Mutations in the Pax6 gene causes developmental defects in the eyes of flies, mice, and humans. This suggests that:
a. Pax6 has been evolutionarily conserved.
b. Pax6 activates multiple genes.
c. Pax6 may be a master regulator of eye development.
d. All of these choices are correct.
Answer: All of these choices are correct.
Explanation:
Paired box 6 also known as Pax6 is an evolutionarily conserved transcription factor that plays a major role in the development of the vertebrate eye and the central nervous system. Mutations in mammalian Pax6 are characterized by eye defects, e.g (congenital cataracts, as well as Central Nervous System disorders like microcephaly, dysgenesis, neuron to glia transformations, etc.) Many studies have demonstrated the importance of Pax6 to vertebrate cell differentiation, proliferation, patterning. etc
Pax6 have the tendency to activate multiple genes. For example, if two fly Pax6 genes, eyeless (ey) and twin of eyeless (toy), (initiate eye specification), and, eyegone (eyg) and the Notch (N) pathway, (independently regulate cell proliferation). Pax6 have the tendency to controls eye progenitor cell survival and proliferation through the activation of eyegone (eyg) , thereby indicating that Pax6 initiates both eye specification and proliferation.
Also, Pax6 is considered to be the master control gene for eye development in all seeing animals. In vertebrates, it is required not only for lens/retina formation but also for the development of the central nervous system.
Throughout the book, the deterioration of Morrie's body is symbolically compared to ____________.
a. The orange cactus plant.
b. The tank of goldfish.
c. The spruce tree outside his window.
d.The pink hibiscus plant.
Answer: option D
Deterioration of Morrie's body is compared to a pink hibiscus plant.
Explanation:
Tuesdays with Morrire is a book written by an American author Mitch Albom. The book was written base on the visits the author made to his form sociology lecturer Morrire Schwartz who was suffering from Amyotrophic lateral sclerosis. The book use a metaphor to describe the deteriorating body of Morrie's using thr pink hibiscus plant. When the pink hibiscus plant is deteriorating, the plant petals start to fall and later dies. Morrie's body also began to deteriorates while he depend on oxygen to breathe, as times goes by, the deterioration continues just as petals of hibiscus plant and he eventually died.
Attached earlobes are recessive to free earlobes. What genotypic ratio is expected when an individual with attached earlobes mates with an individual heterozygous for free earlobes?
a) 2:1
b) 3:1
c) 1:1
d) 1:2:1
Answer:
C
Explanation:
This is a monohybrid cross between involving a gene coding for earlobe. The allele for free earlobe (F) is dominant over that of attached earlobe (f).
N.B: let F represent the gene involved.
A genotypic ratio is the ratio derived from the offsprings based on the actual genetic make-up and not based on their phenotypic appearance as in phenotypic ratio.
A homozygous individual is an individual having the same kind of alleles for a gene while a heterozygous individual is one that has two different alleles i.e. combination of the dominant and recessive allele.
The monohybrid cross is between an individual with attached earlobe (ff) i.e. homozygous recessive individual. (Note that, a recessive trait will only be expressed in a homozygous state) and a heterozygous free earlobeb individual (Ff). (See attached picture). Ff × ff
The resulting four offsprings will
be expected to have only two distinct genotypes i.e heterozygous (Ff) and homozygous recessive (ff) in an equal ratio i.e. 2:2 ~ 1:1
In mating between an individual with attached earlobes and one heterozygous for free earlobes, the expected genotypic ratio for their offspring is 1:1, representing an equal occurrence of heterozygous and homozygous recessive genotypes.
Explanation:When an individual with attached earlobes (which is a recessive trait, aa) mates with an individual who is heterozygous for free earlobes (Aa), the possible genotypes of the offspring can be determined by a Punnett square. The heterozygous individual (Aa) can contribute either the dominant allele (A) or the recessive allele (a), and the individual with attached earlobes can only contribute the recessive allele (a).
A Punnett square matching these contributions results in offspring that are 50% heterozygous (Aa) and possess free earlobes and 50% homozygous recessive (aa) and possess attached earlobes. Therefore, the genotypic ratio expected is 1:1.
Net primary production is: Group of answer choices the same thing as gross primary production. gross primary production plus chemosynthesis. gross primary production plus autotrophic respiration. gross primary production minus au
Answer:
Gross primary production - autotrophic respiration.
Explanation:
Primary production may be defined as the process of formation of organic compound from the photosynthesis. The plants are known as producers in the ecosystem.
More than 90% of the energy is lost from the one trophic level to the next trophic level. The net primary productivity can be calculated by the groos primary productivity produced by the plant minus the productivity lost during the process of respiration of plants.
Thus, the correct answer is option (4)
Net primary productivity refers to the total energy that remains after accounting for the energy consumed by primary producers for growth and reproduction. This energy is what is available for primary consumers at the next trophic level in an ecosystem.
Explanation:Net primary productivity refers to the energy that remains after accounting for the energy consumed for processes like growth and reproduction by primary producers like plants, algae, and photosynthetic bacteria. These entities, known as photoautotrophs, incorporate energy from the sun in a process known as gross primary productivity. However, not all this energy is available for organisms at the higher trophic levels. The primary consumers at the next trophic level can only utilize the energy remaining, which is referred to as the net primary productivity.
The calculation of net primary productivity is done by subtracting the energy utilized for cellular respiration from the gross primary productivity. This measures the net energy accumulation within an ecosystem, thus being a crucial element in the study of energy flow within ecosystems.
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If a cell has completed the first meiotic division and is just beginning meiosis ii
Answer: The correct answer is : It has half the amount of DNA as the cell that began meiosis.
Explanation: Meiosis gives rise to four unique daughter cells, each one having half the number of chromosomes that the mother cell, this is meiosis I. During meiosis II the two cells go through four phases of division again, at this stage there the number of chromosomes in daughter cells is reduced.
How does the simple primary and secondary structure of dna hold the information needed to code for the many features of multicellular organisms?
Answer:
Explanation:
The nitrogenous base sequence of the DNA is responsible for carrying the genetic information needed to code for proteins and many features of multicellular organisms.
What can be interpreted about type 2 diabetes mellitus in monozygotic twins when it affects only one twin 70% of the time and affects both twins 30% of the time? a. The trait is recessive. b. Homologous genes can undergo chromatid exchanges. c. Nongenetic factors can influence expression. d. Mutation repair is incompletely penetrant.
Answer:
The correct option is C. Non-genetic factors can influence expression.
Explanation:
Non- genetic factors can be described as factors other than genetic makeup. These factors include environmental factors and factors arising from the lifestyle of an organism.
Certain studies have shown that both the genetic as well as the non- genetic factors play a role in the development of certain characteristics.
In the above-mentioned scenario, both the monozygotic twins have diabetes but one of them is affected 70% of the time due to his lifestyle. The person might be consuming more sugars and alcohol.
A primigravida is admitted to the birthing unit in early labor. A pelvic examination reveals that her cervix is 100% effaced and dilated 3 cm. The fetal head is at +1 station. In which area of the client's pelvis is the fetal occiput?
The fetal occiput is in the Ischial spines.
Explanation:During labor and delivery, the baby passes through the “pelvic bones to reach the vaginal opening”. The pelvis is located between the hip bones and is wide/flat in females. The pelvis has the uterus, cervix and vagina. The muscles in the uterus push the baby down. The baby’s head presses the cervix releasing oxytoxin. Then it dilates and allows the baby to pass through fetal station.
Fetal station is the fetus/baby is in the pelvis. Occiput is the lower part of the head/skull. The presenting part of the baby passes through the birth canal. Most of the time it would be “baby's head, shoulder, the buttocks, or the feet”. Ischial spines are “bone points” on the “mother's pelvis”. It is the “narrowest part of the pelvis”.
0 station: This is the position when baby's head is at the Ischial spines. The baby is "engaged" when largest part of the head enters the pelvis. If the presenting part lies above the Ischial spines, the station is reported as a negative number from -1 to -5.
how do humans cause ecological issues in ecosystems?
Answer:
Explanation:
Human activities cause serious damage and ecological issues to the ecosystem which is the environmental deterioration and depletion of resources like air, water and good soil. These harmful activities that cause such destruction includes: pollution, extinction of wild life, the demand for food and shelter as the population increases, overfishing, agriculture all these have negative impacts on the ecosystem.
Which three of the following statements are consistent with the images?a. In a sea turtle's flippers, heat is transferred from (3) to (1).b. In a sea turtle, blood warms as it flows from the body (1) to the tip of the flipper (2).c. In a sea turtle's flippers, heat is transferred from (1) to (3). d. At a dolphin's testes, heat is transferred from (2) to (1).e. In a dolphin, blood cools as it flows from the aorta to the testes. f. At a dolphin's testes, heat is transferred from (1) to (2).
Answer:
C
Explanation:
American vultures used to be classified in the same family as African vultures. Which discovery caused scientists to reclassify American vultures as more closely related to storks?
Answer:
DNA evidence revealed the American vultures share more recent ancestor with the Storks
Explanation:
The hooded vultures that is mostly found in the African continent have a close resemblance with the American vultures and were traditionally classified to belong to the Falcon family.
However, it was observed that the American vultures shared a similar behavior with Stork which is not common to the vulture found in Africa, including the hooded vulture. The Stork and the American vulture exhibit the behavior of urinating on their legs when being overheated. When the urine gets evaporated, it helps them to cool their body temperature.
This shared behavior between the storks and the American vultures led scientists into using molecular analysis in analyzing the DNA of the hooded vultures found in Africa, the American vultures, and the stork.
Evidence from the DNA analysis later revealed that the American vultures and the storks share a more common DNA sequences than African vultures and American vultures do.
Salivary amylase enzymatically breaks down glucose in the oral cavity. true or false
Answer:
FALSE
Explanation:
Digestion begins in the oral cavity when food enters in contact with the salivary amylase, the principal enzyme in saliva. Carbohydrates in the form of starches (like potatoes, rice, or pasta) are hydrolized from polysaccharides into disaccharides, amylose and amylopectin are hydrolyzed into smaller chains of glucose (dextrins and maltose). This step in the digestion of sugars is limited due to the brief exposure time of the food to the enzyme.
Therefore we can conclude that the answer is FALSE because salivary amylase enzymatically breaks down polysaccharides into disaccharides, glucose is not broken down at this point.
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In the cross between a female A/a;B/b;c/c;D/d;e/e and male A/a;b/b;C/c;D/d;e/e, (Assume independent assortment of all genes and complete dominance.)What proportion of the progeny will be phenotypically identical to
(1) the female parent?
(2) the male parent,
(3) either parent and
(4) neither parent?What proportion of the progeny will be genotypically identical to
(1) the female parent?
(2) the male parent,
(3) either parent, and
(4) neither parent?
Answer:
Phenotypically identical
(1) the female parent: 9/64
(2) the male parent: 9/64
(3) either parent: 9/32
(4) neither parent: 23/32
Genotypically identical
(1) the female parent: 1/16
(2) the male parent: 1/16
(3) either parent: 1/8
(4) neither parent: 7/8
Explanation:
To find out the answers, we will have to find out probability of each gene separately:
female male
A/a;B/b;c/c;D/d;e/e A/a;b/b;C/c;D/d;e/e
If we will look at only A gene combinations in both the parents the results will be as under:
Parentals : A/a x A/a
The 4 gametes in the progeny will be as under :
AA = 1/4
Aa = 1/4 Combined probability of Aa = 1/4+ 1/4 = 1/2
Aa = 1/4
aa = 1/4
If we will look at only B gene combinations in both the parents the results will be as under:
Parentals : B/b x b/b
The 4 gametes in the progeny will be as under :
Bb = 1/4 Combined probability of Bb = 1/4+ 1/4 = 1/2
Bb = 1/4
bb = 1/4 Combined probability of bb = 1/4+ 1/4 = 1/2
bb = 1/4
If we will look at only C gene combinations in both the parents the results will be as under:
Parentals : c/c x C/c
The 4 gametes in the progeny will be as under :
Cc = 1/4 Combined probability of Cc = 1/4+ 1/4 = 1/2
Cc = 1/4
cc = 1/4 Combined probability of cc = 1/4+ 1/4 = 1/2
cc = 1/4
If we will look at only D gene combinations in both the parents the results will be as under:
Parentals : D/d x D/d
The 4 gametes in the progeny will be as under :
DD = 1/4
Dd = 1/4 Combined probability of Dd = 1/4+ 1/4 = 1/2
Dd = 1/4
dd = 1/4
If we will look at only E gene combinations in both the parents the results will be as under:
Parentals : e/e x e/e
The 4 gametes in the progeny will be as under :
ee = 1/4
ee = 1/4 Combined probability of ee = 1/4+ 1/4 + 1/4 +1/4 = 1
ee = 1/4
ee = 1/4
It is given that genotype of first parent is A/a;B/b;c/c;D/d;e/e or AaBbccDdee .
Also, it is pertinent to mention here that AA and Aa genotype will produce same kind of phenotype. In the progeny, we can calculate the probability of AA & Aa will be 1/4 + 1/2 = 3/4
Similarly with respect to gene B, BB and Bb will produce same kind of phenotype but BB genotype will not get produced so we will only find out probability of Bb alone which is 1/2
Similarly the allelic combinations of gene C which will be similar to first parent will be 1/2
The combinations of gene D which will be similar to first parent will be a combination of DD & Dd which is 1/4 + 1/2 = 3/4
For gene e the combinations which will produce same phenotype as of first parent will be 1 because all the combinations are ee.
CALCULATIONS FOR PROGENY WHICH ARE SIMILAR TO PARENTS PHENOTYPICALLY.
(1) So, the combined probability of resemblance of phenotype of progeny with first parent which is female = 3/4 x 1/2 x 1/2 x 3/4 x 1 = 9/64.
(2) The genotype of male parent is A/a;b/b;C/c;D/d;e/e or AabbCcDdee.
So, in a similar way we can find out the combined probability of resemblance of phenotype of progeny with second parent = 3/4 x 1/2 x 1/2 x 3/4 x 1 = 9/64
(3) The progeny which are similar to either parent will be 9/64 + 9/64 = 18/64 = 9/32.
(4) The progeny which will have phenotype which does not match any parent will be 1 - 9/32 = 32 - 9 /32 = 23/32.
CALCULATIONS FOR PROGENY WHICH ARE SIMILAR TO PARENTS GENOTYPICALLY.
When we will look for progeny which are genotypically similar to parents, we will look for allelic combinations which are exactly similar to parents. While finding genotypes, homozygous dominant and heterozygous will not be same. For example, AA and Aa will not produce same genotype.
(1) The probability of progeny which will be genotypically identical to female parent (AaBbccDdee) is 1/2 x 1/2 x 1/2 x 1/2 x 1 = 1/16.
(2) The probability of progeny which will be genotypically identical to male parent (AabbCcDdee) is 1/2 x 1/2 x 1/2 x 1/2 x 1 = 1/16.
(3) The progeny which will be genotypically similar to either parent will have probability = 1/16 + 1/16 = 1/8
(4) The progeny which will have genotype which does not match any parent will be = 1 - 1/8 = 7/8
Final answer:
The expected proportion of dominant phenotype offspring from a tetrahybrid cross in all loci (A-D) is 81/256. To be phenotypically identical to either parent, offspring must have the dominant phenotype, regardless of homozygosity or heterozygosity. The genotypic likeness to a parent is difficult to calculate without a Punnett square, especially in tetrahybrid crosses.
Explanation:
To answer the question regarding the inheritance patterns from a cross between a female A/a;B/b;c/c;D/d;e/e and a male A/a;b/b;C/c;D/d;e/e, we must consider each gene locus independently due to the assumption of complete dominance and independent assortment.
Phenotypic Proportions
For phenotypic proportions, we are interested in the offspring that exhibit the dominant phenotype for each of the four loci. Using the sum rule and product rule, the probability for each dominant allele being present (either homozygous dominant or heterozygous) is 3/4. So, the expected proportion of offspring with the dominant phenotype at all four loci (A-D) is 81/256 (3/4 x 3/4 x 3/4 x 3/4).
Genotypic Proportions
The genotypic ratio for a cross of Aa x Aa individuals is 1:2:1, which results in a phenotypic ratio of 3:1 for the dominant trait. This calculation can be applied to each locus separately. To be phenotypically identical to either parent, they must express the dominant trait regardless of whether they are homozygous or heterozygous for it.
Phenotypically identical to the female parent: The proportion is 81/256 (dominant phenotype at all loci).Phenotypically identical to the male parent: The proportion is also 81/256 since the dominant phenotype is the same for both parents.Phenotypically identical to either parent: Since both options above are the same, the proportion remains 81/256.Phenotypically identical to neither parent: This case is not possible since all dominant phenotypes will match the parents.Before exploring some of the features of the different types of eukaryotes, we should first review some of the fundamental differences between eukaryotes and prokaryotes. The two groups of organisms differ fundamentally in the structure of their individual cells. For each of the following statements, identify whether it refers to prokaryotes, eukaryotes, or both groups.
Drag each statement into the appropriate bin.
a. Prokaryotes only.
b. Eukaryotes only.
c. Both prokaryotes and eukaryotes.
Answer:
Prokaryotes only:
cell wall contains peptidoglycan or pseudomurein binary fission 70S ribosomes singular circular chromosomeEukaryotes only:
membranous organelles, including mitochondria, lysosomes, endoplasmic reticulum 80S ribosomes nuclear envelope compartmentalizes the chromosomesBoth prokaryotes and eukaryotes:
plasma membrane encloses the cytoplasm has both DNA and RNA includes unicellular cellsFinal answer:
Prokaryotic cells lack a nucleus and have simpler structures, while eukaryotic cells have a defined nucleus and are more complex, with multicellular organisms being exclusively eukaryotes. Both share basic cell components like the plasma membrane and ribosomes.
Explanation:
The distinctions between prokaryotic cells and eukaryotic cells are key in understanding cell structure and function. Prokaryotic cells, which include Bacteria and Archaea, are characterized by lacking a nucleus and membrane-bound organelles, and their DNA generally exists in a single, circular chromosome within a nucleoid region. On the contrary, eukaryotic cells, which make up animals, plants, fungi, and protists, contain a well-organized, membrane-bound nucleus with multiple, rod-shaped chromosomes and other internal membrane-bound organelles.
Both prokaryotic and eukaryotic cells share some fundamental components such as the plasma membrane, which serves as a barrier to the environment, the cytoplasm, which contains organic molecules and salts, a DNA genome that stores genetic information, and ribosomes, where proteins are synthesized. However, eukaryotic cells are generally larger and more complex, with multicellular organisms exclusively falling into the eukaryote category.
If organisms A, B, and C belong to the same class but to different orders and if organisms D, E, and F belong to the same order but to different families, which of the following pairs of organisms would be
expected to show the greatest degree of structural homology?
a. A and B
b. A and C
c. B and D
d. C and F
e. D and F
The greatest degree of structural homology is expected to be found between organisms D and F as they belong to the same order but different families. Higher order taxonomic ranks such as class represent a wider range of organisms, so similar rank organisms from different orders are likely to show less homology.
Explanation:In the biological classification of organisms, the 'Class' is a higher taxonomic rank than 'Order', which is higher than 'Family'. Organisms in the same class can be quite diverse, whereas organisms within the same order tend to have more similarities. Hence organisms belonging to the same order but in different families (D, E, F) would exhibit a greater degree of structural homology than organisms that belong to the same class but in different orders (A,B,C). So, the pair showing the greatest degree of structural homology would be D and F as they belong to the same order but to different families.
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Organisms A and B would be expected to show the greatest degree of structural homology. The correct option is a. A and B
Organisms A and B belong to the same class but different orders, indicating a closer taxonomic relationship compared to the other options. Structural homology refers to similarities in anatomical structures due to shared evolutionary ancestry.
Since organisms within the same class share more recent common ancestors than those in different classes, A and B are more likely to exhibit greater structural homology in terms of anatomical features compared to the other pairs listed.
What is the genotype of the parent with orange eyes and white skin? (Note: orange eyes are recessive.)B=black eyes G=green skinb=orange eyes g=white skina. BbGgb. bbGGc. bbGgd. bbgge BBGG
Answer:
D
Explanation:
Two genes are involved in this case: one coding for eye color and the other for skin color.
According to the question, the allele for Black eyes (B) is dominant over the allele for orange eyes (b) while on the other gene, the allele for green skin (G) is dominant over the allele for white skin (g).
This means both alleles for orange eyes (b) and white skin (g) are recessive i.e masked by their counterpart allele
Since they are recessive, they will only be phenotypically expressed if they are in their homozygous state. i.e same type of allele.
In this case, the parent has orange eyes and white skin i.e.both recessive traits
Therefore, the genotype will be bbgg.
The genotype of the parent with orange eyes and white skin is bbgg.
Explanation:The genotype of the parent with orange eyes and white skin is bbgg. In genetics, lowercase letters represent recessive alleles while uppercase letters represent dominant alleles. Since orange eyes are recessive, the parent must have the genotype bb for eye color. Similarly, since white skin is recessive, the parent must have the genotype gg for skin color.
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Cerebrospinal fluid circulates within the ventricles of the brain and in the subarachnoid space. true or false
Answer:
True
Explanation:
CSF is produced by choroid plexus. It starts flowing from the lateral ventricles and move towards the fourth ventricle via the third ventricle. It then moves towards the subarachnoid space which is the region surrounding the brain and spinal cord. From the brain and spinal cord, CSF is absorbed through the blood vessels and passed into the bloodstream again. It then moves to the kidney and liver for filtration.
Hence, give statement is true
Yes, it's true that Cerebrospinal fluid circulates within the ventricles of the brain and in the subarachnoid space. These areas are part of the central nervous system and the fluid provides crucial protection.
Explanation:The statement provided is true. Cerebrospinal fluid, also known simply as CSF, indeed circulates within the ventricles of the brain and in the subarachnoid space. These are two key areas in the central nervous system. The ventricles of the brain are a connected network of cavities filled with CSF. The subarachnoid space, on the other hand, is the area between the arachnoid mater and the pia mater, also filled with CSF, which serves to provide protection to the brain and spinal cord from injury.
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The hypothalamus is the area where afferent impulses from all senses and all parts of the body are sorted out and then relayed to the appropriate area of the sensory cortex.A.TrueB. False
Answer:
False
Explanation:
Hypothalamus is responsible for performing vital functions such as regulating the release of hormones and body temperature, controlling the appetite and sexual behaviour and also manages the emotional responses and thus is an essential part of the brain.
Structurally it has three main parts and all the three parts have individual nuclei whose major function is to release hormone which then interact with pituitary gland and modulate release of further more hormones. It has no role in receiving the impulses from different parts of the body and then sending it to relevant section of the sensory cortex
Hence, the given statement is false
how do the nervous system and endocrine system work together
Answer:
the endocrine system and the nervous system are the two major regulating and communicating systems of the body. During rest and exercise, the nervous system along with endocrine system regulate the body mechanism to initiate the maintenance of Homeostasis in the body.
both endocrine and nervous system are both together involved in the transmission of chemicals signals which are important for physiological processes in the body . The difference is that endocrine system uses Hormones while nervous system uses neurotransmitters for chemical signalling.
The main area of the brain where these system work together and perform their function collectively is Hypothalamus where they signal each other when they receive proper stimulus.
Answer:
Nerves tell glands when to release chemicals.
Explanation:
A P E X
Coexisting species of wild cats differ in the size of their canine teeth, which corresponds to differences in their preferred species of prey. This outcome is most likely the result of: Competitive exclusion Resource partitioning Ecological release Preemptive competition
Answer:
The correct answer is- Resource partitioning
Explanation:
According to the competitive exclusion principle, two species can not share the same niche because species that share the same niche have the same needs which leads to the interspecies competition.
This competition leads to resource partitioning which means the species who share the same niche evolved by natural selection to occupy different niche by dividing their resource which leads to the coexistence of two different species.
So here coexistence of wild cats differ in the size of their canine teeth is the result of resource partitioning.
Micah visits his doctor complaining of a frontal headache and pressure over his cheekbones and eyes. He is congested and has a nasal discharge. Micah's voice has an odd nasal sound. What is the doctors diagnosis of Micah's condition?
a) tonsillitis
b) rhinitis
c) sinusitis
d) pleurisy
Answer:
The answer would likely be C as it seems that his nasal passage is mostly affected. Based on diagrams of sinusitis, the cheekbones and eyes bones areas are likely to be impacted by the nasal blockage.
Explanation:
Final answer:
Micah's doctor would likely diagnose his condition as sinusitis based on the presence of a frontal headache, pressure over the cheekbones, nasal congestion, and a nasal discharge, which indicates inflammation of the sinuses.
Explanation:
Symptoms and Diagnosis of Micah's Condition
Based on the symptoms described, Micah's doctor would likely diagnose his condition as sinusitis. Sinusitis is an inflammation of the sinuses, marked by headaches, pressure over the cheekbones, nasal congestion, and a nasal discharge. The presence of an odd nasal sound in Micah's voice further supports this diagnosis as it indicates that his sinuses, which are connected to the nasal passages, are affected.
Rhinitis, which is inflammation of the nasal cavity, often accompanies sinusitis but is not characterized by the same type of headache and pressure over the cheekbones. This combination of symptoms, with the addition of sinus involvement, is more aligned with rhinosinusitis. While tonsillitis and pleurisy are conditions that also affect parts of the respiratory system, their symptoms do not match Micah's presentation.
Bacterial rhinosinusitis, which is an infection and inflammation of the paranasal sinuses usually occurring after a viral infection, is commonly caused by pathogens such as Streptococcus pneumoniae, Haemophilus influenzae, and Moraxella catarrhalis.
What type of microscope would be best for studying the structures found inside of cells?
A Transmission Electron Microscope (TEM) is best for studying the structures found inside of cells because it offers high resolution. A Scanning Electron Microscope (SEM) is better for studying the surface of cells.
Explanation:The type of microscope best suitable for studying the structures found inside of cells is a Transmission Electron Microscope (TEM). This is because a TEM uses a high-energy electron beam to illuminate the sample, resulting in much higher resolution than light microscopes. With a resolution of less than 0.5 nanometer, the TEM allows for the visualization of structures inside the cell such as organelles, protein complexes and even individual molecules to be viewed in detail.
A Scanning Electron Microscope (SEM) can also be used, but it is best for studying the surface of cells rather than internal structures. For example, SEM can be used to study the texture and shape of the cell membrane.
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While observing a new client perform the pushing assessment, the personal trainer notices that his head protrudes forward. Which of the following groups of muscles are probably overactive?
A. middle trapezius
B. erector spinae
C. rhomboids
D. levator scapulae
Answer:
The correct answer is D. levator scapulae.
Explanation:
The levator scapulae is the muscle involved in the scapula lifting, as it's located on the sides and back of the neck. As the person's head is protruding forward it's highly possible that the muscle involved is the levator scapulae.
In cocker spaniels, black coat color (B) is dominant over red (b), and solid color (S) is dominant over spotted (s).
(a) If a red solid male was crossed with a black solid female to produce a red spotted puppy, the genotypes of the parent (with male genotype first) would be?
(b) If red spotted male was crossed with a black solid female and all the offspring from several crosses, expressed only the dominant traits, the genotype of the female would be?
Answer:
a) male : bbSs , female : BbSs
b) BBSS
Explanation:
a) Male is red and solid so it can either have bbSS genotype or bbSs genotype. The puppy is spotted though (ss) which means it got one recessive s allele from male parent. Hence, male's genotype will be bbSs.
Female is black and solid. Both of these are dominant traits so she can either be homozygous for both, heterozygous for one or heterozygous for both. Since the puppy is red and spotted (bbss) it got recessive alleles b and s for both the genes from female parent too. Hence, female has genotype BbSs.
b) Black solid female can either be homozygous for both traits, heterozygous for one or heterozygous for both. She mated with a red spotted male (bbss). All the offspring will express only dominant traits when the female has dominant homozygous genotype for both traits.
BBSS X bbss :
BS
bs BbSs
As evident, resultant progeny has BbSs genotype which means all of them will be only black and solid. Hence, genotype of female would be BBSS.
Final answer:
The genotypes for the red solid male and black solid female parents are bbS- and Bb(S)(S) respectively, where the female could be SS or Ss for the solid trait. For subsequent crosses where a red spotted male (bbss) produces offspring with a black solid female showing only dominant traits, the female must be homozygous dominant, with a genotype of BBSS.
Explanation:
Genotypes of Parent Cocker Spaniels
To determine the genotypes of the parent cocker spaniels, we must consider the inheritance of coat color and patterning. Since black coat color (B) is dominant over red (b), and solid color (S) is dominant over spotted (s), the fact that they produced a red spotted puppy means certain things about their genotypes:
The red solid male must be homozygous recessive for coat color (bb) because he expresses the red phenotype. For his solid patterning, he must have at least one solid allele (S-).The black solid female must have one copy of the red allele (Bb) since she was able to pass it to the red spotted offspring. Being solid, she could either be homozygous (SS) or heterozygous for the solid trait (Ss).Therefore, the male's genotype must be bbS-, and the female's could be BbSS or BbSs.Genotype of Female for Subsequent Crosses
Concerning the second cross: A red spotted male (genotype bbss) when crossed with a black solid female resulted in all offspring expressing only the dominant traits. This means that the female must be homozygous for both dominant traits to avoid the expression of any recessive characteristics in any of the offspring:
Therefore, the black solid female would have to be BBSS in genotype.This is an organism that has two different alleles for the same genetic trait.
Answer:
Heterozygous
Explanation:
An organism is said to be heterozygous when he has two different alleles of the same trait. A homozygous individual have the two same alleles for a given trait. For example, height is a trait which has two alleles that are T and t.
So in heterozygous individuals, two different alleles will be present which is Tt and in homozygous individual two same allele of height will be present that can be TT or tt.
In heterozygous condition normally one allele will be dominant and another allele will be recessive and in homozygous condition normally both the allele will be dominant or recessive. Therefore heterozygous is the organism that has two different alleles for the same trait.
The organism that has two different alleles for the same genetic trait is called a heterozygous organism.
The organism that has two different alleles
In genetics, alleles are alternative forms of a gene that occupy the same position (locus) on a pair of chromosomes. When an organism has two different alleles for a particular trait, it means that each of the two copies of the gene it possesses is different.
This could result in the expression of a dominant allele over a recessive allele, or in the case of co-dominance or incomplete dominance, both alleles may contribute to the phenotype of the organism.
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Modern humans have a number of anatomical characteristics that distinguish them from archaic humans. Drag only the correct modern human characteristics to the modern human skull. You will not use all the labels.
Answer:
Modern human are called as Homo sapiens.
Explanation: Following are the characteristics of the modern human skull-
i) They have projecting nose bone and comparatively small face.
ii) Eye sockets are square in shape.
iii) the neck muscles are reduced.
iv)The skull is round at the back.
v) Their cranial capacity is 1350 cc. Earlier they had a cranial capacity of 1500 cc.
vi) There has no narrow constriction behind the orbits.
Which protective covering of the brain provides passageways for cerebrospinal fluid (CSF) to drain into the superior sagittal sinus? a) arachnoid mater b) periosteal part of the dura mater c) membranous part of the dura mater d) pia mater.
Answer:
arachnoid mater
Explanation:
CSF is gradually reabsorbed into the blood at the same rate at which it is formed to maintain the constant pressure. The arachnoid mater has finger-like extensions that project into the dural venous sinuses, especially the superior sagittal sinus. The projections of arachnoid mater are called arachnoid villi and a group of arachnoid villi is called an arachnoid granulation. Arachnoid villi are the structures through which cerebrospinal fluid is drained into the superior sagittal sinus and is reabsorbed into the bloodstream.
Answer:
The answer is A
Explanation:
Arachnoid mater
In bacterial cells, the replication of DNA begins at a specific site called the ____ of replication, and proceeds in both direction around the circular DNA until a specific _______ site is reached
In bacteria, replication of DNA starts at a specific site called the origin of replication (oriC) and proceeds in both directions in the circular DNA until a specific termination site is reached.
What is DNA replication?DNA replication is the biological process of producing two identical replicas of DNA from single original DNA molecule. This process of DNA replication requires a machinery which is assembled on the original DNA molecule.
In bacteria, the DNA replication machinery is assembled at the single origin of replication site in a characteristic location. The origin of replication (oriC) is a particular sequence in a genome at which replication is initiated in bacteria.
Bacterial chromosomes have a single, unique replication oriC sequences.
DNA replication in the circular DNA of bacteria proceeds until a specific site called termination site is reached. The termination is of two types: Intrinsic termination and Rho-dependent termination.
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The following are directly needed for protein synthesis EXCEPT rRNA.a.peptidyl tRNA.b.mRNA.c.amino acyl tRNA.d.DNA
Answer:
The correct answer will be option-D
Explanation:
The process of protein synthesis from DNA takes place through a process known as the translation which translates the mRNA to the amino acid sequence. The process of translation involves three different types of RNA mainly mRNA, rRNA and tRNA.
The rRNA forms the structural component of the ribosomes which acts as a site for the translation. The mRNA is read by the ribosomal subunits which utilize the aminoacyl tRNA used to attach the amino acids corresponding to the codons. The peptidyl tRNA interferes with the process of the protein elongation.
Thus, option-D is the correct answer.
Response to a stimulus, such that the effect of the stimulus is counteracted, is called a ________
Answer:
Negative Feedback
Explanation:
The balancing feedback also called as negative feedback. The negative feedback regulates by stimulus that decreases the function. This mechanism balances the release of substance. The examples of negative feedback are thermoregulation and blood sugar regulation. The excess glucose is decreased by insulin. The changes in the body temperature are balanced and normal temperature is maintained. Thus, Response to a stimulus, such that the effect of the stimulus is counteracted, is called a Negative Feedback.