speeding up the process, RSA requires more computational resources for the same level of security.

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Answer: Option B) loop of Henle

Explanation:

The longitudinal section of the kidney has two distinct regions: an outer cortex and an inner medulla. The outer cortex has the following parts: glomerulus, bowman's capsule, convoluted tubule; while the inner medulla has the loop of Henle, a U-shaped loop that comes from the proximal convoluted tubule found in the medulla.

Answer: Option B.

Loop of henle.

Explanation:

The kidney longidutinal section is divide into inner and outer medulla.

The outer contain the Bowman's capsule, glomerulus e.t.c . The inner medulla of the kidney contains the loop of henle

Loop of henle is a long u shaped like part , found in the Inner medulla of the kidney. It move urine from each nephron of the kidney. It help to reabsorb water and sodium chloride from the urine. It carries filtrate from the proximal tubule to the renal medulla and move it back to the renal cortex.

Answer:

The correct answer is: β1  adrenergic receptor activation. Increased in cAMP lead to increased amounts of Na influx (though if channels) and calcium influx of  pacemaker cells. This increases the frequency of APs of the pacemaker and increases the rate of contraction.

Explanation:

cAMP generated in response to β1 adrenergic receptors,results in excitation contraction coupling by initiating PKA and causing phosphorylation of L type Ca2+ channels and ryanodine receptors. This increases the concentration of intracellular Ca2+ ions in atrial cells, ventricular cells and AV pacemaker cells. This increase in firing rate. As a result rate of contraction increases.

Answer:

C. cannot be larger than its fundamental niche

A species' realized niche, defined by the conditions under which it can survive and reproduce in the face of competition and other contrasts, cannot be larger than its fundamental niche, which represents all the environmental conditions under which the species can exist.

A species' realized niche refers to the conditions under which a species is able to survive and reproduce in a specific environment, in the presence of competition and other environmental constraints. It is essentially the range of abiotic and biotic conditions under which a species can persist. Conversely, the fundamental niche refers to the full set of environmental conditions under which a species can survive and reproduce in the absence of competition from other species.  

Therefore, a species' realized niche cannot be larger than its fundamental niche, since the realized niche is shaped by the presence of competition and other limiting factors, narrowing down the conditions under which the species can exist from the broader possibilities offered by the fundamental niche.

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Answer:false

Explanation: protists are mainly

Unicellular organism( i.e they are single cell organism) and are eukaryote( they have a nucleus) but some protist exist as multicellular example is some Algea, kelp. They can be find in moist environment.

Protist can be grouped into the

1. Animal protist are heterotrophs and they depend on autotrophs or producers for their food or other organism and are mobile( they move)

2. Plant protist are mainly autotrophs they have ability to synthesis there food through photisynthesis.

3.Fungi protist reproduce using spores. They are heterotrophs.

Some protist move by pseudopodia while some uses flagella.

Hence protein are not mainly multicellular eukaryotes but both unicellular and multicellular eukaryotes.

Answer:

false

Explanation:

protists are a diverse group of mainly unicellular eukaryotes.

Answer:

The size of the rotary and stabilizing components of muscle are 129.41 N and 482.96 N

Explanation:

According to the angle produced between the deltoid attached and the humerus  

The Rotary Formula (or Vertical Component) is

FR = 500 N · sin 15° = 129.41 N

and the Stabilizing Formula (or Horizontal Component) is

FS = 500 N · cos 15° = 482.96 N

Final answer:

For a muscle force of 500 N attached at a 15-degree angle, the rotary and stabilizing components are calculated using trigonometry, resulting in a rotary component of approximately 129.4 N and a stabilizing component of approximately 482.95 N.

Explanation:

The question pertains to calculating the rotary and stabilizing components of a muscle force when the total muscle force is given and the muscle is attached at a specific angle. To solve this, we apply knowledge of trigonometry and vector decomposition. Given that the total muscle force is 500 N and it's attached at a 15-degree angle to the humerus, the force can be decomposed into two components: the rotary (perpendicular) component and the stabilizing (parallel) component.

Calculations:

Rotary component (Fr): This is the component of the force acting perpendicular to the lever arm. It is calculated using the sine function (Fr = Total Force × sin(θ)). For a 15-degree angle, Fr = 500 N × sin(15°) = 500 × 0.2588 ≈ 129.4 N.

Stabilizing component (Fs): This is the component of the force acting parallel to the lever arm. It is calculated using the cosine function (Fs = Total Force × cos(θ)). For a 15-degree angle, Fs = 500 N × cos(15°) = 500 × 0.9659 ≈ 482.95 N.

In conclusion, for a total muscle force of 500 N attached to the humerus at a 15-degree angle, the rotary component is approximately 129.4 N and the stabilizing component is approximately 482.95 N.

Answer:

The radial symmetry and mesodermal skeleton in echinodems

Bilateral symmetry and internal skeleton of chordates

Explanation:

The echinoderms have radial symmetry and their skeleton is mesodermal made up of calcites known as ossicles while the chordates have bilateral symmetry and their internal skeleton is made up of bones and cartilages.

Explanation: