During drought years, seeds with thin coats are more likely to desiccate (dry out) and not be able to germinate
when wet conditions return. If a drought persists for several years, what should one expect to result from natural
selection? (application level)
O Plants that produce seeds with thicker seed coats will be able to start making seeds with thinner coats, thus saving energy and
nutrients for future droughts.
O Fewer plants with thin seed coats will be able to germinate, leading to a higher proportion of plants that produce seeds with
thick seed coats.
Plants that produce seeds with thinner seed coats will anticipate the continuation of drought, and will start to produce seeds
with thicker seed coats.
O Drought conditions will mutate the seed coat genes with the result that later-generation offspring have thicker seed coats.

75% of the offspring would be Rh positive (RR or Rr) and 25% would be Rh negative (rr).

When two heterozygous individuals with the dominant rh factor gene (r) reproduce, there is a 75% chance that their offspring will be rh positive (rr or Rr genotype) and a 25% chance that their offspring will be rh negative (rr genotype). This is because each parent has a 50% chance of passing on the dominant r allele, and if both parents pass it on, the offspring will be rh positive. However, if both parents pass on the recessive r allele, the offspring will be rh negative. It is important to note that being rh positive or rh negative has implications for blood transfusions and pregnancy, as an rh negative individual can have an adverse reaction to rh positive blood or the fetus of an rh positive mother if proper precautions are not taken.
Hi! I'd be happy to help with your question. In humans, the Rh factor is inherited as a dominant gene (R). Individuals with this allele are referred to as Rh positive. When two heterozygous individuals (Rr) decide to reproduce, the possible genotypes of their offspring would follow a Punnett square with the following distribution:

- RR (Rh positive): 25%
- Rr (Rh positive): 50%
- rr (Rh negative): 25%

So, 75% of the offspring would be Rh positive (RR or Rr) and 25% would be Rh negative (rr).

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EXPLANATION/ANSWER:

Since no trees are involved, it may be grassland. Grasslands are open areas with mostly grasses maintained by grazing animals like the impala seen in the photo.

Their sagittal crests distinguish primates that are adapted for eating large amounts of plants and leaves.

Many mammalian and reptilian skulls, among others, have a ridge of bone running longitudinally along the midline of the top of the skull (at the sagittal suture). There are incredibly powerful jaw muscles because of the existence of this bone ridge.

What kinds of animals have sagittal crests?

The top of the skull in carnivores and omnivores has a ridge known as the sagittal crest. The sagittal crest, which is where the powerful muscle that regulates the lower jaw's crushing action attaches, is highly noticeable in opossums and fishers but less so in coyotes, raccoons, and otters.

Therefore sagittal crests distinguish primates that are adapted for eating large amounts of plants and leaves

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The most beneficial action to stop the dead zones from reappearing in the Gulf of Mexico in the future would be to restore inland and coastal wetland areas that have been destroyed or degraded (option A).

Deаd zones аre аreаs of wаter where oxygen levels аre so low thаt mаrine life cаnnot survive. They аre cаused by аn excess of nutrients, such аs nitrogen аnd phosphorus, which come from аgriculturаl runoff аnd other sources. These nutrients stimulаte the growth of аlgаe, which then die аnd decompose, consuming oxygen in the process. By restoring wetlаnd аreаs, the excess nutrients cаn be аbsorbed, preventing them from reаching the Gulf of Mexico аnd contributing to the formаtion of deаd zones.

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A. After the statement quiz2 xVar, yVar executes in main, xVar contains 9 (decimal).

B. The statement quiz2 2, 6 will produce an error when the macro is invoked. (False)

C. Upon completion, EDX will contain 90 (decimal).

A. The quiz2 macro multiplies the value of pVar with itself using the MUL instruction and performs the multiplication for the number of times specified by qvar using the LOOP instruction. In this case, xVar is initially set to 3, so when it is passed as the pVar argument to quiz2 along with yVar (also 3) as the qvar argument, the macro will execute the multiplication loop 3 times. Therefore, xVar will be multiplied by itself three times, resulting in 9.

B. The quiz2 macro accepts two arguments, pVar and qvar, and does not impose any restrictions on the values that can be passed to these arguments. As long as the arguments are valid DWORD values, the macro will execute without producing an error. Therefore, the statement quiz2 2, 6 will not produce an error when the macro is invoked.

C. In the given code snippet, the value 10 is assigned to EBX, and the value 3 is assigned to EDX. Then the quiz2 macro is invoked with EBX as pVar and EDX as qvar. The macro multiplies EBX (10) with itself three times, resulting in 1000. Since EDX is not modified within the macro, it retains its initial value of 3. Therefore, upon completion, EDX will contain 3 (decimal).

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Answer:

Chlorophyll

Explanation:

stomata are used for water regulation

roots absorb nutrients and water

stems provide structure

Answer:Chlorophyll

Explanation:

Its A frequently

explantation: the higher the frequency wave  the higher the pitch of the sound

25% of their daughters are expected to have normal color vision, while the other 75% are expected to be carriers of the color blindness gene.

Red-green color blindness is a genetic condition caused by a mutation in a gene located on the X chromosome. The gene responsible for normal color vision is dominant, while the gene responsible for color blindness is recessive.

If two people with normal color vision have two sons, one color-blind and one with normal color vision, it is likely that the parents are carriers of the color blindness gene. This means that they each have one normal X chromosome and one X chromosome carrying the color blindness gene.

When the parents have children, they each pass on one X chromosome to their offspring. The probability of passing on the normal X chromosome is 50%, while the probability of passing on the X chromosome carrying the color blindness gene is also 50%.

For their daughters, the probability of having normal color vision is:

(normal X from mother) * (normal X from father) = 0.5 * 0.5 = 0.25.

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Answer:

STEM CELLS

They serve as a repair system for the body.

The labeling of the five words mRNA, mRNA codons, tRNA, tRNA anticodons, and amino acids in protein synthesis is found in the attachment.

The molecules that are involved in protein synthesis are as follows:

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Answer:

Microbes, salts, and pollution from agriculture and industry all contribute to the problem. Global warming will likely have major impacts on the world's freshwater resources.

Explanation:

Water source protection involves the protection of surface water sources and the protection of groundwater sources from contamination of any kind. Water source protection includes basic measures and rules such as the installation of a fence around the source or the banning of grazing animals in the surrounding area.

Ebola viruses make many copies of themselves by combining host and virally encoded enzymes with host cell architecture rather than by any type of cell division.

The virus completes a set of steps when infecting each individual cell.

The following are the steps during the replication of Ebola Viruses:

The several steps in the life cycle of the Ebola virus, such as host cell attachment and entrance utilizing glycoproteins, absorption by endocytosis, replication of the RNA genome by the viral polymerase enzyme, transcription and translation of the Ebola genes, and finally viral release.

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To determine the possible blood types of the offspring, we need to consider the genotypes of the parents and the inheritance patterns of the alleles.

The woman has type AB blood, which means she must have inherited one A allele (IA) from one parent and one B allele (IB) from the other. Since IA and IB are codominant, both alleles are expressed, resulting in the AB blood type. Therefore, her genotype can be written as IAIB.

The man has type B blood, so he must have inherited the B allele (IB) from both parents. Therefore, his genotype is IBIB.To determine the potential blood types of their offspring, we can combine the possible alleles from each parent:

For the woman (IAIB):

IA from the mother combines with IB from the father, resulting in IAIB (type AB).

IA from the mother combines with i from the father, resulting in IAi (type A).

For the man (IBIB):

IB from the father combines with i from the mother, resulting in IBi (type B).IB from the father combines with i from the mother, resulting in IBi (type B). So the potential blood types of the offspring are AB, A, B, and B.

The probabilities for each blood type depend on the likelihood of each genotype combination. Since we don't have any information about the parents' backgrounds or genotypes, we can assume that all genotypes are equally likely. Therefore, each blood type has a 25% chance of occurring in the offspring.

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Answer:

Prophase

Explanation:

The Prophase characterizes because the chromatin finishes the condensation process, which began during the final stage of the interphase.

Individual chromosomes are now visible to optic microscopy. Each chromosome is composed of two sister chromatids joined by a centromere.

During the prophase, the pares of centrioles separate, each of them migrating to opposite poles. The spindle fibers form. At this point, nucleoles have already disappeared. The nuclear membrane breaks and disappears too, and genetic material gets in contact with the cytoplasm.

Fermentation can only take place in an anaerobic environment because yeast is relatively anaerobic. When there is oxygen present, yeast engages in aerobic metabolism.

Yeast fermentation: When oxygen is present, yeast engages in aerobic respiration, converting carbohydrates (a source of sugar) into carbon dioxide and water. In the absence of oxygen, yeast ferments carbohydrates and produces carbon dioxide.

Through the process of electron transfer and oxygen, yeast mitochondria regenerate NAD+. Alcohol and CO2 are created during the fermentation process, which essentially only takes place in the absence of oxygen (resulting in carbonation). The roommate left the bucket open to the air, which provided the yeast with an ongoing supply of oxygen.

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The notation of this karyotype is 45, X, -23. It means that they have an aneuploidy.

Aneuploidy is a chromosome abnormality where an individual does not have the normal number of chromosomes.

In aneuploidy, chromosomes may be laking (missing) or may be present in extra amounts.

Aneuploidies can lead to differential phenotypes (for example, Down syndrome is caused by a trisomy) and often are associated with health problems.

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The role of a transposon in bacterial transduction involves facilitating the transfer of genetic material between bacteria.

Transposons, also known as "jumping genes," are DNA sequences capable of moving from one location to another within the genome. They play a significant role in bacterial evolution, helping bacteria adapt to their environment by promoting genetic diversity. In bacterial transduction, a bacteriophage (a virus that infects bacteria) transfers DNA between two bacterial cells. During this process, the transposon can "jump" into the phage's genome, ensuring its transfer from the donor bacterium to the recipient. Upon infection of the recipient bacterium, the phage injects the DNA, including the transposon, into the new host cell.

Once inside the recipient bacterium, the transposon can integrate into the bacterial genome, providing new genetic information that can enhance the bacterium's survival, this may include antibiotic resistance, metabolic capabilities, or virulence factors. Additionally, transposons can create mutations or rearrangements in the recipient's DNA, further promoting genetic variation and evolution. In summary, transposons play a crucial role in bacterial transduction by facilitating the movement of genetic material between bacterial cells, promoting genetic diversity, and contributing to bacterial adaptation and evolution.

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Answer:

d. there's nothing for Soundwave to bounce off of I believe

Both the blue sea and the dolphin are made up of atoms.

The term atom has to do with the smallest particle of a substance that could take part in a chemical reaction. We know that all substances in the universe are composed of atoms. Atoms are found in the food we eat, the clothes we wear and even in the air we breathe.

As such atoms are everywhere in the universe. Living things are also composed of different kinds of atoms in differing proportions. Looking at the image as we can see it, there is the blue sea and there is a dolphin. Both the blue sea and the dolphin are made up of atoms.

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Answer:

the process by which green plants and some other organisms use sunlight to synthesize foods from carbon dioxide and water. Photosynthesis in plants generally involves the green pigment chlorophyll and generates oxygen as a byproduct.

Explanation:

AChR type 1 has two binding sites and can be activated by acetylcholine and nicotine. Whereas AChR type 2 has only one binding site and can only be activated by acetylcholine.

Acetylcholine is one of the most commonly known types of neurotransmitters (chemicals that carry nerve stimulation). These neurotransmitter compounds can be found in the nervous system of vertebrate organisms. Acetylcholine plays a role in transmitting received signals or stimuli to be forwarded between adjacent nerve cells or at the neuromuscular junction

The acetylcholine receptor (AChR) is a membrane protein that binds to the neurotransmitter acetylcholine (Ach). AChR is one of the protein molecules in cell membranes in the central and peripheral nervous system that is stimulated by acetylcholine or acetylcholine-like substances.

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The mouse's DNA is entirely distinct from that of the other mice. When one DNA nucleotide in a gene's protein-coding region changes without changing the order of amino acids that make up the protein that the gene codes for, the change is called a silent mutation.

A mutation known as a silent mutation does not affect the amino acid and hence has no effect on how the protein functions. The replacement mutation in the fur color gene was a quiet mutation, therefore it had no effect on the final phenotype.

When a cell wants to synthesize a certain protein, transcription from DNA produces messenger RNA (mRNA).

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Alfred Wegener suggested the Lithospheric plates were broken into Sections that move Around the Mantle. These slabs react with each other in different ways. Some slide, Some converge, Both uplift, Others pull away from each other causing Mid-Oceanic Ridges upon Earth's surface.

Alfred Wegener was a German geophysicist and meteorologist who is best known for his theory of continental drift, which he first proposed in 1912. Wegener suggested that the continents were once joined together in a single landmass he called "Pangaea," which had subsequently broken apart and drifted to their current positions on the Earth's surface.

Although his theory was initially met with skepticism from the scientific community, it was eventually validated through the development of the theory of plate tectonics in the mid-20th century.

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Image transcribed:

Alfred Wegener Suggested the L______ plates were broken into S____ that move A________ the M____________. These slabs react with each other in different ways. S ________, S ________, B ________, O ________ and even M _________ away from each other causing M _______ C ______ upon Earth's surface.

Repairing a single-stranded break is generally more efficient than repairing a double-stranded break due to the nature of the DNA molecule.

DNA is typically double-stranded, meaning that it consists of two complementary strands of nucleotides that are joined together by hydrogen bonds. When a single-stranded break occurs, only one of these strands is affected, leaving the other strand intact to serve as a template for repair. This makes it easier for repair enzymes to accurately restore the damaged section of DNA. In contrast, when a double-stranded break occurs, both strands of the DNA molecule are affected, which makes repair more complex and error-prone. There are two main mechanisms for repairing double-stranded breaks: non-homologous end joining (NHEJ) and homologous recombination (HR). NHEJ is a quick but error-prone process that involves directly joining the broken ends of the DNA, often resulting in small deletions or insertions at the repair site. HR, on the other hand, uses a complementary template to repair the break, but it requires a second intact copy of the DNA molecule to be available for use as a template.

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Answer:

Endocytosis moves substances out of the cell and is a type of active Transport.