Erythrocytes have no nuclei. Without nuclei, erythrocytes can carry a higher amount of haemoglobin explain how this is an advantage in exchanging and carrying oxygen
Answer:
The correct answer is - hemoglobin binds with respiratory gases such as oxygen so more hemoglobin will carry more oxygen .
Explanation:
Red blood cells have adapted the trait of having no nucleus for several reasons but the major reason is to increase the amount of hemoglobin so this can carry more oxygen to various parts of the body from lungs. If there will be more hemoglobin then it will allows RBC to transfer more oxygen.
Hemoglobin is the respiratory pigment that has four heme group, containing iron ion, arround globin group, each iron ion in the heme can bind to one oxygen molecule; So one molecule of hemoglobin can transport four oxygen molecules as it binds with oxygen.
You prepare a gel mobility assay with the following samples:
Lane 1 Radiolabeled lacO DNA + Lac Repressor protein preincubated with allolactose
Lane 2 Radiolabeled lacO DNA + Lac Repressor protein preincubated with lactose
Lane 3 Radiolabeled lacO DNA + Lac Repressor protein missing its DNA binding domain
Lane 4 Radiolabeled lacO DNA + Lac super-repressor (Is) preincubated with allolactose
Lane 5 Radiolabeled lacO DNA + lac repressor protein
On which lanes do you expect to see two bands?
a. 3
b. Lane 1 and 2
c. Lane 2, 4, and 5
d. 2 and 4
e. 1 and 5
Answer:
c. Lane 2, 4 and 5.
Explanation:
Two bands will be seen when a fraction of DNA added to gel is bound by Lac repressor protein. Radiolabeled LacO DNA will have two bands. Lac repressor is encoded with lacl gene. It has binding ability due to allolactose formation.
Match the following description with the appropriate type of respiration:
a. occurs in the mitochondria
b. resting muscles depend on this type of respiration
c. lactic acid builds up in muscle fibers
d. rapidly produces ATP for short time periods
e. produces large quantities of ATP but takes longer to synthesize
1. anaerobic respiration
2. aerobic respiration
Answer:
The correct answer is -
1. c, and d.
2. a, b, and e.
Explanation:
Anaerobic respiration is the short and quick way of producing energy, however, it produces less amount of energy than aerobic respiration and produces lactic acid as a byproduct. It takes place in the cytoplasm of the cell and working or acting muscles depend on such respiration.
Aerobic respiration is the main and important respiration process that takes place in mitochondria and it takes time to produce ATPs that are much more than anaerobic respiration. Resting muscles get energy by this type of respiration.
Which statement describes her hypothesis
Answer:
A hypothesis is a plausible answer to a scientific problem or question that is testable, but not yet tested. Some science textbooks call it an "educated guess.
From a chemical view, how is an amino acid is being recognized by its specific aminoacyl tRNA synthetase?
Answer:
An aminoacyl-tRNA synthetase (aaRS or ARS), also called tRNA-ligase, is an enzyme that attaches the appropriate amino acid onto its corresponding tRNA. It does so by catalyzing the transesterification of a specific cognate amino acid or its precursor to one of all its compatible cognate tRNAs to form an aminoacyl-tRNA.
characteristics between invertebrates and vertebrates in a tabular form.
Answer:
Invertebrates: Do not possess a backbone and doesn't have an internal skeleton. Has an exoskeleton. Are generally smaller than vertebrates. Possess an open circulatory system. The majority have compound eyes. Includes radial or bilateral body symmetry. Presence of a simple and unorganized nervous system. Mode of nutrition includes Autotrophic, Parasitic, and Heterotrophic. 95% of animal species are invertebrates. Flatworms, arthropods, sponges, insects are few examples of Invertebrates.
Vertebrates: Possess a backbone and has an internal skeleton. Does not have an endoskeleton. Larger than invertebrates. Possess a closed circulatory system. Do not have compound eyes. Has bilateral body symmetry. Presence of complex and highly specialized organ systems with specific functions. The mode of nutrition is usually heterotrophic. 5% of animal species are vertebrates. Mammals, fish, reptiles, amphibians, and birds are examples of Vertebrates.
In another experiment, the leu gene was found to be located 5 minutes from the pro gene. In which location(s) in the gene map could the leu gene be located? (Note that the distances depicted between the genes on the map are not to scale.)
Answer: Hello your question has some missing data attached below is the missing data
answer:
positions : A and E
Explanation:
Given that the leu gene was found 5 minutes from pro⁺ gene hence it can be located either after the bio⁺ gene or before met⁺ genes
hence the locations in the gene map = A and E
attached below is the transfer of genes
g The greater tubercle of the humerus is analog to the _______________. Group of answer choices lesser tubercle of the humerus lesser trochanter of the humerus greater trochanter of the femur lesser trochanter of the femur
Answer:
greater trochanter of the femur
Explanation:
Analogous structures refer to those that are similar morphologically and even have a similar function in the organism, but their genetic origin differs.
The origin and embryological development of these organs are not the same.
Both structures are muscles insertion points or surfaces. These muscles are related to the extremity movement.
8.the following one is considered as safety rule in laboratory
1 point
never eat or drink anything in the science lab
always clean up when you have finished the experiment
wear safety goggles while doing experiments
all the above
In eukaryotes, miR-5 is a microRNA with sequence complementarity to the 3’UTR of the mRNA transcript encoded by the UCD gene. The UCD gene encodes for a protein that acts as an allosteric activator of the protein p101. The p101 protein is a DNA methyltransferase (DNMT) which is an enzyme that methylates DNA when the protein is in its active form.
A) In a lung cancer cell line, the p101 protein is overly active leading to misregulation of gene expression. In these lung cancer cells which of the following would be a correct prediction in regards to miR-5? [choose ALL that apply, note that incorrect selections will deduct points]
A) The gene encoding miR-5 is in a region of chromatin with hypoacetylation
B) The gene encoding miR-5 is in a region of chromatin with hyperacetylation
C) The gene encoding miR-5 in a region of euchromatin
D) The gene encoding miR-5 in a region of heterochromatin
E) The miR-5 transcript is translated at high levels
F) The miR-5 transcript is activating translation of the UCD gene
B) A different lung cancer cell line is found to have significantly decreased amounts of DNA methylation. Which of the following could explain why there is decreased DNA methylation in this cell line? [choose ALL that apply, note that incorrect selections will deduct points]
A) increased amounts of the p101 protein due to low levels of miR-5
B) increased amounts of the UCD protein due to high levels of miR-5
C) decreased amounts of the UCD protein due to low levels of miR-5
D) the absence of the UCD protein due to miR-5 inhibiting translation of the UCD mRNA transcript
E) the absence of the p101 protein due to miR-5 inhibiting transcription of the p101 mRNA transcript
Answer:
A) A) The gene encoding miR-5 is in a region of chromatin with hypoacetylation.
D) The gene encoding miR-5 in a region of heterochromatin.
B) D) the absence of the UCD protein due to miR-5 inhibiting translation of the UCD mRNA transcript
Explanation:
MicroRNAs (miRNAs) are small regulatory RNA sequences (approximately 20-24 nucleotides in length) that regulate gene expression by RNA interference (RNAi) mechanisms. These sequences (miRNAs) bind by base complementarity to messenger RNAs and thus inhibit protein translation and/or trigger mRNA degradation. In this case, miR-5 binds to the 3’UTR of the mRNA transcript of the UCD gene, thereby inhibiting/slowing protein UCD synthesis. The UCD protein is an allosteric regulator that binds and activates the expression of the p101 protein, thereby the miR-5 RNAi pathway also indirectly decreases the expression of the p101 gene. Moreover, hypoacetylation is an epigenetic mark generally associated with gene silencing (heterochromatin is a transcriptionally inactive state of chromatin).
Two organisms, AABBCCDDEE and aabbccddee, are mated to produce an F1 that is self-fertilized. If all of these genes are on the same chromosome, and there is no recombination, how many different genotypes will occur in the F2
Answer:
2 genotypes: AABBCCDDEE; aabbccddee
Explanation:
The law of segregation (also called Mendel's First Law) describes how genes on different chromosomes segregate into gametes independently of each other. Moreover, recombination (also known as crossing over) is the interchange of genetic material between non-sister chromatids during meiosis. In this case, the five genes are on the same chromosome (i.e., they are linked together and do not segregate) and there is no recombination, thereby they are inherited as a haplotype block to the next generation:
- Genotypes parental (P) lines: AABBCCDDEE; aabbccddee
- Cross P1: ABCDE X abcde
- Genotype F1: AaBbCcDdEe (100%)
- Gametes F1: ABCDE; abcde
- Genotypes F2: AABBCCDDEE; aabbccddee (ratio 1:1)
in some plants pink flowers are dominant over yellow flowers if Mario crosses two hybrids what happens
What role does mutation play in the evolution of a species and the development of new characteristics?
Answer:
Mutation plays a vital part in evolution; it is the greatest source of genetic variation. In evolution, it creates a new DNA sequence for a particular gene, creating a new allele. Additionally, recombination can produce a DNA sequence (a new allele) for a specific gene through intragenic recombination.
Explanation:
Hope this helps!
click on the link below to see the primate family tree diagram. Which of the following statements is true?
Answer:
biology is amazing it contains some reproductive topics
Orangutans evolved from an ancestor they share with gorillas is true regarding primate family tree diagram. This means that orangutans and gorillas have a common ancestor. Option C is the correct answer.
Evolution is the process through which species change over time. It occurs through the accumulation of genetic variations and natural selection. The primate family tree shows the evolutionary relationships between different primate species, including orangutans and gorillas. Option C is the correct answer.
Orangutans and gorillas share a common ancestor in their evolutionary history. This means that at some point in the past, there was a species that gave rise to both orangutans and gorillas. Over time, as the common ancestor of orangutans and gorillas diverged into separate lineages, each species developed its own unique traits and characteristics. Despite their distinct characteristics, orangutans and gorillas still share certain genetic similarities due to their common ancestry. These similarities can be observed in their DNA sequences.
Learn more about Primate here:
https://brainly.com/question/29214758
#SPJ2
The complete question is, "Which of the following statements is true regarding primate family tree diagram?
A. Orangutans evolved from lesser apes and Old World monkeys.
B. Gorillas evolved from orangutans.
C. Orangutans evolved from an ancestor they share with gorillas.
D. Gorillas are more closely related to lesser apes than to chimpanzees."
Which of the following is an example of an enzymatic cycle?
Answer:
Catabolism
Explanation:
The process of catabolism degrades the bacterial and fungal enzymes into simple inorganic molecules.
A substance, without being a reactant, which speeds up a chemical process is referred to as a catalyst. Enzymes are known as catalysts for biological reactions in living organisms. Although ribonucleic acid (RNA) molecules behave as enzymes, they are usual proteins. Enzymes.
Enzymes carry out the essential role of reducing the activated energy of a reaction — that is, the amount of energy needed to start the process. Enzymes work by attaching and retaining reactant molecules so that the chemical bonding and bonding activities are carried out more easily.
Read the information below then answer the questions which follow:
To grow, plants require water. They cannot get this water unless it is available in the soil. Plants
obtain water from the soil through their roots. It then passes up the stem and to the leaves and
flowers. The plant does not take all the water available in the soil. Much of the remaining water
evaporates into the surrounding air.
In an experiment, a stem which contained several flowers were placed in a beaker of water
containing red ink.
QUESTION 1
(1 mark)
What is the purpose of the experiment?
Answer:
I believe the experiment was too see if the flowers that would grow from the stem would turn out to be red in color. (Which is the reason they put red ink.)
(hope this helped :P)
which primitive organic molecule was essential to form lipid bilayer?
Answer:
autocatalytic RNA is the primitive organic molecules was essential form liquid bilayer .
If you measured the energy content of organisms in each trophic level in an ecosystem, which would have the greatest total energy content
Protein channels allow
molecules in and out of
the cell without using
energy. Therefore, they
are an example of which
type of transport?
Answer:
I think that it is an example of a passive transport
Explanation:
The locus B on the X chromosome of a malaria-carrying mosquito shows a 49% recombination rate with respect to the locus M. Since a recombination rate of 50% is essentially indistinguishable from independent assortment, you might be tempted to look for a locus that falls between B and M. Before you decide to do all that work, you run a chi square test to determine the P value of your experiment.
Required:
What range of P values would tell you that you should accept the conclusion that locus B and locus M are, indeed, 49mu apart and that another locus is not necessary?
Answer:
The correct answer is - P = 0.45
Explanation:
The Chi-square test is a test that determines the assorted genes independently. Here p-values used to make conclusions in significance testing.
If the p value is less than the significance level we expect or choose, then this null hypothesis is rejected in favor of the alternative hypothesis. It p value is greater than or equal to the significance level, then we fail to reject the null hypothesis.
So in this case, as the recombination rate is 49% so the p value which satisfies the conclusion will be around 0.45 and values near to it.
If you were rolling dice in a casino and observed results comparable to those observed in Part 2, would you assume the dice were faulty or rigged? Explain your answer using data from your chi-square analysis.
Answer:
Yes, I would guess that the dice are faulty or rigged.
Explanation:
Yes, I would guess that the dice are faulty or rigged.
This is because the chi-square analysis indicates that there is a less than 5% probability that the difference between the expected and observed results were due to pure chance.
In Drosophila melanogaster the recessive alleles for brown and scarlet eyes (of two independent genes) produce a novel phenotype so that bw/bw;st/st is white. If a pure-breeding brown is crossed to a pure-breeding scarlet, what proportion of the F2 will be white
Kevin's supervisor, Jill, has asked for an update on today's sales, Jill is pretty busy moving back and forth between different store locations. How can Kevin most effectively deliver an update to her ? a) Call with a quick update Ob ) Send a detailed text message c ) Book a one-hour meeting for tomorrow morning d) Send a detailed email
Answer:
d) Send a detailed email
Explanation:
Send a detailed email is the best way to deliver an update to her because in the email he can send all the information in detail form which can satisfy his owner. He can't call or message because it takes too much time to provide information so email is the best way to provide information. Booking a one-hour meeting is not worth it and the reason for this is that there is no big presentation which takes one hour of description one email is enough for it.
is the citric acid cycle the same thing as fermentation
Answer:basically
Explanation:
Alizarin yellow is a pH indicator that transitions from red to yellow when the pH falls from a value of 11 to below 10. Why is phenol red a better pH indicator than alizarin yellow for detecting a change in pH of broth containing pathogenic bacteria such as E. coli
Answer:
The correct answer is - acidic conditions wouldn't trigger a change in the color of Alizarin yellow.
Explanation:
The growth of E. coli generally occurs at neutral pH, however, its growth is normal at acidic conditions as well. The change in the growth of E. coli is not able to detect by alizarin.
The phenol red turns yellow in the presence of an acid, and the change in pH in an alkaline environment can be detected by the red color of phenol red. Growth of E.coli will grow in pH of 10-12 . But, very slowly. The color change in alizarin is also apparent at pH 10.2 to 12 only.
How can algal blooms be
harmful to an aquatic
environment?
A. by blocking all sunlight and killing the bottom plants
causing no oxygen to go into the water
B. by putting oxygen up into the air for the
surrounding plants
C. by absorbing the phosphorus in the atmosphere and
having a symbiotic relationship with cyanobacteria
Answer:
(A) By blocking all sunlight and killing the bottom plants
causing no oxygen to go into the water
EVOLUTION CONNECTION The history of life has been punctuated by several mass extinctions. For example, the impact of a meteorite may have wiped out most of the dinosaurs and many forms of marine life at the end of the Cretaceous period. Fossils indicate that plants were less severely affected by this mass extinction. What adaptations may have enabled plants to withstand this disaster better than animals
Answer:
Dissaster like meteorite impacted all type of living forms, however, animals are impacted the most and plants are impacted least. It is possible due to ability of plants to prepare their food on their own as they are autotrophic and use photosynthesis. The raw material for preparing these are easily available that are sunlight, CO2 and water.
They are not depende on other for their cellular respiration. The other major reason is that the Plant seeds can remain dormant in unfavorable conditions for many years in the soil. The roots always provide nutrition even the stems get effected.
Identify the energy carrier molecule ATP and its importance
Which of the following describes a predator?
A a fish that is killed and eaten.
B a bear that kills and eats fish
C a worm that lives inside a bear
D a bear that has a worms in its gut.
Answer:
b
Explanation:
a predator is a bear that kills and eats fish
Answer:
B. bear that kills and eat fish
Which of the following statements is false? Choose one: A. Some Bcl2 family members promote apoptosis, whereas others inhibit apoptosis. B. Bax and Bak are death-promoting members of the Bcl2 family that induce the release of cytochrome c from mitochondria into the cytosol. C. Some death-inhibiting members of the Bcl2 family inhibit apoptosis by blocking cytochrome c release from mitochondria. D. The death-promoting members of the Bcl2 family include Bcl2 itself.
Answer: The statement that is FALSE is D (The death-promoting members of the Bcl2 family include Bcl2 itself.)
Explanation:
Apoptosis is a process found in multicellular organisms whereby cells follows a order of events which eventually leads to their death without releasing harmful substances into the surrounding area of the cells. Through this process, old cells, unnecessary cells, and unhealthy cells are eliminated from the body.
The process of apoptosis can be regulated by cell regulator proteins such as Bcl-2 family.
Bcl-2 stands for B-cell lymphoma 2 which are encoded in human cells by BCL2 gene. The Bcl-2 family are capable of regulating apoptosis by either inducing it or inhibiting it.
Bcl-2 and Bcl-XL, PREVENT apoptosis by preventing the release of mitochondrial apoptogenic factors such as cytochrome c and AIF (apoptosis-inducing factor) into the cytoplasm
Part of the Bcl-2 family is the Bax and Bak and they are the core regulators of the intrinsic pathway of apoptosis. They PROMOTE apoptosis by inducing the release of cytochrome c from mitochondria into the cytosol. Therefore, the death-promoting members of the Bcl2 family does not include Bcl2 itself.
