Describe the helical and coiled-coil structures of either collagen or α-keratin. What type of covalent bond holds the coiled-coils in staggered arrangements to form larger fibrous structures of this protein?

Answers

Answer 1

Answer:

Collagen is an extracellular matrix protein and one of the most abundant protein in the body. Its triple-helical structure is a result of the 3 coiled chains. Each chain mainly comprises of Glycine, Proline and Hydroxyproline. The Glycine hydrogen bonds with residues of the adjacent chain while the proline and hydroxyproline C-N bonds provide stability in the structure.

Explanation:


Related Questions

What element was pointed out for special attention in the structure of a DNA base?
a. oxygen
b. carbon
c. phosphorus
d. nitrogen
e. calcium

Answers

Answer:

The correct answer will be option-D.

Explanation:

Deoxyribose nucleic acid or DNA is the genetic material of the organisms which is composed of nucleotide monomers.  Each nucleotide monomer is made up of a fiver carbon sugar, four different nitrogenous bases and a phosphate group

The DNA bases can be easily distinguished on the basis of nitrogen element as the structure of purine which is adenine and guanine contains four nitrogen molecule whereas cytosine and thymine contain two nitrogen molecules in their structure.

Thus, option-D is the correct answer.

What type of lipid would most likely aggregate into a micelle rather than a lipid bilayer when mixed with water?
a. glycerophospholipid
b. cholesterol
c. galactolipid
d. shingolipid
e. free fatty acids

Answers

Answer:

e. free fatty acids

Explanation:

Free fatty acids would most likely aggregate into a micelle rather than a lipid bilayer when mixed with water.

Free fatty acids produced by the activity of lipase enzyme on stored triglycerides in the cytoplasm.

Micelles are the molecules of lipid which are arranged in a spherical manner which have a polar head and a hydrophobic tail. Micelles transport the fatty acids and insoluble monosaccharides.

Which of the following statements about DNA isolation from E. coli is not correct?
a. chemical extraction using phenol removes proteins from the DNA
b. RNA is removed from the sample by RNase treatment
c. detergent is used to break apart plant cells to extract DNA
d. lysozyme digest peptidoglycan in the bacterial cell wall
e. centrifugation separates cellular components based on size

Answers

Answer:

The correct answer is c. detergent is used to break apart plant cells to extract DNA.

Explanation:

Here we are talking about DNA isolation in E.coli and detergent is used to break apart the bacterial cell membrane by destabilizing the cell membrane. Detergent disintegrates the proteins present in the cell membrane which causes the rupture of membrane releasing the cell extract outside the cell.  

Detergent is not able to break plant cell because it is made up of cellulose and require blending or mashing. Detergent can only be used to disintegrate the cell membrane because only cell membranes contain lipids and proteins.

Hence, the correct answer is c.  detergent is used to break apart plant cells to extract DNA.

Discuss the relationship between the recombination frequency and the map distance separating two loci on a chromosome.

Answers

Answer:

Linkage chromosomes may be defined as the chromosome present at the same chromosomes and has the ability to undergo the process of recombination. The linkage results in the production of recombinant gametes that differ from their parents.

Map distance may be defined as the distance between the two gene on a chromosome.  The recombination frequency is equal to the map distance. For example: Gene A and B has the recombination frequency of 10%, this means there map distance is 10 mu.

The evolution of archosaurs was heavily impacted by the climate and environmental changes that occurred in the Early Triassic. Which of these best describes the environment in the Triassic?
a. The vegetation was dominated by grasses and angiosperms which were new food sources
b. The equatorial climate became much more arid compared to that seen in the Permian
c. The polar regions were dominated by permanent polar ice caps that were linked to lower sea levels.
d. Oxygen levels were extremely high due an increase in worldwide vegetation

Answers

Answer:

Option (B)

Explanation:

The Triassic period belongs to the Mesozoic era in which the dominant species was the dinosaurs. During this time, the environment and the type of climate was relatively dry and there was no precipitation to support the vegetation. During, this Triassic period, the super-continent Pangaea was present that eventually got split up into different smaller continents by Jurassic and Cretaceous. The climate in the equatorial region during that particular time on earth was comparatively much more arid than that was in the Permian.

There were no polar ice caps during that time and tectonic activities were rapid. The earth experienced a temperature of about 10°C more than the present day temperature and oxygen level was also relatively low then. The plants that existed during this period were mostly gymnosperms.

Hence, the correct answer is option (B).

The structural level of a protein least affected by a disruption in hydrogen bonding is the
a. primary level.
b. secondary level.
c. starch d. carbohydrate
c. tertiary level.
d. quaternary level.

Answers

Answer: a. Primary level.

Explanation:

The primary level is characterized by the sequence of aminoacids of the protein, aminoacids have a carboxyl and an amino group bonded to a carbon atom, and are held together by peptide bonds between the carboxyl end of an aminoacid and the amino end of the next one, making an aminoacid chain.

Secondary, tertiary and quaternary levels determine the shape and thus the function of the protein according to different forces, hydrogen bonds take part in it bending the chain in different ways.

Final answer:

The primary structure of a protein is least affected by a disruption in hydrogen bonding because it is maintained by covalent peptide bonds, which are not broken under conditions that disrupt weaker hydrogen bonds.

Explanation:

The structural level of a protein least affected by a disruption in hydrogen bonding is the primary structure. This level of structure refers to the sequence of amino acids in the polypeptide chain, held together by peptide bonds, which are much stronger than hydrogen bonds. Despite the breaking of hydrogen bonds during denaturation, which will disrupt secondary, tertiary, and quaternary structures, the primary structure remains intact because the peptide bonds are not broken under conditions that disrupt hydrogen bonds. Therefore, the answer is a. the primary level.

The secondary structure is defined by the spatial arrangement of the polypeptide chain's backbone, primarily through hydrogen bonding, forming structures like α-helices and β-pleated sheets. Interruption of hydrogen bonding in this structure affects the correct folding and thus its stability and function.

Tertiary structure is the three-dimensional shape of a single polypeptide chain, stabilized by various interactions, including hydrogen bonds. Quaternary structure involves the interaction and assembly of multiple polypeptide subunits, which is also dependent on hydrogen bonds among other interactions for stability.

While all secondary, tertiary, and quaternary structures can be disrupted by the loss of hydrogen bonding, it is specifically the primary structure that will persist because its stability is owed to covalent peptide bonds.

Photosystem II in photosynthesis, obtains its electrons from
a. Water
b. Oxygen
c. Carbon Dioxide
d. NADPH

Answers

Answer:

a. Water

Explanation:

An arrangement of chlorophyll and other pigments in the chloroplast is known as photosystems. In the photosynthetic eukaryotes, there are two photosystems, Photosystem II and Photosystem I. Photosystem I uses chlorophyll a, with reaction centre as P700. Photosystem II uses chlorophyll a with reaction centre P680.

During light reaction or photochemical reaction of photosynthesis, the light energy causes the removal of an electron from chlorophyll a reaction centre of Photosystem II P680. After losing an electron the P680 is deficient of an electron. P680 takes an electron from photolysis of a water molecule into eletron, H⁺ ions and O⁻² ions.

   

Amino acids differ from one another by the ________________ that is present
a. amino group
b. carboxyl group
c. R group
d. R plasmid
e. core group

Answers

Final answer:

Amino acids, fundamental in biological processes, possess an amino group, a carboxyl group, a hydrogen atom, and a side chain or R group. Amino acids differentiate from each other based on their unique R group. This distinctive R group gives each amino acid its singular characteristics.

Explanation:

Amino acids are organic compounds that play a crucial role in biological processes. They are composed of an amino group (NH2), a carboxyl group (COOH), a hydrogen atom, and a side chain (R group). Each amino acid differs from another by its unique R group. So, the answer to your question is c. R group.

The R group can vary among different amino acids and this variation is what gives each amino acid its unique properties. For example, the R group can be as simple as a single hydrogen atom (as in the case of the amino acid glycine) or a more complex arrangement of atoms (as in the case of tryptophan).

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Final answer:

Amino acids differ from each other by their respective R groups, which represent unique side chains. These side chains could range from simple hydrogen atoms to complex multi-ring structures.

Explanation:

Amino acids differ from one another by the R group that is present. Every amino acid has three main components: a central carbon atom, an amino group (-NH2), and a carboxyl group (-COOH). However, each also has a unique side chain, or R group, that sets it apart from the others.

This R group could be as simple as a single hydrogen atom or as complex as a multi-ring structure. It's due to these distinct R groups that we have 20 different common amino acids in proteins.

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Atoms of which three elements could bond together to form an organic compound?


(a) carbon, iron, and nitrogen

(b carbon, hydrogen, and oxygen

(c) oxygen, hydrogen, and nitrogen

(d) oxygen, iron, and carbon

Answers

Answer:

b

Explanation:

carbon, hydrogen and oxygen

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HEY hun!

your answer is B) Carbon, Hydrogen, and Oxygen

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In organic soil with plenty of oxygen, organic notrogen will be converted to NH4+ by decomposens the NH4+ will be converted to NO3- by bacteria.
a. True
b. False

Answers

Answer:

False NH₄+ is transformed first to NO₂ and then to NO₃

Explanation:

Organic nitrogen which appears in different forms like urine, dead animals or plants, Nitrogen will be converted to ammonium (NH₄+) by either fungi or bacteria, this process is called ammonification.Nitrification is the process through which ammonium (NH₄+) is transformed to nitrites (NO₂) and then into nitrates (NO₃). Each transformation is performed by different types of bacteria species.

The mitochondrial inner membrane form a series of infoldings known as cristae to:
a. Increase surface area
b. Decrease surface area
c. Increase production of H+
d. Decrease production of H+

Answers

Answer: Increase the surface area.

Explanation:

The mitochondria is an important cell organelle that is found in both the plant cell and animal cell.

It basically aids in the production of ATP, hence also known as the power house of the cell.

The mitochondria is a double membrane system in which the inner membrane of mitochondria helps in increasing the surface area for integral proteins.

It helps in embedding the proteins in the folding known as cristae.

Which of these is true regarding prokaryotes?
a. It takes about 6-8 hours to duplicate DNA
b. Microtubules bind to the centromeres and pull the chromosomes apart
c. Separated DNA is attached to the cell membrane before the cell divides
d. The nucleus dissolves

Answers

Answer:c. Separated DNA is attached to the cell membrane before the cell divides.

Explanation:

The prokaryotes are single celled organisms. These are simple organisms which reproduce through asexual mode of reproduction that is cell division. They do not posses well define nucleus. Thus the genetic material remain in the cytoplasm of the cell. On cell division the genetic material (DNA) is distributed into halves for development of two daughter cells. Due to lack of nucleus and it's associated membrane the separated DNA get attach to the membrane before the cell actually divides.

According to the fluid mosaic model of membrane structure, proteins of the membrane are mostly
a. spread in a continuous layer over the inner and outer surfaces of the membrane.
b. confined to the hydrophobic interior of the membrane.
c. embedded in a lipid bilayer.
d. randomly oriented in the membrane, with no fixed insideoutside polarity

Answers

Answer:

Embedded in a lipid bilayer. (Ans. C)

Explanation:

Fluid mosaic model explains different observations such as the structure of functional cell membranes. According to this model, there is a thin polar membrane composed of a two-layer of lipid molecules called lipid bi-layer or phospholipid bi-layer in which proteins molecules are embedded. This layer provides fluidity and elasticity to the membrane structure.

These membranes form a continuous barrier around all cells and they are a flat sheet in structure. The lipid bilayer is working as a barrier that keeps proteins, ions and other molecules where they are needed to be and preventing diffusing them to other areas. They are impermeable to most hydrophilic molecules. They are particularly impermeable to ions, which allow cells to regulate pH and salt concentration in membrane through the transport ions across their membranes using proteins knows as ion pumps.

Final answer:

According to the fluid mosaic model of membrane structure, proteins of the membrane are mostly embedded in a lipid bilayer and have a specific inside-outside polarity.

Explanation:

According to the fluid mosaic model of membrane structure, proteins of the membrane are mostly embedded in a lipid bilayer. They can be found throughout the membrane, both on the inner and outer surfaces, as well as in the hydrophobic interior. However, they are not randomly oriented in the membrane, but have a specific inside-outside polarity.

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Which of the following is not considered evidence supporting the evolutionary relationship among all plant taxa?
a. Plants share a significant number of genes
b. Plants and plant ancestors have chlorophyll
c. Fossils of plants show us transition series and evolutionary patterns of plants
d. Structures/characteristics among most plants are very similar

Answers

Answer:

The correct answer is option d. "Structures/characteristics among most plants are very similar".

Explanation:

There are extensive evidence to support the evolutionary relationship among all plant taxa, however, similarity of structures and characteristics among most plants is not one of them. Plants are one of the most diverse kingdoms having   very different species such as flowering plants, conifers, lichens, cycads, algae, among others. The structures and characteristics of these plants are very different among them, and this diversity makes difficult to classify and clarify the evolutionary relationship that they share.

The functioning of enhancers is an example of
a. a eukaryotic equivalent of prokaryotic promoter functioning.
b. transcriptional control of gene expression.
c. the stimulation of translation by initiation factors.
d. post-translational control that activates certain proteins.

Answers

Answer:

b. transcriptional control of gene expression.

Explanation:

The binding sites of transcription factors (enhancers and silencers) can activate or deactivate, respectively, a gene in specific parts of the body.

Some genes must be expressed in a part of the body. For example, if you need to activate a gene in the spine, skull and toes, but not in the rest of the body, how can transcription factors do this task?

As a gene of this type of pattern it can have several enhancers or silencers, where each one can activate or repress the gene of a specific type of cell or part of the body, joining transcription factors that occur in that specific part of the body.

Final answer:

The functioning of enhancers exemplifies transcriptional control of gene expression in eukaryotes, where they augment the transcription rate by enabling interactions between transcription factors and the transcription machinery at the promoter.

Explanation:

The functioning of enhancers is an example of transcriptional control of gene expression. Enhancers are regions in eukaryotic DNA that can be located quite far from the genes they regulate. They contain specific DNA sequences where transcription factors can bind, thus influencing the transcription of a gene. These enhancers can increase the transcription rate of associated genes by facilitating the interaction between transcription factors and the general transcription machinery, including RNA polymerase at the promoter site. This process involves the folding of DNA to bring enhancers into proximity with promoters, making it possible for the transcription factors and the transcription initiation complex to interact effectively. Unlike prokaryotes which rely heavily on the interaction of activators and repressors with promoters, enhancers provide an additional layer of transcriptional regulation in eukaryotes. The concept of enhancers expands the understanding of intricate gene regulation mechanisms present in eukaryotic cells beyond what is seen in prokaryotic systems.

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Which test provides an operational definition of a gene?
a. complementation test
b. biuret test
c. gram stain
d. neurospora test
e. fluctuation test

Answers

Answer:

The correct option is: A) Complementation test

Explanation:

Complementation test is used for determining that the mutations in the two strains, associated with a particular phenotype, are present in different genes.

The mutations are said to be complementary, if present in different genes.

Therefore, the complementation test is the process for determining the function of the gene in recessive allelism.

A mutant cucumber plant has flowers that fail to open when mature. Crosses can be done with this plant by manually opening and pollinating the flowers with pollen from another plant. When closed x open crosses were done, all the F1 progeny were open. The F2 plants were 145 open and 59 closed. A cross of closed x F1 gave 81 open and 77 closed. How is the closed trait inherited? What evidence led you to your conclusion?

Answers

Answer:

This trait is inherited according to  Mendel's principles and it posses a complete dominance, where open is dominant and closed is recessive.

Explanation:

First, there are closed-flowers plants and open-flowers plants. A cross between them produces only open flowers at F1. With this information, we can say that open condition is dominant over closed, and probably, parental plants were homozygous, like this:  

- Open flowers: FF

- Closed flowers: ff

- F1 open flowers: Ff

When F1 was crossed, the progeny was:

- 145 open

- 59 closed

When we cross two heterozygous, we obtain a phenotypical radio of 3:1, like this:

- (1) FF Open

- (2) Ff Open

-  (1) ff Closed

This ratio is very similar to the F2 above reported (145 open and 59 closed).

When F1 (Ff) plants were crossed with closed plants the progeny was:

- 81 open

- 77 closed

This is alike to a ratio 1:1.

In the Punnett Square, it is possible to notice that in this cross it is expected a ratio of 1:1 closed/open.

In conclusion, this trait is inherited according to  Mendel's principles and it posses a complete dominance, where open is dominant and closed is recessive.

In humans, a. How many sperm develop from 100 primary sper matocytes? b. How many sperm develop from 100 secondary spermatocytes? c. How many sperm develop from 100 spermatids? d. How many ova develop from 100 primary oocytes? e. How many ova develop from 100 secondary oocytes? f. How many ova develop from 100 polar bodies?

Answers

Answer:

In males primary spermatocytes undergo two meiotic divisions. First meiotic division give rise to two secondary spermatocytes and secondary meiotic division give rise to four spermatids from two spermatocytes.  

Therefore a. from 100 primary spermatocytes 400 sperms will develop (100*4= 400)

.

b. From 100 secondary spermatocyteS 200 sperms will develop(100*2= 200).

c. Spermatids do not undergo any further division rather they undergo differentiation to form mature sperms. Therefore from 100 spermatids, 100 mature sperms will develop.

d. One primary oocyte gives rise to one secondary oocyte and one secondary oocyte give rise to one mature ova so only one mature ova will develop from 100 primary oocytes.

e. 100 ova will develop from 100 secondary oocytes.

f. Polar body do not develop into ova and degenerates after some time so no ova will develop from 100 polar bodies.

Approximately 200 sperm develop from 100 primary spermatocytes.

b. Around 400 sperm develop from 100 secondary spermatocytes.

c. Roughly 100 sperm develop from 100 spermatids.

d. Only one ovum develops from 100 primary oocytes.

e. Only one ovum develops from 100 secondary oocytes.

f. No ova develop from 100 polar bodies.

What is the numbers of sperm

In people, spermatogenesis makes sperm and oogenesis makes ova, but they make different amounts of each. Here are the estimated numbers of sperm and eggs that develop from certain cell stages:

When 100 cells called primary spermatocytes go through a process called meiosis, they turn into 200 cells called secondary spermatocytes. Each cell called primary spermatocyte breaks into two smaller cells named secondary spermatocytes.

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tRNAs:
a. have a "cloven hoof" secondary structure
b. serve as adapters between amino acids and DNA substitutions
c. have their corresponding amino acid covalently attached to their 5' end
d. are postranscriptionally modified to include abnormal bases such as dihydrosphinctosine
e. have an acceptor stem

Answers

Answer:

Have an acceptor stem.

Explanation:

RNA is present as genetic material in some viruses only. RNA is the single stranded molecules that are formed from the DNA template by the process of translation.

Different types of RNA are rRNA, mRNA and tRNA. Transfer RNA (tRNA) contains the anti codons that are complementary to the codon present on mRNA. The tRNA consists of the acceptor stem that acts as the amino acid attachment site during the process of translation.

Thus, the correct answer is option (e).

Energy pyramids, which display graphically the relative amounts of energy in each trophic level of an ecosystem, are very similar across a wide range of different ecosystems, but biomass and organism abundance at different trophic levels are more variable between different ecosystems.
a. True
b. False

Answers

Answer:

a. True

Explanation:

Energy pyramids are a simple representation of the flow of energy at each trophic level in an ecosystem, based on the well known fact that a lot of energy gets lost between consumers (approximately only around 10% is made into tissues, while 90% is used up and is lost as heat).

While the biomass and abundance of the organisms at a trophic level, even the identity of that organisms (species), is highly dependent on the characteristic of the environment. So very different environments (for example, a dessert and a rain forest) will have also very different species with different biomass and abundance fulfilling the same trophic role.

Base-pair substitutions involving the replacement of a purine with a pyrimidine and vice versa are called:
a. isomers
b. transitions
c. transversions
d. inversions
e. translations

Answers

Answer:

The answer is: c. tranversions

Explanation:

A mutation is the result of changes in the structure of a gen, these variations  happen in the nucleotide sequence of a genome, and can be as a consequence of DNA damage.

One type of DNA mutation is the substitution of one base pair for another. Two type of substitutions can happen. One of them are the transversions, which are interchanges of a purine (A or G) for a pyrimidine (C or T) and vice versa.

The other type of substitution are the transitions that are interchanges of one purine (A) for another purine (G) or a pyrimidine (C) to another pyrimidine (T).

Describe the structure and function of haemoglobin (Hb). You should include a description of both tertiary and quaternary structures of the protein and the location and function of non-protein components.

Answers

Answer:

Heamoglobin is the structure present in red blood cells which is involved in carrying of oxygen from the blood to all parts of the body.

Explanation:

Tertiary Structure :  A molecule of heamoglobin consists of four groups of heme and a group of globulin.A molecule of heamoglobin contains four iron atoms to which oxygen molecules get attached. As there are four iron atoms in a molecule of heamoglobin so a single heamoglobin molecule can carry four oxygen atoms.

Quaternary structure: Multi sub unit globular proteins combine to make the Quaternary structure of heamoglobin. Alpha helices are made by the amino acids present in heamoglobin. Non helical segments, short in length, connect the alpha helices. The helical segments are stabilized by hydrogen bonds which results in folding of the polypeptide into the specific Quaternary shape.

 Function:

1.The basic function of heamoglobin is that it transfers oxygen from the lungs to different parts of the body.

2. CO2 ions and hydrogen ions are also transported by the heamoglobin.

Hypothesize how over-washing of hands can affect the population of ""good"" bacteria that resides on human skin.

Answers

Answer:

The human body inhabits various microbes, which coexist and does not cause any kind of harm, these bacteria are considered as good bacteria and are known as microbiota. Even the skin comprises various categories of bacteria, that is, more than 500 species. These microbes demonstrate skin microbiota, that is, an effective system of safeguarding, which takes place in immune defenses.  

The beauty of the skin is nearly associated with the balance of the microbes, which populate it, these are considered as the first line of defense from outside threats. When the bacterial ecosystem is differentiated and balance, the skin stays healthy.  

It has been hypothesized that over-washing of hands diminishes the population of good bacteria from the skin. As it is considered that over-washing extricates the majority of the beneficial bacterial cells from the human skin. Hence, may result in the loss of natural microflora from the section of the skin.  

Bacteria are single-celled prokaryotes that can be beneficial and harmful to organisms. Overwashing hands can harm the good bacteria on the skin surface.

What are good bacteria?

The body of the organism is full of bacteria and microflora that can be beneficial to the organism. Good bacteria are helpful to the organism as they help in maintaining health.

The skin surface of the organism is home to various microbiota that are good and helpful. They help in safeguarding and protecting the skin cells and tissues. They are part of the first line of defence as they help to regulate the immune check.

When the skin surface is overwashed then the population of the good bacteria gets affected and their habitation gets destroyed. Overwashing can result in the loss of the natural and good microflora of the skin surface.

Therefore, overwashing can affect the population of "good" bacteria.

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One oak tree cell with 14 chromosomes undergoes mitosis. How many daughter cells are formed, and what is the chromosome number in each cell?

Answers

Answer:

The correct answer will be option- 2 daughter cells with 14 chromosomes.

Explanation:

Mitosis is a type of cell division which is used by the organism to divide the somatic cells. This method is therefore considered an asexual mode of reproduction.

Mitosis proceeds in two stages: karyokinesis (a division of the nucleus) and cytokinesis (a division of the cytoplasm). Mitosis results in two daughter cells with the same amount of DNA content that is clone.  

In the given question, since the tree cell contains 14 chromosomes therefore on the mitotic division will produce two daughter cells with the same amount of chromosomes that are 14 chromosomes.

Thus, 2 daughter cells with 14 chromosomes are the correct answer.

A researcher wanted to study the effect of low light environments on the growth of alfalfa plants, so she grew six plants and measured their heights after two weeks. She calculated a sample mean of 5.35 cm and a sample standard deviation of 0.90 cm. If the heights of the first five plants in her sample were 6.0 cm, 5.9 cm, 4.0 cm, 5.5 cm, and 6.2 cm, what is the height of the other plant?

Answers

Answer:

Height of the other plant is [tex]4.5[/tex] cm

Explanation:

- Given -

Mean of the sample set of plants [tex]= 5.35[/tex]cm

Mean of a given set of data is equal to the sum of integral values of each data divided by the number of such data points

[tex]M = \frac{x}{n}[/tex]

Where M signifies the Mean values

X signifies the sum of all unit data with in a sample

and n signifies total number of data units with in a sample

On substituting the given values in above equation, we get -

[tex]\frac{6.0+5.9+4.0+5.5+6.2+A}{6} = 5.35\\5.35 * 6= 27.6 + A\\A = 32.1-27.6\\A=4.5[/tex]

Hence, height of the other plant is [tex]4.5[/tex] cm

The height of the sixth alfalfa plant is 4.5 cm, calculated by subtracting the total of the first five plant heights from the overall sum required to achieve the sample mean.

The student requires the calculation of the height of the sixth alfalfa plant given the sample mean, standard deviation, and the heights of the first five plants. To find the height of the sixth plant, we will use the formula for the sample mean: mean = (sum of all values) / number of values. Given that the sample mean is 5.35 cm and there are six plants, the sum of all the plant heights is 5.35 cm * 6 = 32.1 cm. The sum of the heights of the first five plants is 6.0 cm + 5.9 cm + 4.0 cm + 5.5 cm + 6.2 cm = 27.6 cm. Subtracting this from the total sum we get the height of the sixth plant: 32.1 cm - 27.6 cm = 4.5 cm.

The technique used when a biologist separates tissues by gently placing the closed tips of a pair of blunt-ended scissors between the two tissue layers and separating them by opening the scissors is called __________.
Select one:
a. sharp dissection
b. blunt dissection
c. normal dissection
d. tissue dissection

Answers

The correct answer is B. Blunt dissection

Explanation:

The word dissection refers to the process of separating tissues or different anatomical structures to analyze them or as part of medical processes such as surgeries. Although there are multiple techniques in dissection the two main ones are sharp dissection that uses sharp objects to cut through tissues and blunt dissection in which blunt-ended object such as special scissors or fingers are used to separate tissue without cutting through them. For doing this, fingers or scissors are placed between tissues and then open gently. According to this, the technique described is blunt dissection not only because a blunt-ended object is used but because this is used to separate tissue layer rather than to cut through them.

The correct answer is b. blunt dissection.

Blunt dissection is a technique used in surgical and laboratory settings to separate tissues or layers within tissues without cutting them. This method involves the use of blunt instruments, such as the closed tips of blunt-ended scissors, to gently pry apart the layers. The process minimizes damage to the tissues and preserves their integrity, which is particularly important when the tissues need to remain viable for further study or when preserving the function of the tissues is crucial, as in surgery.

In contrast, sharp dissection involves cutting through tissues with a scalpel or sharp scissors, which can cause more damage and is typically used when a precise incision is needed or when the tissue being cut will not be used for further study or function. Normal dissection and tissue dissection are not specific terms used to describe a particular technique like blunt dissection. Therefore, the correct term for the technique described in the question is blunt dissection.

Compare the means of sex determination in different organisms.

Answers

Answer:

Sex determination may be defined as the phenomena of determining the sex whether the organism is male, female or hermaphrodite. Different methods of sex determination are present in different organism.

In case of humans, the sex chromosome of an individual with XY is male whereas XX is female. Humans male are heterogametic. In case of birds the chromosome with ZZ are males (homogametic) whereas ZW is female. In case of Drosophila XY and XO flies are males , while XX, XXY, and XXYY flies are females.

The tissue(s) that is/are considered excitable because of the ability to generate electrical signals is/are called ________ tissue. The tissue(s) that is/are considered excitable because of the ability to generate electrical signals is/are called ________ tissue. muscle neural epithelial muscle tissue and neural muscle tissue, neural tissue, and epithelial

Answers

Answer:

Muscle tissue and neural tissue

Explanation:

Excitability refers to the ability of muscle and nerve cells of the respective tissues to respond to a stimulus and generate an action potential. Both muscle cells and neurons respond to a stimulus and convert it into the action potential.

Action potential refers to the electrical signal. Propagation of action potential along the membranes of these cells results in muscle contraction and functioning of neurons.

The membrane potential of these cells changes in response to the stimulus and the changed potential is propagated to the other cells.

Which of the following innervates the superior, medial and inferior rectus muscles of the eye and is a constrictor of the pupil?
a. Trochlear
b. Abducens
c. Oculomotor
d. Trigeminal
e. Facial

Answers

Answer:

The correct option is: c. Oculomotor

Explanation:

Oculomotor nerve is one of the twelve pairs of cranial nerves, emerging from the brain. Oculomotor is the third cranial nerve that originates from the third nerve nucleus in the midbrain.

The oculomotor cranial nerve (CN III), innervates the extrinsic eye muscles including the superior rectus, inferior rectus, and the medial rectus; that are responsible for raising the eyelid and movements of the eye.

The fibers of this nerve innervates the intrinsic eye muscles that are responsible for the constriction of the pupil.

Briefly describe the functional recovery of brain after brain injury or stroke.

Answers

Answer:

The brain is made up of billions of neural connections that transit and gather signals that are processed into information. Brain injury or stroke may cause damage to some of these connections. Neurons have a limited ability to repair themselves, unlike normal cells hence making brain injury difficult to recover fully from. The brain recovery after such as injury is are mainly driven by the brain finding new neural pathways around the damaged ones.

However difficult, there is a chance that a neuron can be repaired especially if the cell body is not damaged in the injury. If the axon is regenerated, it needs to make the right connections to re-establish signaling. The problem is that glial cells, that are integral to neurons structure and function, reduce the chances of recovery because they clean up damaged neurons and produce molecules that inhibit recovery  

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