Males have hemophilia when they are hemizy-gous for a nonfunctional recessive mutant allele of the X-Iinked gene for clotting factor Vlll. Factor Vlll is normally secreted into the blood serum by cells in the bone marrow that produce it. a. Do you think that females heterozygous for the hemophilia disease allele could have hemophilia in some parts of their bodies and not others? b. If such a female "carrier" of hemophilia suffered a cut, would her blood coagulate (form clots) faster, slower, or in about the same time as that of an individual homozygous for a normal allele of the factor Vlll gene? Would the rate of clotting vary significantly among heterozygous females?

Answer:

Let's organize this with the four biomolechules:

-NUCLEIC ACID (Nucleotides, Polynucleotides, Phosphodiester linkages)

-LIPIDS (Fatty acids, Triacylglycerols, Ester linkages)

-PROTEINS (Polypeptides, Pepptide bonds, Aminoacids)

-CARBOHYDRATES (Monosaccharides, Polysaccharides, Glycosidic linkages)

Explanation:

Answer: Answers are detail in the attachment

Explanation:

Answer:

Explanation:

(a) aabbccdd : 0.4%

(b) AaBbCcDd : 6.3%

(c) AABBCCDD : 0.4%

(d) AaBBccDd : 1.6 %

(e) AaBBCCdd: 0.8 %

doing the tetrahybrid cross can be awfully difficult. You can always search for online sites that can help you with it.

ATP is essential in a cell as it stores and transfers energy required for various cellular activities. Its energy is stored in high-energy phosphate bonds, which are broken to facilitate energy release. It does not contribute to the cell membrane structure due to its polar, hydrophilic characteristics.

ATP, or Adenosine Triphosphate, plays a crucial role in the cell as it acts as a medium for energy storage and transfer. It is the main source of energy for most cellular processes. ATP doesn't store energy in carbonyl groups, rather it stores energy in the high-energy phosphate bonds. These bonds are broken to release the energy needed for various cellular activities.

ATP is not an important component of cell membranes, nor does it have a long hydrocarbon tail. It doesn't contribute to the structure of the cell membrane because it is polar and hydrophilic, unlike the major constituents of the cell membrane - phospholipids, which are both hydrophobic and hydrophilic.

The process of ATP releasing energy involves hydrolysis where a phosphate group is removed, allowing the stored energy to be used for various metabolic and physiological processes within the cell. It doesn't add phosphate groups but rather loses them in the process of energy release.

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ATP is the primary energy currency of the cell, storing energy in high-energy phosphate bonds and releasing it to power cellular processes. Cells produce ATP through cellular respiration, and it is used in various essential activities like metabolism and muscle contraction. ATP is continually recycled within the cell to maintain energy balance.

The Importance of ATP in Cellular Processes

Adenosine triphosphate (ATP) is often described as the energy currency of the cell. It is a molecule that stores and transports chemical energy within cells. ATP's energy is stored in its phosphate ester bonds, particularly the two high-energy phosphate anhydride linkages. When one of these bonds is broken, usually through the removal of a terminal phosphate group, energy is released. This energy is utilized in a variety of cellular processes, including metabolism, active transport, muscle contraction, and cell division.

Cells produce ATP during cellular respiration, a process which involves the oxidation of carbohydrates, proteins, and fats. The oxidative reactions that produce ATP occur in the cytoplasm and mitochondria of the cell. Once produced, ATP can be used immediately or stored for short periods until the energy is required for cellular reactions or for performing mechanical work, such as muscle contraction.

Phosphorylation is the process by which ATP transfers a phosphate group to another molecule. This process requires energy and is a key mechanism by which cells convert ATP's stored energy into a form that can do biological work, powering a vast array of endergonic reactions in the cell. ATP is continually recycled within the cell; it is constantly being made from ADP and inorganic phosphate and used up in cellular activities, maintaining the energy homeostasis of the cell.

Answer:

b. red algae—eukaryotes that acquired plastids by secondary endosymbiosis

Explanation:

Answer:

The results can be explained if the gene that determines color of feathers is located in the Z chromosome.

Explanation:

Unlike mammals that have X and Y sex chromosomes, birds have Z and W sex chromosomes. Females are ZW and males are ZZ.

The gene that determines color of feathers has the alleles:

B_ = brown color

bb = yellow color.

yellow female  x   brown male

Gametes

Female: [tex]Z^b[/tex] , [tex]W[/tex]

Male: [tex]Z^B[/tex]

brown female   x   yellow male

Gametes

Female: [tex]Z^B[/tex] , [tex]W[/tex]

Male: [tex]Z^b[/tex]

The results of crossing pure-breeding yellow and brown canaries suggest a sex-linked inheritance of feather color, with brown being the dominant allele. A hypothesis could be that the gene for feather color is on the X chromosome, resulting in sex-specific phenotypes among the offspring. This follows Mendelian inheritance principles, as demonstrated in monohybrid crosses with peas resulting in predictable phenotypic ratios.

To explain the results of crossing the two pure-breeding lines of canaries, one could hypothesize that the feather color is sex-linked and that brown is a dominant allele over yellow. The fact that all crosses result in brown offspring suggests that the allele for brown feathers is dominant. Since the crosses from yellow female canaries to brown male canaries results solely in brown offspring, but the cross from brown female canaries to yellow males results in brown sons and yellow daughters, it can be hypothesized that the gene controlling feather color is located on the sex chromosomes, possibly the X chromosome considering the sex-specific inheritance pattern. In this case, the brown allele is dominant, which is denoted by 'B', whereas the yellow allele is recessive, denoted by 'b'.

Monohybrid cross analyses using a Punnett square help determine probable offspring genotypes and phenotypes. When two true-breeding parents are crossed, such as YY and yy in pea plants, all F1 offspring are Yy showing the dominant phenotype. Similarly, in canaries, the same principles apply where the feather color gene could follow dominant or recessive patterns.

Using Mendelian genetics and large numbers of crosses, genotypic and phenotypic ratios can be calculated to predict other genetic crossing outcomes. For example, when F1 offspring with genotypes Yy are crossed among themselves, there is a 3 in 4 probability of offspring with a yellow phenotype due to the dominant yellow allele. This explains the common 3:1 phenotypic ratio seen in Mendel's F2 generation of pea plants. Drawing parallels, similar principles could be attributed to the canary crossing results mentioned in the question, suggesting that the inheritance of feather color follows a Mendelian pattern of inheritance assuming complete dominance and sex linkage.

Answer:

Explanation:

Transcription takes place in nucleus in the presence of the enzyme RNA polymerase.

During transcription, the code present in DNA gets transcribed into the code in mRNA.  (messenger RNA)

RNA even though is a single stranded shows base pairing.

The pre-mRNA gets created from the DNA.

DNA has two strands, one of these strands acts as coding and another non coding strand.

The DNA shows base pairing with AGTC.   [A-Adenine, G- Guanine, T-Thymine and C- Cytosine]

Whereas mRNA base pairing with AGUC.  [ U- Uracil ]

Here in mRNA, instead of thymine, uracil gets paired.

According to base pairing of coding strand/sense strand of DNA, complementary bases pairs in 5'---->3' direction in mRNA.

The mRNA created looks exactly like the non-coding strand/sense strand of DNA except instead of thymine, uracil gets replaced.

The pre-RNA undergoes splicing, capping and tailing to form mature mRNA.

Final answer:

During transcription's initiation phase, RNA polymerase binds to a promoter, unwinds DNA, and catalyzes complementary base pairing between RNA nucleotides and DNA template, forming mRNA. Elongation continues this process, and termination occurs when specific base-pairing signals the end of transcription.

Explanation:

In transcription, the formation of the mRNA strand from a DNA template involves specific steps where complementary base pairing is essential. Initiating the process, RNA polymerase binds to a gene's promoter region during the initiation phase and unwinds the DNA helix, allowing one of the DNA strands to be 'read'. The new mRNA strand grows by the addition of RNA nucleotides that form hydrogen bonds with the complementary deoxyribonucleotides of the unwound DNA template strand. Here, the base cytosine (C) pairs with guanine (G), guanine (G) with cytosine (C), adenine (A) with uracil (U), and thymine (T) with adenine (A).

The elongation phase of transcription includes the continued addition of RNA nucleotides. Complementary base pairing ensures the correct sequence of nucleotides is added to the growing mRNA strand. The termination phase concludes transcription;

Complementary base pairs also play a role in the termination of transcription. Unique sequences in the DNA signal the end of the mRNA transcript, and following the base-pairing rules, these sequences can form structures such as hairpins that signal RNA polymerase to stop transcription.

Answer:

The correct answer is option B. likely give dominant males and females a reproductive advantage .

Explanation:

Dominance hierarchy is a social hierarchy in which members of a group of animals interact with each other to form a ranking system or superiority and inferiority of the organism. In such social groups, members have to compete for access for the food and resources which are limited.

Mating is also decided on the basis of the dominance of males and females in such social groups. Dominant male-female have always advantage for mating in comparison to lower members in the ranking.

For instance, female baboon mated to a dominant male or high ranked male so she can get high rank as well for the time of period.

Thus, the correct answer is option B.  likely give dominant males and females a reproductive advantage .

Gastric ulcer is nothing more than an injury to the tissue lining the stomach. It may also be called peptic ulcer or stomach ulcer.  

The most common cause of ulcer formation is the proliferation of a bacterium called Helicobacter pylori.  There are other less common factors such as diet, overuse of drugs, smoking, stress and alcohol.

The best way to prevent an ulcer contraction is to maintain hygiene habits to prevent Helicobacter pylori infection. It is also good to avoid excessive consumption of alcohol, smoking and medications such as pain killers.

Answer:

Broca's area refers to a motor speech region, which assists in movements needed at the time of the production of speech. When an injury takes place in the frontal parts of the left hemisphere, it can give rise to various kinds of language issues. This section of the brain plays an essential role in putting words together to produce complete sentences. The injury to the left hemisphere is known as Broca's aphasia, also known as expressive aphasia.  

Wernicke's aphasia also called receptive aphasia, posterior aphasia, or sensory aphasia is a kind of aphasia in which individual encounters with difficulty in understanding spoken and written language. Thus, the two kinds of aphasia are expressive aphasia in which one faces difficulty in conveying thoughts via writing and speech. The other is receptive aphasia, in which one finds difficulty in understanding the written or spoken language.  

By analyzing the symptoms that whether the patient exhibits difficulty in understanding speech and using accurate words to express thoughts or the movements that are needed to generate speech, one can find the site of destruction.  

Answer:

d. Ligand-gated ion channels

Explanation:

a.Intracellular receptor: it's activated through second messengers since its activation and actions happen only inside the cell there's no distribution of ions on opposite sides of the membrane.

b.G-protein-coupled receptor: these receptors bind specific ligands at the cell surface (hormones and neurotransmitters) to relay the signal across the membrane. This means they act as a messager, there's no distribution of ions.

c.Phosphorylated receptor tyrosine kinase dimer: works as a messenger as well, it propagates a signal through the plasma membrane when it allows tyrosine to be trans phosphorylated.

d. Ligand-gated ion channels open to allow ions such as Na+, K+, Ca2+ or Cl- in response to the binding of a chemical messenger, one example of this kind of channel in action is when a neurotransmitter binds to a receptor located on the postsynaptic neuron, opening ion channels, this lead to a flow of ions across the cell membrane that turns into a depolarization or a hyperpolarization by changing the distribution of ions on opposite sides of the membrane.

I hope you find this information useful! good luck!

Final answer:

The answer is d) ligand-gated ion channel. This receptor, upon binding to a signaling molecule, allows the passage of ions through a membrane pore, altering ion distribution.

Explanation:

The binding of a signaling molecule to a d) ligand-gated ion channel leads directly to a change in the distribution of ions on opposite sides of the membrane. This type of cell-surface receptor works by having an extensive membrane-spanning region, and upon binding to a ligand, it undergoes a conformational change that allows specific ions to pass through the pore. Ligand-gated ion channels facilitate the almost instantaneous passage of millions of ions, which drastically change the interior environment of the cell.

Answer:

Stem cells may be defined as the potipotent cells that has the ability to differentiate into different cells. The stem cells are now used in research as well.

The main purpose of the stem cells are as follows:

The damaged tissue or organ can be replaced by growing stem cells in the laboratory.

The genetic defects can be corrected by the stem cells discovery.

The new drugs can be tested on the stem cell.

The reason of the development of specific cancer cells.

The correct answer is option d, as RNA splicing is indeed catalyzed by spliceosomes.

d. RNA splicing can be catalyzed by spliceosomes.

In RNA processing, the process of removing introns and joining exons together is called RNA splicing. This process is facilitated by the presence of ribozymes, which are RNA molecules that catalyze chemical reactions. One type of ribozyme involved in RNA splicing is the spliceosome, a large and complex assembly of proteins and RNA.

Therefore, the correct answer is option d, as RNA splicing is indeed catalyzed by spliceosomes.

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The incorrect statement about RNA processing is that 'Exons are cut out before mRNA leaves the nucleus.', as it is actually the introns that are removed, not the exons. The correct answer is a.

The student has asked which statement is not true about RNA processing. The correct answer to this question is 'a. Exons are cut out before mRNA leaves the nucleus.' In fact, during RNA processing, it is the introns that are cut out while the exons are joined together. This process, known as splicing, is essential for converting the pre-mRNA transcript into a mature mRNA molecule that can be translated into protein.

The modifications that occur during RNA processing in eukaryotic cells include the addition of a 5' cap and a poly-A tail to the pre-mRNA, and the removal of introns by splicing which can be catalyzed by spliceosomes or ribozymes. Different splicing patterns can lead to the production of various mRNA transcripts from the same DNA sequence, a phenomenon known as alternative splicing.

Therefore, the statement that 'Exons are cut out...' is not correct since exons are the sections that remain in mRNA after the introns have been removed. RNA processing includes the removal of introns, addition of nucleotides at both ends, and splicing by spliceosomes or ribozymes.

Answer:

True

Explanation:

Gross primary productivity (GPP) of a system is defined as the rate by which the primary producers trap the solar radiation during the process of photosynthesis. The gross primary productivity is represented as total photosynthesis per unit area in a given time.

The net primary productivity (NPP) refers to the energy available to the plants for their growth after cellular respiration and is represented as plant growth per unit area in a given time.

Plants use part of the gross primary productivity in cellular respiration to produce the energy required to perform other vital functions. Hence, the energy consumed in cellular respiration differentiates the gross primary productivity and the net primary productivity.

NPP = GPP - Plant respiration

Net biomass productivity is the difference between gross productivity and respiration. It is influenced by factors like climate and soil nutrients. Forests tend to have high net biomass productivity while deserts have low productivity.

Net biomass productivity is the difference between gross productivity (production of plant material by photosynthesis) and respiration. Net productivity is closely related to a number of environmental factors like climate, soils, and available nutrients. Net biomass production will be highest where there is an ample supply of moisture to meet the needs of plants. Biomass productivity is also high where soils are rich in nutrients and have a positive soil moisture balance. Forest, especially tropical forest have high net biomass productivity while deserts have low productivity.

The statement that the difference between gross primary productivity and net primary productivity is determined by the respiration of primary producers is true. Gross primary productivity (GPP) is the rate at which photosynthetic primary producers, such as plants, incorporate energy from the sun. However, not all of this energy remains within the plants because they must use a portion of it for their own metabolic processes, primarily through cellular respiration. The energy that remains after accounting for this respiration is known as the net primary productivity (NPP). It is the energy that is actually available to primary consumers in an ecosystem. The NPP can be influenced by various environmental factors, such as climate, availability of nutrients, and moisture levels, leading to variations in the productivity of different ecosystems, like tropical forests having high NPP while deserts have relatively low NPP.

Answer:

Virus may be defined the acellular organism that contain DNA or RNA as their genetic material. Thee virus may contains double stranded or single stranded DNA or RNA molecule.

The DNA has composition of 20% A, 35% T, 25%G, and 20% C. The virus DNA does not follow Chargaff's rule in which the amount of thymine is not equal to the amount of adenine. The amount of guanine is not equal to the amount of cytosine. The DNA virus consist of single stranded DNA.

Answer:

The correct answer will be option-C.

Explanation:

The lymphocytes are a type of immune cells which provides immunity to the organism. The lymphocytes differentiate into two types: B cell and t cell.

T cell lymphocytes are programmed to recognize and respond to kill the antigen containing cell.  The T cells provide immunity wither by directly killing cell containing antigen or by producing lymphokines which induce other cells to kill the affected cell. This type of immune response is known as cell-mediated immune response.

Thus, Option-C is the correct answer.

Sperm are first produced in the seminiferous tubules inside the testes. They mature in the epididymis where they acquire motility and the ability to fertilize. They are released during ejaculation, travelling through the ductus deferens and exiting via the urethra.

Sperm are first produced in the seminiferous tubules that are coiled inside the testes. The least mature cells, the spermatogonia, dwell along the basement membrane inside the tubule and undergo division to produce primary and secondary spermatocytes, and then spermatids. These spermatids finally mature into spermatozoa, or formed sperm, in a process known as spermatogenesis.

The sperm cells mature further in the epididymis, where they gain the ability to move and fertilize an oocyte. The mature sperm are stored in the epididymis until ejaculation. During ejaculation, sperm exit the epididymis through the ductus deferens, where secretions from the prostate and bulbourethral glands help form semen, assisting the sperm in their journey through the urethra and into the female reproductive tract.

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Answer:

d. a single nucleotide insertion downstream of, and close to, the start of the coding sequence.

Explanation:

A single nucleotide insertion downstream and close to the start of coding sequence will produce the most harmful effect among all these given options because it will cause the entire reading frame to shift in the beginning itself leading to insertion of wrong amino acids in the entire polypeptide chain, hence the entire protein will have extremely different amino acids than the original protein was supposed to have and such protein will be dysfunctional.

In option (a), it is given that the three nucleotides near the middle of the gene will be deleted. This scenario will produce comparatively less harmful effect because in this case only one amino acid which is coded by this deleted triplet codon will be absent in the polypeptide.

In option (b), a single nucleotide deletion has occurred that too in an intron. It will produce no harmful effect in case of eukaryotes because introns are ultimately removed before translation so the amino acids in the polypeptide will remain the same. In case of prokaryotes such deletion may cause some harmful effect in case of mRNA only because introns in prokaryotic mRNA are not removed. But, if the mRNA is for prokaryotic tRNA and rRNA then there will be no harmful effect because they undergo processing in which introns are removed.

In option (c) also there will be very less harmful effect because deletion is occurring near the end of coding sequence which will lead to insertion of wrong amino acid in the polypeptide at the end of polypeptide chain that means very less abnormal amino acids will be present so effect will be less harmful.

Answer:

D. a single nucleotide insertion downstream of, and close to, the start of the coding sequence.

Explanation:

Just wanted to confirm. The explanation above me is correct.

Answer:

1/2 or 50%

Explanation:

Given that the extra digit is a dominant trait, the man with an extra digit may be homozygous or heterozygous dominant for the trait.

However, having a daughter with his normal wife makes him heterozygous dominant. Let's assume that the genotype of the heterozygous dominant male is "Dd" and that of his wife is "dd".

The heterozygous dominant male and homozygous recessive female would have progeny in the following ratio= 1/2 Dd (child with extra digit): 1/2 dd (child with no extra digit).

The fraction of this couple's children expected to have extra digits is 1 (all of the children) out of 1 (total number of children), which is equivalent to 1/1 or simply 1.

The man in this scenario has six fingers on each hand and six toes on each foot, a condition caused by a dominant gene. His wife and their daughter have the normal number of digits. Since having extra digits is a dominant trait, the man must be homozygous dominant (DD) for the gene responsible for the extra digits.

When the couple has children, all of their offspring will inherit one dominant allele from the father, resulting in the expression of the trait. Therefore, the fraction of their children expected to have extra digits is 1, meaning that all of their children will have the additional digits.

Learn more about dominant trait, here:

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Answer:

Translation is the synthesis of protein synthesis from RNA where genetic information encoded in mRNA, which is then translated into a sequence of amino acids present in a polypeptide chain. These are the steps in translocation:

1) Ribosomes bind to mRNA present in the cytoplasm and move with the molecule in a 5'-3' direction until it reaches a start codon AUG.

2) Anticodon present on tRNA molecule binds with appropriate opposite codon according to complementary base pairing. (translocation depends on complementary base pairing, e.g. AUG=UAC between codons present on mRNA and anticodons on tRNA)

3) Each tRNA molecule has specific amino acids, according to the genetic code.

4) Ribosomes catalyze the formation of peptides bonds with the help of condensation reaction between adjacent amino acids.

5) Ribosomes move with the mRNA molecule which is synthesizing a polypeptide chain until it reaches a stop codon.

6) At this stage, translation comes to stop and the polypeptide chain is released.

Answer:

Biogeographic realms refer to the huge spaces or regions on the surface of the Earth within which the development of ecosystems takes place. The animal and plant species witnessed in a biogeographic realm share identical features all the way through.  

The eight biogeographic realms are the afrotropical realm, Indomalayan realm, Antarctica realm, Neotropical realm, Nearctic realm, Palaearctic realm, Oceanian realm, and Australian realm.  

The biodiversity signifies the variability between the living species in every possible manner. Each of the biogeographic realm exhibiting identical environmental conditions or identical mediators of modification has been found to comprise an identical kind of biota.  

However, diversity can be found within it too. For example, the tropical moist rainforest found in any section of the globe will share identical procedures and vegetation, however, its specific structure will vary on the basis of the biogeographic realm in which it is witnessed.  

Analogous adaptation to the similar kind of environmental conditions between the distinct species of the distinct biogeographic realm can be witnessed at a global level.  

Answer:

A definitely

Explanation:

Us human are evil.

that's my honesty

Final answer:

Population growth and Earth's resources are at odds, with concerns over carrying capacity and sustainability. Technological and planning efforts can mitigate the situation, aiming for sustainable equilibrium and zero population growth in developed nations. Careful management of resources is essential to avoid a population catastrophe.

Explanation:

The question of whether an ever-increasing human population can be sustained on Earth revolves around notions of carrying capacity, resource depletion, and sustainability. The limits of growth debate considers whether societies can grow indefinitely or face a limit based on Earth's resources. With the current ecological footprint, some argue that we have already surpassed Earth's capacity to support our population, leading to environmental damage and extreme poverty. Nonetheless, the prospect of a sustainable equilibrium depends on human actions, including the use of technology, planning, and global cooperation to manage resources wisely.

Sustainability efforts and technological advancements in agriculture, energy, and water management are vital to expanding resources and reducing the ecological footprint. Nevertheless, environmental alterations, ranging from climate change to biodiversity loss, have resulted from technological interventions, casting uncertainty over the capacity for continued support of human populations without significant changes to consumption and production patterns.

Achieving zero population growth in some economically developed countries suggests that stabilizing populations is feasible. However, the disparities between developed and developing nations in reaching such a balance remain a critical challenge. The path to avoiding a population crash involves mindful stewardship of Earth's resources and equitable distribution to prevent drastic outcomes for humanity.

Answer:

In the given case, the FAST or face, arms, speech, and time test can be done to evaluate the function of the brain. In supplementation, imaging tests, and blood tests can also be performed. In the given case, ischemic stroke is determined.  

For the treatment of ischemic strokes, a tissue plasminogen activator also known as tPA is administered intravenously in the arm. The tPA functions by dissolving the clot and bettering the flow of blood to the section of the brain, which is being deprived of blood flow.  

Apart from this, the antiplatelet agents like aspirin and anticoagulants like warfarin can be provided to the patient. In addition, the antihypertensives can also be administered on the basis of patient's condition.  

Answer:

False

Explanation:

When one of the glands is removed, the other continues to perform its function because they are independent entities that can continue to secrete hormones. This happens with the adrenal glands for example.

Answer:

The correct option is: e. Both B and D

Explanation:

Type I alveolar cells are the thin squamous epithelial cells that lines the alveolar surface of the pulmonary alveoli.

These cells are involved in the gas exchange process. The thin lining of the type I alveolar cells, allows the rapid diffusion of gas and thus enabling the exchange of gases between the blood present in the capillaries and the air present in the pulmonary alveoli.

Answer:

1/3 of black F2 progeny will be homozygous.

Explanation:

Here, BB = black coat color = true breeding dominant black strain

bb = white coat color =  true breeding recessive white strain

When true breeding black and white guinea pigs are mated to give F1 progeny:

BB X bb = Bb ( all black guinea pigs )

When two of the F1 guinea pigs are mated to give F2 progeny:

Bb X Bb = BB, Bb, Bb, bb

F2 progeny has 3/4 progeny as black (BB and Bb) . Out of them 1/3 are homozygous and 2/3 are heterozygous.

Hence, 1/3 of black F2 progeny will be homozygous.

In the F2 generation of guinea pigs, where the first generation was all black from a true-breeding black and white strain, 25% of the black progeny is expected to be homozygous.

In guinea pigs, coat color is determined by a single gene with two alleles. If a guinea pig from a true-breeding black strain is mated with a guinea pig from a true-breeding white strain, the F1 progeny would be all black, indicating that black is the dominant color. If two of these F1 progeny are mated together, their offspring, or the F2 generation, would exhibit a phenotypic ratio of 3 dominant (black):1 recessive (white) according to Mendelian genetics. Therefore, in this case, about 3 out of 4, or 75%, of the black F2 progeny would be expected to be heterozygous (Bb), and 1 out of 4, or 25%, would be homozygous (BB). Thus, the proportion of the black F2 progeny expected to be homozygous is 25%.

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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.

Answer:

Lysosome

Explanation:

Freezer burn occurs due to the sublimation of ice in unprotected meat rich in muscles during long-frozen storage. It appears after thawing. The ice crystals rupture the lysosomes which in turn contain hydrolyzing enzymes.

Leakage of the hydrolytic lysosomal enzyme causes freezer burn and affects the appearance of the frozen meat. The hydrolytic enzymes of lysosomes partially digest the cells which in turn impart bad flavor to the stored food.

Answer:

The correct answer is E. All of the above

Explanation:

The body contains different physical and chemical barriers to protect itself from the invasion of any pathogen. These barriers act as the first line of defense to the body and include skin, mucous membrane, acids, and antimicrobial enzymes.

Skin acts as a mechanical barrier to microbes and does not allow any pathogen to invade through it until it gets punctured.

Saliva contains antimicrobial enzymes like lysozyme which kills bacteria by destroying their cell wall. Hydrochloric acid released in the stomach is highly efficient to kill pathogens coming with food.  

Mucous membranes in the digestive and respiratory tract contain WBCs, antimicrobial enzymes and proteins which kills the invaded pathogen like Vibrio.

Answer:

In the progeny, six kittens were short haired and two were long haired. The ratio of short to long haired kittens is 3:1 which is a classic monohybrid cross ratio. Here, short hair is inherited dominantly and long hair is inherited recessively.  Both the parents are heterozygous for short hair trait (Ss). Cross between them:

Ss X Ss :

   S      s

S  SS   Ss

s   Ss    ss

Genotypically: 1    :    2    :    1

                      SS      Ss       ss

Phenotypically:   3         :      1          = 6 short and 2 long haired kittens

                       short hair   long hair

Hence short hair is dominant over long hair in these cats.

Final answer:

Hair length in cats is inherited according to Mendelian genetics, with short hair being dominant and the observed offspring ratio suggesting both parent cats are heterozygous for the trait.

Explanation:

The mating of two short-haired cats resulting in both short-haired and long-haired kittens suggests that the trait for hair length in cats is inherited in a Mendelian fashion, with the allele for short hair (S) being dominant over the allele for long hair (s). Since both parents are short-haired but have produced long-haired offspring, it implies that each parent is heterozygous (Ss), carrying one allele for short hair and one for long hair. The distribution of short-haired to long-haired kittens (a roughly 3:1 ratio) fits the expected outcome of a Mendelian monohybrid cross, where both parents are heterozygous for a trait.

Answer:

Explanation:

Glucose is a sugar, highly soluble in water because it has lots of polar hydroxyl groups (-OH) which can form hydrogen-bonds with water molecules. These types of bonds are intermolecular forces which are present in other macromolecules like the DNA or proteins.

Cholesterol and triglycerids are lipids. These have long chains of hydrocarbons, which are non polar and therefore insoluble in water.

Blood is made of a solid and a liquid part. The solid part contains cells. The liquid part (plasma), is made of water, salts, and proteins. Glucose and cholesterol/triglycerids travel in the liquid part of the blood, in an aqueous environment. Because the lipids are not soluble, they need to travel with a carrier protein that keeps them protected from the surrounding water.