In Drosophila, males from a true-breeding stock with raspberry-colored eyes were mated to females from a true-breeding stock with sable-colored bodies. In the F1 generation, all the females had wild-type eye and body color, while all the males had wild-type eye color but sable-colored bodies. When F1 males and females were mated, the F2 generation was composed of 216 females with wild-type eyes and bodies, 223 females with wild-type eyes and sable bodies, 191 males with wild-type eyes and sable bodies, 188 males with raspberry eyes and wild-type bodies, 23 males with wild-type eyes and bodies, and 27 males with raspberry eyes and sable bodies. Explain these results by diagramming the crosses, and calculate any relevant map distances.

Answer:

food webs are drawn in a diagram similar to a spider web

Answer:

The main substrate of chymotrypsin includes tryptophan, tyrosine, phenylalanine and methionine

Explanation:

1.

Histidine yields a proton to aspartate and recovers it from serine.

Seen in another way: aspartate captures a proton from the serine through histidine.

2.

a) The serine (deprotonated) is thus capable of attacking the peptide bond (nucleophilic carbonyl attack) and forms a tetrahedral intermediate; The substrate is thus covalently bound to the enzyme (now it is a transition state).

b) The peptide bond is broken and the released amino terminus (R) recovers a proton from histidine.

c) Histidine, in turn, recovers it from aspartic.

3.

a) Aspartate captures a proton of histidine again, so that it can capture it in turn from water.

b) This generates a hydroxide anion that attacks the ester intermediate between the serine and the carboxyl part (R ′) of the substrate peptide.

c) A new tetrahedral intermediate bound to the enzyme is formed (via serine residue).

4.

a) The carboxyl group of the peptide is regenerated, the serine being separated and the other peptide fragment being free (the R ′ part with a free carboxyl end)

b) The serine recovers the proton at the expense of histidine, which in turn captures it from aspartic acid.

c) The catalytic triad (Asp, His, Ser) has been regenerated in its original state.

The net reaction is:

R–NH – CO –R ′ + H2O ⟶ R– NH2 + HOOC –R ′ ⟶ R – NH3 + + −OOC –R ′

The active site or catalytic center of chymotrypsin is formed by several amino acid residues, among which the essential role corresponds to the "catalytic triad".

Answer:

The correct answer will be- homologous chromosomes fail to pair.

Explanation:

The seedless variety of the banana are triploid that is each cell contains three copies of the chromosomes.  The triploid variety of the organism is considered sterile or not able to produce offspring as they are unable to generate gametes for fertilization.

The formation of gametes relies upon a type of cell division called meiosis which proceeds in four phases of karyokinesis stage.

At metaphase of the meiosis, chromosomes fail to pair at the equatorial plate as two chromosomes pair leaving one extra chromosome which is not paired. The gametes are produced with unequal chromosomes and on fertilization zygote is not formed.

Thus, homologous chromosomes fail to pair is the correct answer.

Answer:

-The skeletal muscle has long and cylindrical cells. The cardiac muscle cells are short and together they form a branch.

-The heart muscle fibers have only one nucleus in the center. The skeletal muscle fibers are multinucleated.

-The skeletal muscle is voluntary and the cardiac is involuntary.

- Sarcoplasma and glycogen are more abundant in the heart muscle.

-The cardiac muscle is longitudinal striated and the skeletal is transverse striated

Answer:

a. No, it is not possible.

A heterozygous female carries one copy of functional gene which is enough for the production of clotting factor. Hemophilia does not show continuous variation or polygenic inheritance and thus, its level does not depend on the number of normal alleles.

It that was the case, then all males would show hemophilia in some parts of the body as they only carry one X chromosome and thus, only one functional gene.

Thus, heterozygotes are only the carriers of the disease, they do not show any symptom of the disease.

b. In perspective of homozygosity or heterozygosity, the rate of blood clotting should be the same as both of them have functional gene. As mentioned above, it does not show continuous variation so, it will not show any increased or decreased rate of clotting in homozygotes or heterozygotes.

However, in reality, the rate of clotting depends on the concentration of clotting factor present in blood plasma. This percentage depends on the physiology of a person but not on the number of alleles present. For example, proteins or enzymes required for gene expression, et cetera.

Answer:

The correct answer will be option-B.

Explanation:

Deoxyribose nucleic acid or DNA is the genetic material of the organism which is made up of nucleotide monomer.  Each monomer is made up of a five-carbon deoxyribose sugar, a phosphate group and four different nitrogenous bases.

The nitrogenous bases- purines (adenine and guanine) bond with pyrimidines (cytosine and thymine) via hydrogen bond between them. The hydrogen bonds are weak bonds but establish the DNA structure.

Thus, Option-B is the correct answer.

The bases of DNA are connected by hydrogen bonds.

The bases of DNA are connected to each other by hydrogen bonds. These bonds form between specific complementary base pairs: adenine (A) with thymine (T) and guanine (G) with cytosine (C). Hydrogen bonds are relatively weak compared to covalent bonds, but they are crucial for maintaining the structure of the DNA helix.

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Answer:

[tex]30,000[/tex] mL

Explanation:

Given -

Volume of buffer [tex]= 500[/tex] mL

Concentration of solution [tex]= 60[/tex] mM[tex]= 0.06[/tex]M

New concentration of the solution [tex]= 0.001[/tex]M

As we know

[tex]C_1* V_1= C_2*V_2[/tex]

Where

[tex]C_1, C_2[/tex] are the concentration of the solution before and after dilution.

And [tex]V_1, V_2[/tex] are the volume of solution before and after dilution.

Substituting the given values in above equation, we get -

[tex]500* 0.06 = V_2* 0.001\\V_2 = \frac{30}{0.001} \\V_2 = 30,000[/tex]mL

Answer:

The correct answer will be option-B.

Explanation:

Transcription is a process which produces a complementary transcript molecule of DNA called RNA.  The RNA formation takes place in three steps: initiation, elongation and termination.

The initiation step involves a complex enzyme called RNA polymerase which synthesize a complementary strand using one strand of DNA. the enzyme adds a new nucleotide at 3' end of the strand thereby proceeding the reaction in 5' to 3' direction.

Thus, Option-B is the correct answer.

Answer:

RNA polymerase.

Explanation:

Transcription may be defined as the process of formation of RNA molecules from the DNA template. The process of translation occurs in the 5' to 3' direction.

The main enzyme responsible for the transcription is DNA dependent RNA polymerase or RNA polymerase. This enzyme is important for the synthesis of the RNA molecule complementary to the DNA template. RNA pol II is the main transcription enzymes of the mRNA molecule in case of eukaryotes.

Thus, the correct answer is option (b).

Answer:

The correct answer is a. suppression of LH release from the pituitary.

Explanation:

Anabolic steroids are synthetic steroids made up of mainly androgenic hormone like testosterone. These anabolic steroids are usually taken by young male bodybuilders because it helps in gaining muscles.

These anabolic steroids have many side effects and infertility in males is one of its side effects. Excessive use of these steroids stops the secretion of gonadotropic hormones such as LH in males from pituitary glands by inducing negative feedback to the pituitary.

LH is important in the production of testosterone from Leydig cells and testosterone is important for sperm production. Therefore if LH is stopped no sperm production will occur and it leads to infertility in males.

Answer:

The correct answer will be option-C.

Explanation:

The functional groups are the group of atoms which gets attached to the skeleton of carbon in a molecule and provides certain characteristics to the molecule.  The main functional group includes hydroxyl, carboxyl, amino, phosphate, methyl carbonyl and sulfhydryl.

These groups could be acidic or basic depending upon them if their structure will be able to take H⁺ ions from solution or donate H⁺ ions.  In the given question, the basic group among these is an amino group which has the ability to accept  H⁺ ions from the solution and form NH₃⁺ ions.

Thus, Option-C is the correct answer.

Answer:

IV: Amount of water

DV: Height of the strawberry plant

CG: The fourth strawberry plant

Explanation:

The Independent Variable is the amount of water given to the strawberry plants; the Dependent Variable is the height of the strawberry plants; the Control Group is the fourth strawberry plant that only receives natural ways of water, without any extra water treatment.

In an experiment with strawberry plant clones, different amounts of water are provided to observe their effect on plant height. The amount of water is the independent variable (IV), while the height of the plants is the dependent variable (DV).

One plant receiving no extra water serves as the control group (CG) for comparison. Constants include the experiment's duration, environmental conditions, plant type, measurement method, and initial plant height. This investigation aims to understand the impact of water availability on plant growth.

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Answer:

The unearthed Humerous bones don't belong to the species A.

Explanation:

Hello!

You are studying the Humerus bones species A, who is known to have a mean ratio of 8,5. This value corresponds to the population mean of the length-to-with ratio of bones of species A, symbolized as μ.

The hypothesis you want to study is "The Humorous bones unearthed belong to the species A" if you assume this to be true, then the mean of the length-to-with ratio should be equal to the known population mean of the length-to-with ratio.

Symbolized:

H₀: μ = 8,5

H₁: μ ≠ 8,5

Significance level: α: 0,05

You are asked to use a Z-test, since you don't know the value of the population variance, but have the sample values, the sample size is big enough (more than n=30). Assuming that the sample values are independent, the statistic test of choice is the approximation:

[tex]Z= \frac{x_bar-μ}{S\sqrt[]{n} }[/tex]≅N(0;1)

The critical region, in this case, it's a two-tailed test (remember the type is determined by the null hypothesis) so you'll have two critical values.

Left value [/tex][tex]Z_{\alpha/2} = Z_{0,025} = -1,96[/tex]

Right value [tex]Z_{1-\alpha/2} = Z_{0,975} = 1,96

So you'll reject the null hyphotesis if the calculated [tex]Z_{obs}[/tex] value is ≤-1,96 or ≥1,96 and you'll support it if -1,96<[tex]Z_{obs}[/tex]<1,96

Now we calculate the statistic by replacing the formula with the data:

x_bar = 9,26

S = 1,20

n = 41

[tex]Z_{obs}= \frac{x_bar-μ}{S\sqrt[]{n} }[/tex]

[tex]Z_{obs}= \frac{9,26-8,5}{1,20\sqrt[]{41} }[/tex]

[tex]Z_{obs}[/tex]= 4,0553

Since the calculated value falls in the rejection region, this means, you have statistically significant results. In other words you can reject the null hipothesis (H₀: μ = 8,5) and asume that the unearthed Humerous bones don't belong to the species A.

I hope you have a SUPER day!

Reject null hypothesis; mean ratio of fossil bones significantly greater than Species A mean, based on z-test at 0.05 level.

To check if the fossil humerus bones belong to Species A, we can perform a z-test for a population mean.

Given:

- Population mean [tex](\( \mu \))[/tex] of Species A = 8.5

- Sample mean [tex](\( \bar{x} \))[/tex] of the fossil humerus bones = 9.26

- Sample standard deviation [tex](\( s \))[/tex] of the fossil humerus bones = 1.20

- Sample size [tex](\( n \))[/tex] = 41

- Significance level [tex](\( \alpha \))[/tex] = 0.05

The null hypothesis ([tex]\( H_0 \)[/tex]) is that the mean length-to-width ratio of the fossil humerus bones is equal to the mean ratio of Species A [tex](\( \mu = 8.5 \))[/tex].

The alternative hypothesis ([tex]\( H_1 \)[/tex]) is that the mean length-to-width ratio of the fossil humerus bones is greater than the mean ratio of Species A [tex](\( \mu > 8.5 \))[/tex].

We'll use the z-test formula:

[tex]\[ z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \][/tex]

Substituting the given values:

[tex]\[ z = \frac{9.26 - 8.5}{\frac{1.20}{\sqrt{41}}} \][/tex]

[tex]\[ z \approx \frac{0.76}{\frac{1.20}{\sqrt{41}}} \][/tex]

[tex]\[ z \approx \frac{0.76}{\frac{1.20}{\sqrt{41}}} \][/tex]

[tex]\[ z \approx \frac{0.76}{0.187} \][/tex]

[tex]\[ z \approx 4.07 \][/tex]

Now, we compare the calculated z-value with the critical z-value at [tex]\( \alpha = 0.05 \)[/tex] for a one-tailed test.

From the z-table, [tex]\( z_{\alpha} = 1.645 \)[/tex] (approximately).

Since the calculated z-value (4.07) is greater than the critical z-value (1.645), we reject the null hypothesis.

Therefore, we can conclude that the mean length-to-width ratio of the fossil humerus bones is significantly greater than the mean ratio of Species A, at the 0.05 level of significance. Thus, we cannot conclude that these bones belong to Species A.

D

Explanation:

With the help T cells , B cells make special proteins called antibodies which stick to antigens on surface of germs stopping them in their track.

Answer:

The correct answer is "Yes, β-galactosidase would be produced at induced levels".

Explanation:

The L-arabinose operon, also known as ara operon, is activated in the absence of glucose and in the presence of arabinose. In this study the ara operon was fused with the lac operon, therefore arabinose would active not only the genes of ara operon but also lac operon. Once the researchers provide E. coli with arabinose, the bacteria would produce β-galactosidase as a result of lac operon activation.

Answer: c. During photosynthesis, the hydrogen carrier is NAD

Explanation:

The photosynthesis is a process that occurs in the chloroplast organelle of the plant cells. The typical light reaction of the photosynthesis takes place inside the thylakoid membrane of the chloroplast. The chief reactants such as water and carbon dioxide participate in the reaction. The water gets oxidized in the reaction to release the oxygen as a product. The carbon dioxide get reduced.

The light accepted by the pigment chlorophyll in the chloroplast basically transfers the electron and hydrogen from water to the acceptor that is called as NADP⁺.

Answer:

c. During photosynthesis, the hydrogen carrier is NAD

Explanation:

The co-factor  during photosynthesis is  Nicotinamide Adenine Dinucleotide Phospahte .(NADP), and not NAD.The latter is a coenzyme in cellular respiration  for ferrying H+(as NADH+) together with FADH+ from Kreb's Cycle into the matrix of mitochondria for oxidative phospshorylation, needed  for ATPs synthesis,

NADPH  is a co-factor, needed as  reducing agent in photosynthesis's cyclic and non cyclic photophosphorylation reactions.Therefore option C is not true.

Answer: The element with the greatest influence on cholesterol levels is the fat content of food. Not only the amount of cholesterol, but also the type of fat.

Explanation:

Cholesterol is a type of fat found in the body, which is used to make hormones and vitamin D. The liver makes cholesterol to handle these tasks, but cholesterol can also be incorporated into the body through foods such as meat, dairy, and poultry. If you consume a lot of these, the cholesterol levels could become too high.

There are two types of cholesterol:

If there is a predominance of saturated fatty acids in the fat of the diet, a rise in serum cholesterol and LDL-cholesterol is caused.  To lower the bad cholesterol, you should limit foods such as milk fats, fatty meats or sausage and include more fibre in your diet.  On the other hand, polyunsaturated fatty acids - for example, those present in seed oils or fish - produce their reduction. Monounsaturated fatty acids, such as those provided by olive oil, act like polyunsaturated ones and, in addition, tend to raise HDL-cholesterol, so their use entails a greater benefit.

Answer:

Tubulin are the monomers of the microtubules that plays an important role in the formation of cytoskeleton. The cytoskeleton helps in providing structural framework to the cytoplasm and maintains the cell mobility.

Tubulin is numerous inside the cell. The centrosomes is the main organizing center of the microtubules that plays an important role in the cell division. The golgi bodies also constitute of the microtubules that maintain its structure. The flagella and cilia are made of tubulin that helps in the cell mobility.

Answer:

e. none of the above

Explanation:

Prophase-I of meiosis I include the pairing of homologous chromosomes followed by alignment of these pairs on the cell's equator in metaphase I.

During anaphase-I, the homologous chromosomes are separated from each other and begin to move towards opposite poles. Each homologous pair of chromosomes consists of one maternal and one paternal chromosome carrying corresponding alleles for the genes.

The separation of these homologous chromosomes during anaphase-I is random which means that each member of a homologous pair is randomly distributed to one of the poles. The independent separation of homologous chromosomes to the opposite poles results in unique allelic combinations in gametes.

Answer:

B) Helicase uses energy from ATP Hydrolysis

Explanation:

Helicase's primary function is to separate the annealed nucleic acid strands. It is a motor protein and moves directionally along the phosphodiester backbone. It usually separates strands of double helix DNA or self annealed RNA. It used the energy from ATP hydrolysis and breaks hydrogen bonds between nucleotide bases.

In human body 95 types of helicases are found. They have sequence motifs required for ATP binding, ATP hydrolysis and translocation along nucleic acid phosphodiester backbone. The variable portion in their amino acid sequence imparts specific feature to each helicase.

Starch is broken down into maltose by the enzyme amylase. Maltose is then further hydrolyzed into glucose by the enzyme maltase, primarily in the small intestine.

Starch is broken down into maltose by the enzyme amylase. Initially, starch is hydrolyzed by salivary amylase in the mouth and later by pancreatic amylase in the small intestine, producing shorter polysaccharides, dextrins, and disaccharides like maltose. These smaller sugars are easier for the body to absorb. Following this, the enzyme maltase, which is produced by the cells in the small intestine, further breaks down maltose into individual glucose molecules, which are used by the body for energy.

Answer:

The correct answer will be Option-C.

Explanation:

The pancreas is an organ which plays an important role in the digestion of the food.  Pancreatic juice is secreted by the pancreas which contains digestive enzymes as well as the sodium bicarbonate (base).

The sodium bicarbonate is required for the digestion of the food as after the acidic digestion of the food in the stomach, the food chyme enters small intestine. The inner lumpen is not structured to handle the acidic chyme so sodium bicarbonate neutralizes the acid chyme that entered the lumen of the small intestine.

Thus, Option-C is the correct answer.

Answer:

b. allele

Explanation:

An allele is a variant form of a gene. In this problem, there is one gene (that determines seed color) with two possible alleles (green and yellow).

The other options are wrong because:

The genotype is the combination of alleles an individual has.

The karyotype is an individual's collection of chromosomes, paired and ordered.

A homozygote individual has the same alleles for a particular gene.

A heterozygote individual has different alleles for a particular gene.

Answer:

A and B

Explanation:

A and B are correct answers.

C is incorrect because Botulism does not produce stiff muscle, in fact it produces flaccid paralysis. Stiff muscles and lockjaw is produced by C. Tetani.

D is somewhat incorrect because, although you can kill the toxins via heat, you would need to heat your food to over 100°C for around 10 minutes, which no one does. Thus, reheating beans would not protect you from the bacteria and/or its toxins.

Answer:

Explanation:

1. In prokaryotes , the reproduction takes place by asexual means such as binary fission whereas in eukaryotes the reproduction takes place by sexual means.

2. In eukaryotes mates are required which allows the fusion of males and female gametes, whereas in prokaryotes no mates are required offspring produce from single parent.

3. In prokaryotes many offspring produce from a single parent whereas in eukaryotes only few offspring are produced.

4. Growth of offspring development is complex in eukaryotes whereas the offspring development is simple in prokaryotes.

Final answer:

Eukaryotic reproduction is more complex than prokaryotic due to the presence of membrane-bound organelles, a larger and more complex genome with multiple linear chromosomes, a detailed process of cell division called mitosis, and a slower replication rate requiring tight regulation.

Explanation:

Eukaryotic reproduction is more complex than prokaryotic reproduction for several key reasons.

Answer: C. Radical global climate change.

Explanation:

The entire taxa of the species can be wiped out due to an large catastrophic event associated with the change in the climate. As the climate becomes warm the chances of undergoing the global precipitation rate will change. The climate being an important abiotic factor under unstability is likely to affect the amount of rainfall, snow fall, drought and conditions of desertification. All these factors can be responsible for the removal of the entire taxa of the species in short time.

Answer:

b. most new species accumulate their unique features relatively rapidly as they come into existence, then change little for the rest of their duration as a species

Explanation:

Species can evolve fast or slowly. When species accumulate changes at a slow rate this is called gradualism.

If a species suddenly has a an enormous change for example through mutations in the genes of a few individuals or a single individual the species can has evolve and may become suddenly a new one. A good example can be found in the following link: https://blog.education.nationalgeographic.org/2018/02/07/cloned-crayfish-are-taking-over-the-world/

the article provides evidence of sudden changes that may lead to speciation in at a really fast rate.

The most accurate statement is: 'most new species accumulate their unique features relatively rapidly as they come into existence, then change little for the rest of their duration as a species'. So, option b is true.

According to the punctuated equilibria model, the most accurate statement is: 'most new species accumulate their unique features relatively rapidly as they come into existence, then change little for the rest of their duration as a species'. This model suggests that evolution happens in rapid bursts, interspersed with long periods of stability, often called stasis. Essentially, a species experiences little evolutionary change for most of its existence until a sudden shift in its environment or population pressures spur a rapid period of adaptation and evolutionary change, leading to the formation of a new species. Hence, this model disputes the idea that evolution is a slow, gradual process happening over a long period, a theory known as gradualism.

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Answer: Homologous chromosomes are randomly distributed to daughter cells, this means different chromosomes segregate independently of each other. And they exchange segments of DNA during crossing over. This recombination creates genetic diversity because genes from each parent are exchanged.

Explanation:

Meiosis is a type of cell division that produces gamete cells, which are sex cells (egg and sperm)

Chromosomes that form a pair and are found together are called homologous chromosomes, and they are inherited from each parent. During prophase of meiosis I, the homologous chromosomes exchange segments of DNA in a process called crossing over. This recombination creates genetic diversity because genes from each parent are exchanged. It results in new combinations of genes on each chromosome.

After that, during the anaphase of meiosis I, the two chromosomes line up on the equatorial plane of the cell. Then, they are separated and each will go to a new daughter cell. So homologous chromosomes are randomly distributed to daughter cells, this means different chromosomes segregate independently of each other.

Answer:

Sex linked inheritance may be defined as the type of inheritance in which the trait or mutation is present on X chromosomes rather than on autosomes. Sex linked trait expression will depend upon the type of sex.

Two type of inheritance of sex linked trait are X linked dominant and X linked recessive trait. In X linked recessive trait, the male are more affected as they have only single X chromosome. Females should have both X chromosome affected for the expression of the trait. In sex linked dominant trait, a single X chromosome is enough for the expression of trait.

For example: Color blindness is an sex linked recessive trait, the mother generally passes the trait to their sons. Fragile X syndrome is a sex linked dominant trait in which the affected father will have all the affected daughter.

Answer:

Innate immunity includes non-specific immune responses to prevent the entry of pathogens and to kill the entered pathogen by first line of defenses.

Humoral immunity kills the entered pathogens by production of antibodies specific to the particular type of antigen. B lymphocytes are central to the humoral immunity.

Cell mediated immunity is another type of adaptive immune response wherein the killer T cells directly kill the infected cells.

Explanation:

Cell-mediated immunity is provided by killer T cells against virus-infected cells, foreign cells, and cancer cells. The cytotoxic T cells directly kill the entered pathogens.

Antibody-mediated immunity refers to the specific resistance to disease-causing agents. It includes the production of specific antibodies by B lymphocytes. it is also called humoral immunity.

Innate immune responses are the nonspecific immune responses that prevent the entry of all the diseases causing pathogens and antigens into the body. It includes the first line of defenses such as skin and mucous membranes as well as phagocytes and natural killer cells that kill all the pathogens and antigens non-specifically.