Basidia within a basidiocarp are obvious for mushrooms, but this may not be as obvious for other examples that you have observed. For each of the examples of basidiomycetes that you have observed, speculate about the location of structures analogous to the basidia observed in mushrooms. Think about the formation of spores by meiosis.

a. bird's-nest fungi
b. common puffballs
c. earthstar puffball
d. shelf fungi
e. corn smut

Answers

Answer 1

Answer:

Birds nest fungi: Basidicarp, a cup shaped with several lenticular bodies appear as bird's nest with eggs. The eggs are called the peridioles and the nest serves release the eggs onto the twigs or branches and spores are liberated .

common puffball: The young fruit body is covered with whitish spines that become mark of scar after maturity.

Earth star puff balls: The fruiting body is in the shape of an egg, it has an outer part that separates away from the inner basidiocarp and splits into several rays which curve backwards and resembles a star.

shelf fungi: The fruiting body develops on the tree trunks and termed conk which are tough and leathery often plate-like growing out of the tree trunks.  

corn smut: The diploid teliospores germinate and undergo meiosis to result in four haploid basidiospores.


Related Questions

Which of the following statements about dinoflagellates is true? Select one: A. All known varieties are autotrophic. B. Their walls are usually composed of silica plates. C. Many types lack mitochondria. D. Some produce toxic agents in the water E. Their dead cells accumulate on the seafloor, and are mined to serve as a filtering material.

Answers

Answer: The correct option is D(Some produce toxic agents in the water).

Explanation:

Dinoflagellates are found in the phylum Dinoflagellata which are single celled organisms. They are mostly found in the ocean while few of them can live in streams, freshwater ponds and rivers. The following characteristics can be used to identify a Dinoflagellates, these include:

--> they are golden brown

--> they possess assimilative cell with indented waist.

--> they possess large nucleus with visible chromosomes.

It has been reported that many dinoflagellate species can produce various natural toxins. These toxins can be extremely toxic and many of them are effective at far lower dosages than conventional chemical agents. Therefore the option, Some produce toxic agents in the water, is a correct statement.

Which of the following statements best represents the theory of pangenesis developed by Hippocrates?

a. Pregnancy is a spontaneous event, and the characteristics of the offspring are determined by the gods.
b. Particles called pangenes, which originate in each part of an organism's body, collect in the sperm or eggs and are passed on to the next generation.
c. Offspring inherit the traits of either the mother or the father, but not both.
d. Fertilization of plants is dependent on an animal.
e. Heritable traits are influenced by the environment and the behaviors of the parents.

Answers

Answer:

b. Particles called pangenes, which originate in each part of an organism's body, collect in the sperm or eggs and are passed on to the next generation.

Explanation:

Particles called pangenes, which originate in each part of an organism's body, collect in the sperm or eggs and are passed on to the next generation. The belief of Hippocrates states that "the semen was derived from the whole body." which means the whole body contributes the formation and creation of semen that is responsible for the fertilizing the female's egg with its sperm.

Transport proteins———-

Answers

Answer:

A transport protein is a type of protein that helps an organism move other materials around. Transport proteins are essential for all living organisms' development and survival. Transport proteins come in a variety of shapes and sizes.

OAmalOHopeO

The answer above me is correct the answer above me is correct….yes

A 55-year-old male is sent to your office to have an esophagogastroduodenoscopy. He is a heavy drinker and smoker who also consumes spicy foods on a daily basis. He is quite anxious about having the procedure performed and is afraid he might choke during insertion of the scope. His father and younger brother were both diagnosed with esophageal cancer. His father died, but the brother was just recently diagnosed and is taking chemotherapy.
A. List the risk factors the patient has for developing esophageal cancer.
B. How might you calm the patient's fears about the procedure?

Answers

Answer and Explanation:

A. The risk factors for this patient are age, family history, and smoking and alcoholism. That's because, most cases of esophageal cancer are detected at age 55 years. We must not forget that cancer is a disease with great genetic potential, which means that, as the father and brother of this patient had this type of cancer, this patient has a high probability of developing the same cancer. Furthermore, the constant consumption of cigarettes doubles the risk of developing esophageal cancer. This risk is also increased by alcoholism.

B. To calm the patient down, I would say that there is a very small and almost non-existent risk that he will choke or even feel the endoscopy insertion, as the entire procedure is done with the patient anesthetized and sedated.

The risk factors the patient has for developing esophageal cancer as well as how you can calm the patients fears about the procedure are as follows

A) Risk factors include

People between the age of 45 and 70 have the more risk of esophageal cancer.

If the patient has genetic signs of cancer from the past generations may have highest risk of esophageal cancer.

Males have more risk than women that is 3 to 4 times to develop esophageal cancer.

Alcohol consumption also increases the risk of esophageal cancer.

Too much weight or obesity increases the risk of esophageal cancer.

Deficiency in nutrients, vitamins and minerals can increase risk of esophageal cancer.

B) Calming the patients fears

Make the patient be happy and stress free.

Do exercise which produces mood boost endorphins that get rid of worries and stress.

Stay positive, being friendly keep good health.

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Psychometric scores for anxiety, depression, negative self, somatization, and hostility were combined into a single Global Severity Index (GSI). Global Severity Index was compared for 80 adolescents who were seen in the outpatient clinic for 7 different physical complaints (short stature, obesity, lack of weight gain, breast problems, genital problems, menstrual problems, and hirsutism). Report the degrees of freedom for the relevant F statistic in the numerator and the denominator. (Round your answers to the nearest whole number.)

Answers

Answer:

The correct answer is -

Numerator = 6

Denominator = 73

Explanation:

In F static test or table there are two sets of degrees of freedom: one for the numerator and one for the denominator. The degrees of freedom are equal to the number of observations minus one. Thus, if the sample size were 7 then the degree of freedom would be 7 minus 1

Numerator df = 7 - 1 = 6

Thus, the numerato would be - 6.

Denominator is calculated by subtracting the observation or variables from the total number of sample size.

Denominator df = 80- 7 = 73

Whats bigger, fungi or virus?

Answers

Answer:

fungi

Explanation:

bcuz it is plant-like organism with lack of chlorophyll

what is the source of food for germinating seed ?​

Answers

Endosperm is the reservoir of nutrients in the developing embryo or germinating seed.

Regards
BLACKSHARK

An ecologist counted the number of eggs in randomly selected nests in a population of birds. To test the hypothesis that the number of eggs per nest follows a Poisson distribution at the 5% level of significance, you will need to compare a calculated value with a critical value.
Eggs Number of nests Prob Expected
nest, xi per fi
0 15 0.2845 19.9
1 29 0.3576 25.0
2 19 0.2248 15.7
23 7 0.1331 9.3
SUM: 70 1.0000 70.0
What are the calculated and critical values for this test?
a. calculated = 45.6; critical = 5.991.
b. calculated = 3.10; critical = 5.991.
c. calculated = 5.89; critical = 5.991.
d. calculated = 3.10; critical = 7.815.

Answers

Answer:

d. calculated = 3.10; critical = 7.815.

Explanation:

To test the hypothesis Given in the question above, we use the Chisquare goodness of fit :

The test statistic is defined as :

χ² = Σ(Observed - Expected)² / Expected

Observed = number of nests

χ² = ((15 - 19.9)^2 / 19.9) + ((29 - 25)^2 / 25) + ((19 - 15.7)^2 / 15.7) + ((7 - 9.3)^2 / 9.3)

χ² = 3.1089

The critical value :

df = n - 1 = 4 - 1 = 3

The critical value ;P(χ² at 0.05, df = 3 ) = 7.815

White a paragraph comparing the organices in a to the organs in your body.

Answers

Answer:

An organ is a part of the body of a living organism perform a specific role in the body such as the stomach, the liver and many more. Organelles on other hand are like organs of the cells that are contained in the cytoplasm of cell these also have a certain specialized roles to play for the cell, and they all depend on each other. The example of the organelles are nucleus, mitochondria and many more.

* Organelles are structures inside the cytoplasm of cells

* Organs are amde up of tissues composed of group of specialized cells that, has a particular role in the body

Glycogen is an important and quickly mobilized source of stored glucose. Glucose is mobilized for ATP generation in muscle in response to epinephrine, which activates Gs. How would inhibitors of cAMP phosphodiesterase affect glucose mobilization in muscle

Answers

Answer:

Glucose mobilization would continue

Explanation:

The GTP-bound alpha subunit (Gαs) is a subunit of the heterotrimeric G protein which is required for stimulating the cAMP-dependent pathway through the activation of the membrane-associated enzyme adenylyl cyclase, which in turn catalyzes the synthesis of the second messenger cyclic-3′,5′-adenosine monophosphate (cAMP) from ATP. Moreover, cAMP phosphodiesterase is an enzyme that hydrolyzes cAMP, thereby negatively regulating the levels of this second messenger, which is fundamental for the termination of the signal. In this case, the epinephrine signal initiates glycogen breakdown (glucose mobilization) in muscle cells. In consequence, if the cAMP phosphodiesterase enzyme is inhibited, the concentration of cAMP will persist high even beyond the termination of the signal, thereby glucose mobilization would continue.

What causes fingers to look wrinkled after soaking in water?
O A. The skin cells lose all their water.
O B. Water enters the cells in the skin.
O C. The skin cells shrink in size.
.
D. Salt enters the skin cells.

Answers

Answer:

B. Water enters the cells in the skin.

Explanation:

Which one of the following physical quantities is formed after dividing mass of volume?
I} Weight
II} Velocity
III} Density
IV}Work​

Answers

Answer:

Density

Explanation:

Density is refer to how much space an object in Mass take in volume.

It is mass if an object divided by volume.

It usually in g/cm3

What is gylcosylation

Answers

Glycosylation is the reaction in which a carbohydrate, i.e. a glycosyl donor, is attached to a hydroxyl or other functional group of another molecule in order to form a glycoconjugate. In biology, glycosylation usually refers to an enzyme-catalysed reaction, whereas glycation may refer to a non-enzymatic reaction.

The answer is:

Glycosylation is a biochemical process where a glycan attaches to a protein, a lipid, or other organic molecule, especially through the catalytic action of certain enzymes.


If the time interval is 55 seconds, what is the distance to the epicenter?

Answers

the answer to this question is 800

Schistosoma becomes infectious to humans during which of the following stages?
After cercariae migrate to the liver
After the eggs are released in urine
After it is released from the snail
After mating of male and female worms
Before the eggs hatch

Which of the following groups includes a protozoan that is transmitted sexually?
Chromalveolata
Excavata
Amoebazoa
Archaeplasteda
Rhizaria

Schistosoma becomes infectious to humans during which of the following stages?
After cercariae migrate to the liver
After the eggs are released in urine
After it is released from the snail
After mating of male and female worms
Before the eggs hatch

Which part of a mushroom would be eaten?
Thallus
Sporangium
Mycelium
Septum
Hyphae

Answers

Answer:

After it is released from the snailExcavata    Sporangium

Explanation:

Schistosoma life cycle:

1. Human excretes schistosomes eggs through feces and/or urine. These eggs eventually reach the water.

2. Once in the water, schistosomes eggs hatch and release immature larvae named miracides.  

3. Miracides swim and penetrate a freshwater snail, which is an intermediary host.  

4–5. Once inside the snail, miracides turn into sporocites and then into cercariae. Cercariae characterize by having a bifurcated tail. Snails release the cercariae in water, and they swim until they get in touch with a mammal. Once they reach the mammal, they penetrate the skin and get into the body. This is the stage in which the parasite becomes infectious to humans. Just a few cercariae are enough to infect a person.  

6. When cercariae penetrate their host body, they lose their tails and turn into schistosomules. These new forms travel to the liver, where they end their maturation process.  

7. Males and females mate in the liver and migrate to veins of the intestine or bladder. Once there, the females lay the eggs, which are then excreted by the host. And the cycle begins again.

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Excavata are asymmetrical protists. The group receives its name because of the appearance of the feeding groove. Excavata is composed of many protists. One of them is Parabasalid.

Parabasalids are parasitic protists, and one of them is Trichomonas vaginalis, which is transmitted sexually.

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Sporangium. This is the reproductive structure of the mushroom and the one that is usually eaten. Spores develop inside the sporangium. Its surface is covered by a protector cuticle. It can be eaten when it is closed or when it is already open. However, when open is tastier.

Explain factors that impact North Carolina ecosystems. (Examples: acid rain effects in mountains, beach erosion, urban development in the Piedmont leading to habitat destruction and water runoff, waste lagoons on hog farms, Kudzu as an invasive plant, etc.)

Answers

Answer:

Temperature, moisture, topography, elevation, soils, hydrology and disturbance.

Explanation:

Temperature, moisture, topography, elevation, soils, hydrology and disturbance are the factors that impact North Carolina ecosystems. Pollution, urbanization and invasive species inversely affected the ecosystem of North Carolina. Industries produces harmful gases that causes pollution as well as make the rainfall acidic which adversely affected the infrastructure as well as plants and animals. The introduction of invasive species also has a negative impact on the plants and animals of that ecosystem.

I want answers please ​

Answers

Explanation:

1.arm&base

2.a shutter controlled by a lever

3.Nosepiece,objective lenses,stage clips

6. 100X.....(this means that the image being viewed will appear to be 100 times its actual size

Lectins often bind their ligands via multiple weak interactions. bind their ligands with relatively low specificity. prevent viruses from binding to their target cells. are carbohydrates that bind to receptor proteins.

Answers

Answer:

The correct answer is - B.often bind their ligands via multiple weak interactions.

Explanation:

Lectins are specific types of proteins that identify and bind to specific carbohydrates present on the cell surfaces. They have an essential role in interactions and communication between various cells for identification and recognition.

Binding sites of lectins on the surface of one cell bind to the Carbohydrates on the surface of another cell. A lectin usually has two or more binding sites for carbohydrate units.

Macronutrients is the most readily available for energy production

Answers

Answer:

The correct answer is - true.

Explanation:

All the macronutrients carbohydrates, fats, and protein all three produce energy. Carbohydrates are the most preferred source of energy for the human body as it is the macronutrient that your system most requires.

The human body easily breaks down most carbohydrates and provides a significant amount of energy. The energy-providing process is called cellular respiration starts with glucose as a substrate. Glucose is the simplest carbohydrate.

Muscle cells that are worked often in exercise will require more mitochondria. If we use more mitochondria, it will require us to use more carbohydrates to build ATP’s. What if we run out of carbohydrates stored in our body? What macromolecule can we break down and use in cellular respiration?
Group of answer choices

A) sucrose

B) lactose

C) glycogen

D) lipids

Answers

Answer:

Lipids

Explanation:

Apply logarithm to evaluate T given 644 204=400 000 (1+10%)^T,by first simplify the equation.​

Answers

The value of T after applying logarithm to evaluate the function is 4.9988

Given the expression

[tex]644 204=400 000 (1+10 \%)^T[/tex]

Since 10% = 0.1, the expression becomes;

[tex]644 204=400 000 (1+0.1)^T\\644 204=400 000 (1.1)^T\\\frac{644204}{400000} =1.1^T\\1.61051=1.1^T[/tex]

Take the logarithm of both sides

[tex]log1.61051=log1.1^T\\log1.61051=Tlog1.1\\Swap\\Tlog1.1=log1.61051\\T=\frac{log1.61051}{log1.1}\\T=\frac{0.2069}{0.04139} \\T=4.9988[/tex]

The value of T after applying logarithm to evaluate the function is 4.9988

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You are studying an enzyme that is inactivated by phosphorylation and create a mutant in which the threonine that is normally phosphorylated is replaced with glutamate. Predict the impact of this change on the activity of this enzyme. Group of answer choices

Answers

Answer:

always active  

Explanation:

Phosphorylation is a posttranslational modification that consists of the addition of phosphate groups to specific amino acids on the protein. Phosphorylation acts as a molecular switch for proteins that are phosphorylated (i.e., in some situations phosphorylation acts to activate protein function, whereas in other situations phosphorylation can inactivate protein function). Phosphorylation modifies the three-dimensional structure of the protein, thereby affecting, for example, the accessibility of the active site of a phosphorylated enzyme to its substrate. Phosphorylation can occur only at the side chains of three amino acids: Serine, Threonine and Tyrosine. In this case, the enzyme is inactivated by phosphorylation on the Threonine residue, so it is expected that the mutant enzyme cannot be phosphorylated, remaining in an active state.

5. Which of the following professionals has the authority to prescribe medication and treatment?
A. Nurse practitioner
B. Clinical dietician
C. Nutritionist
D. Home health aide

Answers

Answer: A. Nurse practitioner

Explanation: Hope it helps

Answer:

A. Nurse practitioner

Explanation:

Have a great day!

the valency of flourin is 1 what is the valence electron​

Answers

Answer:

Fluorine has 7 valence electrons and would need another one to have a full outer valence electrons.

Bleach For me!!!!!!!!!​

Answers

Answer:

answer: c

Explanation:

A hemoglobin molecule is made up of four

polypeptide chains, two alpha chains of 141

amino acid residues each and two beta

chains of 146 amino acid residues each . In the complete molecule, four subunits are closely joined, as in a three-dimensional jigsaw puzzle, to form a tetramer.

it also have a structure called quaternary structure, in which two or more polypeptide chains folded in to tertiary structure become associated in the final structure of the protein.

hope this will help !

Answer:

Option C

Explanation:

4 subunits (2 alpha and 2 beta)

Given the enormous heterogeneity of antigen receptors expressed on the populations of naive B and T lymphocytes, the adaptive immune response relies on a process whereby the rare lymphocyte that binds to the antigen is first induced to proliferate, before it can perform its effector function. For B cells, there is a clever mechanism that ensures that the specificity of the antibody secreted by the plasma cell will recognize the same pathogen that initially stimulated the B cell antigen receptor and induced B cell proliferation. This mechanism is:

Answers

Answer:

The naïve B cell expresses a membrane-bound form of the antibody as a receptor and secretes that same antibody when it differentiates into a plasma cell.

Explanation:

Naive B cell is a type of B cell that has still not been exposed to the antigen. The B cell receptors (BCRs) are transmembrane proteins composed of 1-a surface immunoglobulin molecule capable of recognizing the antigen, and 2-two transmembrane subunits that transduce the signal. When a B cell binds to antigen with its BCR, the B cell will proliferate and differentiate into a plasma cell. This plasma cell then releases antibodies, which are soluble forms of the BCR where the transmembrane domain that anchors the antibody protein to the membrane of the B cell is eliminated.

Which of the following groups includes a protozoan that is transmitted sexually?

Chromalveolata

Excavata

Amoebazoa

Archaeplasteda

Rhizaria

Which of the following is a distinction between Ascomycetes and Basidiomycetes?
Ascomycetes form a sac around the sexual spores.
Ascomycetes have zygotes.
Ascomycetes release basidiospores.
Basidiomycetes form a shell around the sexual spores.
Basidiomycetes release conidiospores.

Which of the following is a specialized structure that would contain food ingested by phagocytosis?
Saprozoic
Cytostome
Cytoproct
Oral groove
Holozoic

Which part of a mushroom would be eaten?

Thallus

Sporangium

Mycelium

Septum

Hyphae

Answers

Answer:

Ascomycetes have zygotes

Mycelium

Explanation:

Ascomycetes have zygotes which is the main distinction between Ascomycetes and Basidiomycetes. Mycelium is a part of mushroom that is edible and can be eaten. Mycelium is the vegetative part of a fungus that consist of a mass of branching, thread-like hyphae. Hypha is a long, branching, filamentous body of mushroom as well as oomycete and actinobacteria. Hyphae are the primary mode of vegetative growth of fungi which is generally known as mycelium so mycelium is the edible part of mushroom.

How do bacteria develop resistance to drugs

Answers

Answer: Bacteria gain resistance to drugs because of mutations (permanent and random changes to their DNA) which means they have changed DNA coding, giving them the ability to resist the drug fighting them off. As a result, they survive and reproduce. Over time, more and more bacteria are generated as the DNA code for resistance is passed on over generations. This results in bacteria having the ability to resist drugs. This is particularly prevalent with antibiotics.

Bacteria develop drug resistance through genetic mutations, acquisition of resistance genes, production of inactivating enzymes, efflux pumps, and biofilm formation.

Bacteria can develop resistance to drugs through several mechanisms. One common way is through genetic mutations or acquisition of resistance genes. Mutations can occur in the bacterial DNA, leading to changes in the target site of the drug, rendering it ineffective. Resistance genes can be transferred between bacteria through horizontal gene transfer, allowing the recipient bacteria to acquire resistance traits.

Another mechanism is the production of enzymes that can inactivate the drug. Bacteria can produce enzymes, such as beta-lactamases, that break down antibiotics like penicillin, preventing them from functioning properly. Bacteria can develop efflux pumps that actively pump out drugs from their cells, reducing their concentration and effectiveness. This mechanism helps bacteria evade the lethal effects of antibiotics. Biofilm formation provides a protective environment for bacteria, making them less susceptible to drugs and immune system attacks.

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The correct question is:

How do bacteria develop resistance to drugs?

Cells control or regulate the flux through metabolic pathways by means of I. allosteric control of enzymes. II. covalent modification of enzymes. III. genetic control of the concentrations of enzymes. IV. genetic expression of allosteric regulators.

Answers

Answer:

I, II, III

Explanation:

Allosteric regulation is a common mechanism of regulation of enzyme activity, which generally involves key enzymes in metabolic pathways. Allosteric modulation occurs when a substance/molecule called 'allosteric regulator' binds to an enzyme at a site other than the enzyme's active site, thereby triggering a conformational change in the enzyme and thus affecting its activity. Moreover, covalent modifications such as phosphorylation, acetylation, methylation, etc, are also involved in the regulation of enzymatic activity in metabolic pathways. For example, phosphorylation is a type of reversible covalent modification of proteins consisting of the addition of phosphate groups at specific amino acid residues (i.e., serine, threonine, and tyrosine) by specific enzymes known as protein kinases. Phosphorylation alters the three-dimensional (3D) structure of proteins, thereby turning the substrate or cellular pathway to active and/or inactive. Finally, both transcriptional and post-transcriptional mechanisms are also able to control the expression of enzymes involved in signaling pathways. Transcriptional mechanisms are capable of limiting the amount of mRNA that is produced from a given gene (in this case, an enzyme encoding gene), whereas post-transcriptional mechanisms such as, for example, RNA interference pathways, control the translation of messenger RNA (mRNA) molecules into proteins.

6.Two apparently normal parents have a daughter who suffers from agammaglobulinemia (AGG), an inherited defect of the immune system. Use a Punnett square to show how two normal parents could have a child afflicted with an inherited disease. What are the parents' genotypes

Answers

Answer:

See the answer below

Explanation:

For two normal parents to have produced a daughter who suffers from AGG, it means the disorder is an autosomal recessive trait. In order words, it means the parents are both carrying the allele for the disorder in recessive form.

Let us assume that the allele for the disorder is represented by b, meaning that the alternate form would be B. Both parents will have the genotype Bb and will both appear normal.

      Bb    x    Bb

    BB   Bb   Bb   bb

The affected child would be the one with bb genotype while the remaining children would appear normal. See the attached image for the Punnet's square analysis of the cross.

The genotypes of the parent would be heterozygous.

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Yr.1 Yr. 2 Yr. 3 Capital expenditure 250,000 125,000 75,000 Cash dividends 140,000 140,000 140,000 Cash flow from operations before interest expense 750,000 780,000 800,000 Assuming the Prime rate = 8% each year. What will be the interest payment at year 2? a. 25,000.b. 50,000.c. 100,000.d. 150,000.e. 200,000. What rhetorical device isevident in the followingpassage from Thomas Paine's"The Crisis, No.1"?Wars, without ceasing, willbreak out till that periodarrives, and the continentmust in the end beconqueror; for though theflame of liberty maysometimes cease to shine,the coal can never expire.A. repetitionB. parallelismC. emotional argumentD. establish a common ground The direction of applied force has to be .............. to the distance in order to say work is done (Economics)Which of the following best explains why the money supply is increased whenthe Fed buys Treasury bonds?A. When the Fed buys Treasury bonds, it increases the amount ofdeposits in people's bank accounts.O B. 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