Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
1.1 Introduction
Cell Theory
1. Outline the 3 statements of cell theory
The three statements of cell theory are as following;
All living organisms are composed of cells
Cells are the smallest unit of life
Cells come from pre-existing cells
2. Deduce how scientific evidence is used to support the three statements of cell
theory
Scientific evidence is used to support the three statements by looking for a trend within
general ideas, such as statements made about the cell theory. Such trends were first
experimented by Robert Hooke in 1665, after examining cork and other plant parts.
After describing cells in the cork, he wrote:
Nor is this kind of texture peculiar to cork only, for upon examination
with my microscope, I have found that the pith of the Elder, or almost
any other tree, the inner pith of the Cany hollow stems of several
other vegetables: as of Fennel, Carrets, Daucus, Bur-docks, Teasles,
Fearn, some kind of Reeds, etc. have much such a kind of Schematisme
as I have lately shown that of cork.
This basically means that Hooke had looked at multiple different plant tissues and
discovered a general trend; since then, biologists have looked at tissues from many
different living organisms. Many of these tissues have been found to consist of cells–
hence, the cell theory cannot be discarded.
However, there are some discrepancies–organisms or parts of organisms that do not
consist of typical cells. Nevertheless, it is very unlikely that the given cell theory will be
discarded due to the fact that the majority of examined tissue follows the trend.
3. Outline some exceptions to cell theory
In order to have exceptions to the cell theory, you must ask the question “Does the
organism or tissue fit the trend stated in the cell theory by consisting of one or more
cells?”
Three well known atypical example (these must be known for the IB) are:
a. Striated Muscle
It is a type of issue that we use to change the position of our body; it is
made out of muscle fibres, which, in some ways, are similar to cells in
the fact that they are surrounded by a membrane, and are formed by
pre-existing cells. Along with that, they have their own genetic
material and their own unique energy release system.
However, the fibres are far from typical, as they are much larger
than most animal cells; in humans they have an average length of
30mm, whereas other human cells are mostly smaller than 0.03mm
in length. Along with that, instead of having a single nucleus, they
might have as many as several hundred within one “cell”.
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
b. Aseptate Fungal Hyphae
These consist of narrow thread-like structures called hyphae. These
are usually white in colour and have a fluffy appearance. They have a
cell membrane as well as a cell wall. In some types of fungi, the
hyphae are divided into small cell-like sections by cross walls called
septa. However, in aseptate fungi there are no such septa; each
hypha is an uninterrupted tube-like structure with many nuclei
spread along it (they are not compartmentalised).
c. Giant Algae
They are organisms that feed themselves by photosynthesis and store
their genes inside nuclei, but they are simpler in their structure and
organisation than plants. Many algae consist of one microscopic cell.
There are vast numbers of these unicellular algae in the oceans and
they form the basis of most marine food chains. Less common are
some algae that grow to a much larger size, yet they still seem to
be single cells. They are known as giant algae. Acetabularia is one
example; it can grow to a length as much as 100mm, despite only
having one nucleus. (Any organism with a length of 100mm is
expected to have many cells).
You do not need to know the specifics of all of these; the IB usually asks people to list the 3
atypical examples with a single reason why they’re atypical.
4. Evaluate the validity of cell theory, given that there are exceptions
Refer back to question 2.
NOS: Know that there are exceptions to most theories
Magnification
5. Calculate the magnification of drawings and the actual size of structures and
ultrastructures shown in drawings or micrographs
To find the magnification of a micrograph or a drawing we need to know two
things: the size of the image (in the drawing or micrograph) and the actual size of the
specimen. With that information, we can use the following equation:
𝑠𝑠𝑠𝑠 𝑠𝑠 𝑠𝑠𝑠𝑠𝑠
magnification = 𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠 𝑠𝑠 𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
When you use the formula make sure that the units for the size of the image and the
size of the specimen are the same; they could be in millimeters (mm), micrometers
(µm) but they must not be different or the calculations will be wrong. Millimeters can
be converted into micrometers by multiplying by 1000; micrometers can be converted
into millimeters by dividing by 1000.
6. Use scale bars to calculate the actual sizes in drawings and micrographs
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
Scale bars are sometimes put on micrographs or drawings, or just alongside them.
These are straight lines, with the actual size that the scale bar represents. For
example, if there was a 10mm long scla ebar on a micrograph with a magnification of
x10,000 the scale bar would have a label of 1µm.
Example:
The length of an image is 30mm. It represents a structure that has an actual size of
3µm. Determine the magnification of the image.
Either
30 mm = 30 𝑠 10 −3 𝑠
3 µm = 3 𝑠 10 −6 𝑠
Magnification =
30 𝑠 10 −3
3 𝑠 10 −6
= 10,000 x
Or
30 mm = 30,000 µm
Magnification =
30,000
3
= 10,000 x
Functions of life
7. State that all of the functions of life are carried out by all living organisms,
including unicellular organisms
The functions of life are things that all organisms must do to stay alive. Some
organisms consist solely of a single cell; this cell, therefore, has to carry out all
functions of life. Because of this the structure of unicellular organisms is more complex
than most cells in multicellular organisms.
Unicellular organisms carry out at least seven functions of life
a. Nutrition: obtaining food, to provide energy and the materials needed for growth
b. Metabolism: chemical reactions inside the cell, including cell respiration to
release energy
c. Growth: an irreversible increase in size
d. Response: the ability to react to changes in environment
e. Excretion: getting rid of the waste products in metabolism
f. Homeostasis: keeping conditions inside the organism within tolerable limits
g. Reproduction: producing offspring either sexually or asexually
8. Outline how Paramecium and Chlamydomonas perform each of the functions of
life
Paramecium is a unicellular organism that can be cultured quite easily in the
laboratory. Alternatively, collect some pond water and use a centrifuge to concentrate
the organism in it to see if Paramecium is present.
- The nucleus of the cell can divide to produce the extra nuclei that are needed
when the cell reproduces. Often the reproduction is asexual with the parent
cell dividing to form two daughter cells
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
-
-
-
-
-
Food vacuoles contain smaller organisms that the Paramecium has
consumed. These are gradually digested and the nutrients are absorbed into
the cytoplasm where they provide energy and materials needed for growth
The cell membrane controls what chemicals enter and leave. It allows the
entry of oxygen for respiration. Excretion happens simply by waste products
diffusing through the membrane
The contractile vacuoles at each end of the cell fill up with water and then expel
it through the plasma membrane of the cell, to keep the cell’s water content
within tolerable limits
Metabolic reactions take place in the cytoplasm, including the reactions that
release energy by respiration. Enzymes in the cytoplasm are the catalysts that
cause these reactions to happen
Beating of the cilia moves the Paramecium through the water and this can
be controlled by the cell so that it moves in a particular direction in response to
changes in the environment
Chlamydomonas is a unicellular algae that lives in soil and freshwater habitats. It has
been used widely for research into cell and molecular biology. Although it is green in
colour and carries out photosynthesis it is not a true plant and its cell wall is not made
out of cellulose.
- The nucleus of the cell can divide to produce genetically identical nuclei
for asexual reproduction. Nuclei can also fuse and divide to carry out
asexual form of reproduction
- Metabolic reactions take place in the cytoplasm, with enzymes present to
speed them up
- The cell wall is freely permeable and it is the membrane in side that controls
what chemicals enter and leave; oxygen is a waste product of
photosynthesis and is excreted by diffusing out through the membrane
- The contractile vacuoles at the base of the flagella fill up with water and then
expel it through the plasma membrane of the cell, to keep the cell’s water
content within tolerable limits
- Photosynthesis occurs inside the chloroplasts in the cytoplasm; carbon
dioxide can be converted into the compounds needed for growth here, but in
the dark, carbon compounds from other organisms are sometimes absorbed
through the cell membrane if they are available
- Beating of the two flagella moves the Chlamydomonas through the water.
A light sensitive “eyespot” allows the cell to sense where the brightest light is
and respond by swimming towards it
Surface area: volume ratio
9. State the relationship between size and surface area: volume ratio
The given relationship between the size and surface area:volume ratio is
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
As the cell gets bigger, the surface area:volume ratio decreases
10. Explain the importance of the surface area to volume ratio as a factor limiting
cell size
a. Heat Production vs Heat Loss
The surface area to volume ratio is very important in relation to heat production
and loss because if the ratio is too small, the cells may quickly overheat due to
the fact that the metabolism produces heat faster than it is lost over the cell’s
surface
b. Resource Consumption vs Excretion of Waste
It is also very important in relation to waste production and excretion because
if the ratio is too small, then substances will not enter the cell as quickly as they
are required and waste products will accumulate because they are produced
more rapidly than they can be excreted.
11. Recognize the role of root hairs, microvilli, and the biconcave shape of red blood
cells in increasing the surface area to volume ratio
Root hairs, microvilli, and the biconcave shape of red blood cells play a role in
increasing the surface area to volume ratio by increasing the surface areas of
the area that they play a role in. Root hairs are part of the plants’ root system,
and increase the overall surface area, increasing absorption. Microvilli do the
same in the small intestine, by lining themselves completely along the wall of
the intestine. Red blood cells, having a biconcave shape, increase their own
surface area:volume ratio by taking on that shape.
Stem cells
12. Outline the key properties of stem cells
Stem cells have two key properties that have made them one of the most active areas
of research in biology and medicine today:
- Stem cells can divide again and again to produce copious quantities of new
cells; they are therefore useful for the growth of tissues or the replacement of
cells that have been lost or damaged
- Stem cells are not fully differentiated; they can differentiate in different ways to
produce different cell types
13. Outline the usefulness of embryonic stem cells
Embryonic stem cells are potentially very useful. They could be used to produce
regenerated tissue, such as skin for people who have suffered burns. They could
provide a means of healing diseases such as type 1 diabetes, where a particular cell
type has been lost or is malfunctioning. They might also be used in the future to grow
whole replacement organs–hearts and kidneys for example. These types of uses are
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
called therapeutic, because they provide therapies for diseases or other health
problems
Some non-therapeutic uses could be to produce large quantities of striated muscle
fibres, or meat, for human consumption; the beef burgers of the future may therefore
be produced from stem cells, without the need to rear and slaughter cattle.
It is the early stage embryonic stem cells that are the most versatile. Gradually during
embryo development, the cells commit themselves to a pattern of differentiation; this
involves a series of points at which a cell decides whether to develop along one
pathway or another. Eventually, each cell becomes committed to develop into one
specific cell type; once committed, the cell may still be able to divide, but all of these
cells will differentiate in the same way and therefore will no longer remain stem cells.
14. Outline the limited repair of non-embryonic stem cells
Small numbers of cells remain as stem cells, however, and they are still present in the
adult body. They are present in many human tissues, including bone marrow, skin, and
liver. They give some human tissues considerable powers of regeneration and repair;
the stem cells in other tissues only allow limited repair–brain, kidney, and heart for
example.
15. Describe the therapeutic uses of stem cells
There are few current uses of stem cells to treat diseases, and a huge range of possible
future uses, many of which are actively being researched. Two examples are given;
one involving embryonic stem cells, and one using adult stem cells.
a. Stargardt’s Disease
Stargardt’s Disease, or also known as Stargardt’s Macular Dystrophy is a
genetic disease that develops in children between the ages of six and twelve.
Most cases are due to a recessive mutation of a gene called ABCA4. This
causes a membrane protein used for active transport in retina cells to
malfunction; as a consequence, photoreceptive cells in the retina degenerate.
These are the cells that detect light, so vision becomes progressively worse, to
the point where the patient could be registered as blind.
Researchers, however, have developed methods for making embryonic stem
cells develop into retina cells. This was initially done with mouse cells, which
were then injected into the eyes of the mice that had a condition synonymous
to Stargardt’s Disease; these cells were not rejected, nor did they develop into
tumours or cause any other problems. Encouragingly, they attached to the
retina and improved the mice’s vision.
In November of 2010, researchers in the US got permission to test it on humans
and were delighted by the results; the cells attached to the retina and remained
there, gradually improving the patient’s vision without inducing harmful side
effects.
b. Leukemia
This well-known disease is a type of cancer. As all cancers start, Leukemia
develops when mutations occur in specific genes that control cell division. This
is usually very unlikely, but due to the vast amounts of cells in the body, this
chance becomes much larger. More than a quarter of a million cases of
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
Leukemia are diagnosed each year globally, and there are over 200,000 annual
deaths from the disease.
Once the mutations have occurred, it grows and divides repeatedly, producing
more damaged cells. Leukemia involves the production of abnormally large
numbers of white blood cells; in other types of cancer, it usually forms a lump
or tumour, but remains unseen with Leukemia. White blood cells are produced
in the bone marrow and are then released into the blood, both in normal
conditions and when excessive numbers are produced with Leukemia. A
normal adult white blood cell is between 4,000 and 11,000 per mm cubed, but
in a person with Leukemia, this number can rise up to 30,000 - 100,000 (acute)
per mm cubed.
To cure Leukemia, the cancer cells in the bone marrow that produce the large
numbers of white blood cells must be destroyed. This can be done by treating
the patient with chemicals that kill dividing cells; chemotherapy. However, to
remain healthy, patients must be able to produce the white cells needed to fight
the disease; stem cells that can produce blood cells must be present, but they
are killed by chemotherapy. Therefore, the following procedure is used:
- A large needle is inserted into a large bone–usually the pelvis–and fluid
is removed from the bone marrow
- Stem cells are extracted from this fluid and are stored by freezing them.
They are adult stem cells and only have the potential for producing red
blood cells
- A high dose of chemotherapy drugs is given to the patient, to kill the
cancer cells in the bone marrow. The bone marrow loses its ability to
produce blood cells
- The stem cells are then returned to the patient’s body; they re-establish
themselves into the bone marrow, multiple, and start to produce red and
white blood cells
16. Outline the ethics of stem cell research, along with the sources of stem cells
The research into stem cells has been very controversial as many ethical objections
have been raised. Scientists should always consider the ethical implications of their
research before doing it; some of the research that was carried out in the past would
not be considered ethically acceptable today, such as medical research carried out on
patients without their informed consent. These decisions about whether research is
ethically acceptable or not must be based on a clear understanding of the science
involved; some people dismiss all stem cell research as unethical, but this shows a
misunderstanding of the different possible sources of the stem cells being used.
Stem cells can be obtained from a variety of sources
- Embryos can be deliberately created by fertilising eggs with sperm and allowing
the resulting zygote to develop for a few days until it has between 4 and 16
cells; all the cells are embryonic stem cells
- Blood can be extracted from the umbilical cord of a newborn baby and stem
cells obtained from it can be frozen and stored for possible use later in the
baby’s life.
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
-
Stem cells can be obtained from some adult tissues such as bone marrow
Embryonic Stem Cells
●
●
●
●
●
●
Almost unlimited
growth potential
Can differentiate into
any type in the body
More risk of
becoming tumour
cells than with adult
stem cells, including
teratomas that
contain different
tissue types
Less chance of
genetic damage due
to the accumulation
of mutations than
with adult stem cells
Likely to be
genetically different
from an adult patient
receiving the tissue
Removal of cells
from the embryo kills
it, unless only one or
two cells are taken
Cord Blood Stem Cells
●
●
●
●
●
●
Easily obtained and
stored
Commercial
collection and
storage services
already available
Fully compatible with
the tissues of the
adult that grows from
the baby, so no
rejection problems
occur
Limited capacity to
differentiate into
different cell types–
only naturally
develop into blood
cells, but research
may lead to
production of other
types
Limited quantities of
stem cells from one
baby’s cord
The umbilical cord is
discarded whether or
not stem cells are
taken from it
Adult Stem Cells
●
●
●
●
●
●
Difficult to obtain as
there are very few of
them and they are
buried deep in
tissues
Less growth potential
than embryonic stem
cells
Less chance of
malignant tumours
developing than from
embryonic stem cells
Limited capacity to
differentiate into
different cell types
Fully compatible with
the adult’s tissues,
so rejection
problems do not
occur
Removal of stem
cells does not kill the
adult from which the
cells are taken
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
1.2 Ultrastructure
Prokaryotes
1. Draw and label a diagram of the ultrastructure of Escherichia coli (E. coli) as an
example of a prokaryote. The diagram should show: the cell wall, pili, flagella,
plasma membrane, cytoplasm, 70S ribosomes, a nucleoid with naked DNA
2. Annotate the diagram with the functions of each named structure
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
Sorry for messy handwriting, I was tired :-)
3. Describe the cells of prokaryotes.
Prokaryotes have simple cell structures without compartments. They were the first
organisms to evolve on Earth and remain to have the simplest cell structure. Along
with that, they can be found almost everywhere; in soil, in water, on our skin, in our
intestines, and eve in pools of
hot
water in volcanic areas.
All cells have a cell membrane, but some cells also have a cell wall perimating the
plasma membrane; this serves as an extra protective shell around the prokaryote to
protect it and prevent it from bursting.
As no nucleus is present in the prokaryotic cell, its interior is completely filled with
cytoplasm. Along with that, the only other organelle that prokaryotes have is the 70s
ribosomes that fill up the inner cavity of the cell. However, some areas on micrographs
may seem lighter–this region contains the DNA of the cell, usually in the form of one
circular DNA molecule. This DNA is not associated with proteins, which explains the
lighter appearance compared with other parts of the cytoplasm that contain enzymes
and ribosomes. This lighter area is called the nucleoid.
4. Describe the 3 stages by which binary fission occurs
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
The prokaryotic cells do divide in their own special way; instead of dividing through
mitosis as eukaryotic cells do, they divide by binary fission, and is used for asexual
reproduction.
The single circular chromosome is first replicated- these two copies of the chromosome
then move to the opposite ends of the cell. After they reach the edge of the cell, the
cell quickly divides, leaving each daughter cell with one copy of the chromosome,
making them genetically identical.
5. Interpret electron micrographs of prokaryotic cells to identify features such as
cell wall, pili and flagella, plasma membrane, cytoplasm, 70S ribosomes, and
nucleoid with naked DNA
Eukaryotes
6. Describe the eukaryotic cell
Eukaryotic cells have a much more complicated internal structure than prokaryotic
cells. Whereas the cytoplasm of a prokaryotic cell isn’t one undivided space, eukaryotic
cells are compartmentalised. This means that they are divided up by partitions into
compartments; these partitions are single or double membranes.
The most important of these organelles is the nucleus; it contains the cell’s
chromosomes. Each organelle in the eukaryotic cell is specialised to perform a
particular role.
There are several advantages in being compartmentalised:
- Enzymes and substrates for a particular process can be much more
concentrated than if they were spread throughout the cytoplasm
- Substances that could cause damage to the cell can be kept inside the
membrane of an organelle; for example, the digestive enzymes of a lysosome
could digest and a kill a cell if they were not stored within the lysosome
- Conditions such as pH can be maintained at an ideal level for a rather particular
process, which may be different to the levels needed to perform other
processes in the cell
- Organelles with their contents can be moved around within the cell
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
7. Draw and label the ultrastructure of an exocrine gland cell of the pancreas. The
diagram should show: plasma membrane, cytoplasm, 80S ribosomes, nucleus,
mitochondria, rough endoplasmic reticulum, Golgi apparatus, vesicles,
lysosomes
8. Annotate the diagram with the functions of each named structure
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
9. Draw and label the ultrastructure of a palisade mesophyll cells of the leaf. The
diagram should show: plasma membrane, cytoplasm, 80S ribosomes, nucleus,
mitochondria, rough endoplasmic reticulum, Golgi apparatus, vesicles,
lysosomes, vacuoles, chloroplasts, cell wall
10. Annotate the diagram with the functions of each named structure
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
11. Interpret electron micrographs of eukaryotic cells to identify features listed
above
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
Microscopes
13. Electron microscopes vs light microscopes
Electron microscopes have a much higher resolution than light microscopes.
14. Know that developments in scientific research often follow improvements in
apparatus
Robert Hooke was the first person to ever document seeing a cell, but didn’t know what
they were. However, when Anton van Leeuwenhoek built the first microscope in 1674,
he was able to see living bacteria. It was not until light microscopes were then brought
into play in the late 19th century that plant and animal tissues were seen as groups of
individual cells, as documented by Schleiden and Schwann in 1838.
Animal cells were originally colourless, until staining techniques were used in the end
of the 19th century. Then, in the 1940s, the first electron microscopes were starting
to be used; these led scientists to a greater understanding of cell structure.
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
1.3 Membranes
Phospholipids
1. Outline the structure of a phospholipid molecule, to include hydrophilic head
(phosphate group, glycerol) and hydrophobic tails (2 fatty acids)
Phospholipid bilayers consists of hydrophilic (water-loving) polar heads that are made
up of glycerol bonding to a phosphate group and non polar hydrophobic (water-hating)
tails that are made up of two fatty acids. In the bilayer, the polar heads and the nonpolar
tails are arranged so that the nonpolar tails face each other while the polar head face
outward. The hydrophilic heads of the molecules always appear on the outside of the
membrane because that is where water is present, while the hydrophobic tails face
inward in the bilayer, away from water. As such, phospholipid bilayers are known as
amphiphilic compounds because they contain both hydrophilic and hydrophobic
regions.
2. Explain how the hydrophobic and hydrophilic properties of phospholipids help
to maintain the stability of a phospholipid bilayer.
Phospholipid bilayers are amphiphilic, meaning they have hydrophilic and hydrophobic
regions. The hydrophobic side is anchored inside the membrane while the hydrophilic
side is facing outward towards the water thereby maintaining stability as contact with
water in the hydrophobic region will cause destabilisation to the construct of the bilayer.
The fluid state it’s in allows individual phospholipids to move within the bilayer, thereby
allowing the cell to change its shape easily and thus is what makes the cell membrane
very stable and allows flexibility. This is important as it is crucial for the exocytosis and
endocytosis of the cell.
Models
3. Knowing that scientists use models as representations of the real world, outline
how evidence from electron microscopy supported the Davson-Danielli model
The existence of a lipid bilayer was originally proposed and outlined by Gorter and
Grendel, in 1925. Their ideas were developed and improved by Hugh Davson and
James Danielli, who proposed in 1935 a membrane model, described as a 'lipo-protein
sandwich’, in which two layers of proteins sandwiched the phospholipid bilayer of the
membrane. The Davson– Danielli model was new and it attempted to explain their
observations of the surface tension of lipid bilayers. Since that time, the phenomenon
of surface tension in bilayers has been better explained by studying the properties of
the phospholipid heads. High magnification electron macrophages of membranes were
made in the 1950s which showed two dark lines, representing the protein bands, with
a lighter band, representing the phospholipid bilayer between. Ultimately this was the
evidence that supported the Davson-Danielli model..
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answer IB questions in this much detail.
4. Knowing that theories can be falsified and superseded by other theories when
new evidence becomes available, describe how the falsification of the DavsonDanielli model led to the Singer-Nicolson model
However, in the 1960s, after some experimentation, it was discovered that the DavsonDanielli's model of a phospholipid bilayer sandwiched between two layers of globular
protein was incorrect. Improvements in the biochemical techniques allowed proteins to
be extracted from the membranes which showed that the proteins were varied in size
and shape which was unlike the uniform shape proposed by Davson and Danielli. In
addition, a method involving rapid freezing cells and the fracturing them revealed
irregular touch surface inside the phospholipid bilayer, now known as transmembrane
proteins. This again did not support the Davson-Danielli's model. Through fluorescent
antibody tagging it was also discovered that the membrane proteins were free to move
within the membrane instead of being fixed. membrane proteins from two different cells
were tagged with red and green fluorescent markers and when the two cells were
fused, the markers became mixed revealing the fluidity of the membrane. All this led
to a new membrane model being proposed in 1966 by Seymour Singer and Garth
Nicolson. In this model, the proteins were embedded within the lipid bilayer, occupied
a variety of positions in the membrane and had the ability to move around freely within
the phospholipid bilayer. This model came to be known as the fluid-mosaic model
which we currently use today.
Fluid mosaic model
5. Draw and label a 2D diagram to show the structure of membranes including
phospholipid bilayer (using the symbol of a circle with 2 parallel lines attached)
and should show: cholesterol, glycoproteins, integral proteins (i.e. channels,
pumps, hormone receptors) and peripheral proteins (i.e. enzymes)
6. Describe the function of cholesterol in membrane fluidity
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
Cholesterol is a type of lipid that is often found in animal cells specifically the plasma
membrane. Like phospholipid bilayers, cholesterol is also an amphipathic molecule
that contains a hydrophilic area, the hydroxyl group of cholesterol, and a hydrophobic
area, the remainder of the molecule.
Cholesterol interacts with the fatty acid tails of phospholipids to moderate the
properties of the membrane. It functions to immobilise the outer surface of the
membrane, reducing fluidity and makes the membrane less permeable to very small
water-soluble molecules that would otherwise freely cross. Furthermore, it separates
the phospholipid tails to prevent crystallisation of the membrane, thereby helping
secure peripheral proteins by forming high density lipid rafts capable of anchoring the
protein.
Roles of proteins
7. State the position, structure and function of membrane proteins
Phospholipid bilayers are embedded with proteins that are either permanently or
temporarily attached to the membrane. These proteins can either be integral proteins,
which are permanently attached to the membrane attached or peripheral proteins.
Integrals are transmembrane proteins that have hydrophobic surfaces and are
embedded in the centre of the membrane. They span across the membrane with the
hydrophilic parts going through to the regions of the phosphate heads on either side of
the bilayer. peripherals on the other hand are hydrophilic on their surface and are thus
not embedded in the membrane, but rather attached to one side of the bilayer.
Although the primary function of the cell membrane is to form a barrier through which
ions and large molecules cannot pass through easily, the proteins of the membranes
have a wide range of functions. These functions vary from being pumps used for active
transport to the adhesion of cells to form netted connection in tissues and organs. As
such, because of theses functions, the more active a cell membrane is the more
proteins will be embedded in it.
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
1.4 Transport
1. What is the difference between diffusion and endo/exocytosis?
Diffusion is the passive movement of particles from an area of high concentration to
an area of low concentration (through a concentration gradient) through the cell’s semipermeable membrane. This process does not require energy and transports particles
rather than water.
Exocytosis and endocytosis are energy-driven processes where vesicles containing
fluid enter or exit the cell through the membrane. A vesicle is a small sac of membrane
with a droplet of fluid inside; they’re spherical and normally present in any eukaryotic
cell. To form a vesicle, a small region of membrane is pulled from the rest of the
membrane and is pinched off; proteins in the membrane carry out this process, and
require energy from ATP. Vesicles are formed inside the cell after the cell accepts
material from outside of the cell; this is endocytosis.
Vesicles taken in by endocytosis contain water and solutes from outside the cell, but
also often contain larger molecules needed by the cell that cannot pass across the
plasma membrane through diffusion. An example of this is in the placenta, whereby
proteins from the mother’s blood are absorbed into the fetus by endocytosis.
On the other hand, exocytosis is then the release of vesicles from the cell into their
surroundings. This happens when a vesicle fuses with the cell’s plasma membrane
and the materials are released into the surrounding. Digestive enzymes are often
released from the gland cells by exocytosis; the polypeptides in the enzyme are
synthesized by the rER, processed in the Golgi apparatus and then carried to the
membrane by vesicles for exocytosis. However, in this case, it is referred to as
“secretion”, due to the fact that a useful substance was released, not a waste product.
Smaller Particles
2. Distinguish between diffusion, simple diffusion, facilitated diffusion and active
transport
Diffusion is the spreading out of particles in liquids and gases which happens
because the particles are in continuous random motion. More particles move from an
area of high concentration to an area of low concentration- there is therefore a net
movement from the higher to the lower concentration (movement down the
concentration gradient). This is a passive process.
Simple diffusion across membranes involves particles passing between the
phospholipids in the membrane; this can only occur if the phospholipid bilayer is
permeable to the particles. Non-polar particles such as oxygen can diffuse through
easily; if the oxygen concentration inside a cell is reduced due to aerobic respiration
and the concentration outside is higher, oxygen will pass through natural, passive
diffusion, however. The centre of the membranes are hydrophobic, so ions with
charges cannot easily pass through. Polar molecules, such as those with partial
charges, can diffuse at lower rates between the layers. Small polar particles such as
urea and ethanol can pass through more easily than large particles.
Facilitated diffusion is the movement of ions and other particles that
cannot diffuse between the phospholipids and instead move through channel
proteins. They are called channel proteins due to the fact that the walls of these
channels consist of proteins. The diameter and the chemical properties of the channel
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
ensure that only one type of particle passes through, such as sodium ions, or
potassium ions, but not both. Because the channels help such particles to pass through
down the concentration gradient, it is called facilitated diffusion; cells can control which
types of channels are synthesized and placed in the plasma membrane. This also
means that cells have complete control over what substances diffuse in and out.
Osmosis is the net movement of water in and out of cells. Water is able to move
in and out of cells freely; sometimes the number of water molecules moving in and out
is the same, and there is no net movement, but at other times, more molecules move
in one direction or the other. Osmosis occurs due to differences in concentration of
substances dissolved in water (aka solutes). Substances dissolve by forming
intermolecular bonds with the water molecules; these bonds restrict the movement of
water molecules. Regions with a higher solute concentration therefore have a lower
solute concentration; because of this, there is a net movement of water from regions
of lower solute concentrations to regions of higher solute concentration. This
movement is passive because is no energy is needed to make it happen. Osmosis
happens in all cells because water molecules, despite being hydrophilic, are small
enough to pass through the bilayer. Some cells have water channels called
aquaporins, which greatly increase membrane permeability to water; examples are
kidney cells that reabsorb water and root hair cells that absorb water from the soil.
Active transport is the movement of particles from an area of low concentration
to an area of high concentration with the assistance of energy, or ATP. Cells
sometimes take in substances, even though there is already a higher concentration
inside than outside. The substance is absorbed against the concentration gradient.
Less commonly, the cells may pump substances out, even if the concentration on the
outside is higher. Active transport is carried out by globular proteins in membranes,
also known as protein pumps. A visual below shows how pumps and channels work
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
3. Explain passive transport across membranes by simple diffusion and facilitated
diffusion
An example to help explain passive transport could be the diffusion of potassium in
axons. The structure and function of sodium-potassium pumps aid the facilitated
diffusion in axons. Seeing as a nerve impulse involves rapid movement of sodium and
potassium channels; whilst the sodium channels are part of the active transport, the
potassium channels can be described as a special example for facilitated diffusion.
Each potassium channel consists of four protein subunits with a narrow pore between
them that allows potassium ions to pass in either direction; this pore is 0.3 nm wide at
its narrowest. Potassium ions are smaller than 0.3 nm, but dissolved, bond to water
molecules and become too large to pass through the pore. In order to pass, the bonds
holding the ion and the water together must be broken, and temporary bonds must
form between the ion and the amino acids in the narrowest part of the pore. Once the
potassium ion passes through this part of the channel, it can be bonded with water
again.
To explain the specificity, it has been shows that the other positively charged ions that
attempt to fit through the pump are either too large or too small to form bonds with the
amino acids, so they cannot break their water shell.
Potassium channels in the axons are gated through voltage; this means that the
membrane is charged due to imbalance of positive and negative charges, which create
voltages that travel along the axon. If an axon has relatively more positive charges
outside than inside, potassium channels are closed. At one point during a nerve
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answer IB questions in this much detail.
impulse, there are more positive charges inside than outside; this causes the channel
to open, which allows the potassium ions to diffuse through.
4. Explain the role of protein pumps and ATP in active transport across membranes
In opposition, we could then use the transport of sodium to explain active transport.
Seeing as the nerve impulse involves rapid movements of sodium, active transport is
carried out by the sodium-potassium pump protein. The sodium-potassium pump
follows a repeating cycle of steps that result in three sodium ions being pumped out of
the axon and two potassium ions being pumped in. Each time the pump goes round
this cycle, it uses one ATP. The cycle consists of the following steps:
a. The interior of the pump is open to the inside of the axon; three sodium ions
enter the pump and attach to their binding sites
b. ATP transfers a phosphate group from itself to the pump; this causes the pump
to change shape and the interior is closed
c. The interior of the pump opens up to the outside of the axon and the three
sodium ions are released
d. Two potassium ions from outside can then enter and attack to their binding
sides
e. Binding of potassium causes the release of the phosphate group; this causes
the pump to change shape again, so that it is again open to the inside
f. The interior of the pump opens up to the inside of the axon and the two
potassium ions are released; sodium ions can enter and bind to the pump again
Osmosis
5. What is Osmosis?
The passive movement of water molecules, across a partially permeable membrane,
from a region of lower solute concentration to a region of higher solute concentration.
6. Explain changes in tissue mass following immersion in hypotonic and
hypertonic solutions
A hypotonic solution is one with a lower concentration of solutes than another solution.
When a tissue is placed in a hypotonic solution, the osmotic pressure will force water
to enter the tissue, making it gain mass.
A hypertonic solution is one with a higher concentration of solutes than another
solution. When a tissue is placed in a hypotonic solution, the osmotic pressure will
force water out of the cell, making it lose mass.
The mass change can be tested by weighing the tissue mass before and after soaking
in the solution.
In animal cells, the cell will shrivel if placed in a hypertonic solution and will burst if
placed in a hypotonic solution for too long.
In plant cells, the cytoplasm will decrease (but the cell will remain the same shape due
the cell wall) when placed in a hypertonic solution, but will increase when placed in a
hypotonic solution (but the plasma membrane cannot burst due to the cell wall.
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
7. Explain changes in cell structure following immersion in hypotonic and
hypertonic solutions
Animal cells can be damaged by osmosis. In a solution with higher osmolarity, water
leaves the cells by osmosis so their cytoplasm shrinks in volume. The area of plasma
membrane does not change, so it develops indentations, which are sometimes called
crenellations. In a solution with lower osmolarity, the cells take in water by osmosis
and swell up. They may eventually burst, leaving ruptured plasma membranes called
red cell ghosts.
8. Explain why tissues or organs to be used in medical procedures must be bathed
in a solution with the same osmolarity as the cytoplasm
Both hypertonic and hypotonic solutions damage human cells, but in a solution with
same osmolarity as the cells (isotonic) water molecules enter and leave at the same
rate so they remain healthy. It is therefore important for any human tissues and organs
to be bathed in an isotonic solution during medical procedures. Usually, an isotonic
sodium chloride solution is used, which is called normal saline. It has an osmolarity of
about 200 mOsm. Normal saline is used in many medical procedures. It can be:
- safely introduced to a patient's blood system via an intravenous drip
- used to rinse wounds and skin abrasions
- used to keep areas of damaged skin moistened prior to skin grafts
- Used as the basis for eye drops
- Frozen to the consistency of slush for packing organs that must be transported
Potato investigation
9. Estimate the osmolarity in tissues from experimental data
Osmosis is due to solutes that form bonds with water; these solutes are then osmotically
active. Glucose, sodium ions, potassium ions, and chloride ions are all osmotically active and
solutions of them are often used in osmosis experiments. The osmolarity of a solution is the
total concentration of osmotically active solutes; the units for measuring this is osmole or
milliosmole (mOsm). The normal osmolarity of human tissue is about 300 mOsm.
An isotonic solution has the same osmolarity as a tissue. A hypertonic solution has a higher
osmolality and a hypotonic solution has a lower osmolarity. If samples of a tissue are bathed
in hypertonic and hypotonic solutions, and measurements are taken to find out whether water
enters or leaves the tissue, it is possible to deduce what concentration would be isotonic, and
therefore find out the osmolarity the.
Check the Osmosis in Plant Tissues data-based question on page 42 for extra help.
Larger particles
10. Describe the processes of endocytosis and exocytosis, including how the
plasma membrane changes shape, breaks and reforms
See Question 1
11. State the importance of membrane fluidity in enabling endocytosis and
exocytosis
The importance of membrane fluidity with regards to endocytosis and exocytosis is that without
it, the cell wouldn’t be able to excrete/absorb vesicles and in turn, not receive nor excrete
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
materials and waste. The cell would either die due to “starvation” or burst due to amounts of
waste.
12. Explain how vesicles are used to transport materials within a cell between the
RER, Golgi apparatus and plasma membrane
Refer to the end of Question 1
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
1.5 Origin of Cells
Louis Pasteur
1. Outline the theory of spontaneous generation
Spontaneous generation is the formation of living organisms from nonliving matter. A
Greek philosopher and botanist by the name of Theophrastus reported that a plant
called Silphium had sprung up from the soil where it was not previously present and
described this as an example of spontaneous generation.
Louis Pasteur’s experiments helped to form the theory of spontaneous generation. He
had made a nutrient broth by boiling water containing yeast and sugar. He showed that
if this broth was kept in a sealed flask, it remained unchanged, and no fungi or other
organism appeared. He then passed air through a pad of cotton wool in a tube, to filter
out microscopic particles from the air, including bacteria and the spores of fungi. If the
cotton wool placed in broth in the sealed flask, within 36 hours, there were large
numbers of microorganisms in the broth and mould would grow over its surface.
One of his most famous experiments was the swan-neck flasks whereby he would
place a sample of broth in flasks with long necks, then melted the glass and bent it into
an S shape. Then, he boiled the broth in some flasks to kill the organisms present but
left others unboiled (controlled). Fungi and other organisms soon appeared in the
unboiled flasks but not in the boiled ones, even after long periods of time. The broth in
the flasks was in contact with air, which it has been suggested was needed for
spontaneous generation, yet no generation occurred.
2. Describe how Pasteur's experiments provided evidence against spontaneous
generation
See Question 1
The 1st cell
3. Describe where the first cell originated
The first cells must have arisen from nonliving chemicals via abiogenesis. Abiogenesis
is the process by which life arises from nonliving matter, such as organic compounds.
However, this process can no longer occur due to the fact that there is no longer a
sterile enough environment.
This is, however a step to answering the big question: how could a structure as
complex as the cell have arisen by natural means from nonliving material? There have
been hypotheses for how some of the main phases could have occurred.
a. Production of carbon compounds such as sugars and amino acids
i.
Stanley Miller and Harold Urey passed steam through a mixture of
methane, hydrogen and ammonia. The mixture was thought to be
representative of the atmosphere of the early Earth. Electrical
discharges were used to simulate lightning. They found that amino
acids and other carbon compounds needed for life were produced
b. Assembly of carbon compounds into polymers
i.
A possible site for the origin of the first carbon compounds is around
deep-sea vents. These are cracks in Earth’s surface, characterised by
gushing hot water carrying reduced inorganic chemicals such as iron
sulphide. These chemicals represent readily accessible supplies of
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
energy, a source of energy for the assembly of these carbon
compounds into polymers
c. Formation of membranes
i.
If phospholipids or other amphipathic carbon compounds were among
the first carbon compounds, they would have naturally assembled into
bilayers. Experiments have shown that these bilayers readily form
vesicles resembling the plasma membrane of a small ecll. This would
have allowed different internal chemistry from that of the surroundings
to develop
d. Development of a mechanism for inheritance
i.
Living organisms currently have genes made of DNA and use enzymes
as catalysts. To replicate DNA and be able to pass genes on to offspring
enzymes are needed. The solution to this conundrum may have been
an earlier phase in evolution when RNA was the genetic material. It can
store information in the same way as DNA, but it is both self-replicating
and can itself act as a catalyst. Genetic code provides powerful
evidence that living species are descended from one common
ancestor.
Endosymbiosis
5. Outline endosymbiotic theory:
ENDO = within
SYMBIOSIS = a mutually beneficial relationship between two organisms
The endosymbiotic theory helps to explain the evolution of eukaryotic cells. It states
that mitochondria were once free-living prokaryotic organisms that had developed the
process of aerobic cell respiration. Larger prokaryotes that could only respire
anaerobically took them in by endocytosis. Instead of absorbing and digesting these
smaller organisms, they allowed them to continue thriving and respirate inside their
cytoplasm. As long as the smaller cells grew and divided as fast as the larger ones,
they could persist indefinitely inside of the larger cells. According to the theory of
endosymbiosis, they have persisted over hundreds of millions of years of evolution to
become the mitochondria inside eukaryotic cells today. Natural selection would end up
favouring this symbiosis of cells.
Along with that, it helped explain the origin of chloroplasts in plant cells; if a prokaryote
that had developed photosynthesis was taken in by a larger cell and was allowed to
survive grow, and divide, it could have developed into the chloroplasts of
photosynthetic eukaryotes. Both of the cells would have benefited.
6. Know at least 3 pieces of evidence for the endosymbiotic theory:
Some pieces of evidence for the endosymbiotic theory are as followed:
- mitochondria and chloroplasts contain their own circular DNA
- mitochondria and chloroplasts both have 70S ribosomes
- mitochondria and chloroplasts are about the same size as bacteria
- new mitochondria and chloroplasts are formed only through a process similar
to binary fission
- both mitochondria and chloroplasts have a double membrane
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answer IB questions in this much detail.
-
both mitochondria and chloroplasts appear to resemble free-living organisms
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
1.6 Cell Division
The Cell Cycle
1. List 4 biological scenarios in which mitosis is required
Mitosis is the splitting of a cell to form two genetically identical nuclei; the cell divides
into two daughter cells each with one of the nuclei. Mitosis is generally involved
whenever cells with genetically identical nuclei are required in eukaryotes, such as
during:
- Embryonic development
- Growth
- Tissue repair
- Asexual reproduction (meiosis is used for sexual)
2. Outline the sequence of events that occur during the four phases of mitosis
Given that mitosis involves the division of the nucleus into two genetically identical
daughter nuclei, we can conclude that there are multiple stages or phases for mitosis
to occur. However, before mitosis begins, the cell goes through a phase called
interphase.
Interphase is a very active phase of the cell cycle with many processes occurring in
the nucleus and cytoplasm. The cell cycle is the sequence of events between one cell
division and the next. During interphase, the numbers of mitochondria that reside in
the cytoplasm increase; this is due to the growth and development as well as the
splitting of the organelles. In opposition, the chloroplast count in plant cells increase.
Along with that, they synthesise cellulose and use vesicles to reinforce the cell wall.
Interphase itself consists of 3 stages, the G1 stage, the S stage, and the G2 stage.
During G1, the cells continue to grow and synthesise enzymes and nutrients that are
later needed for DNA replication. During S, the cell replicates all the genetic material
in its nucleus, so that the new cells have a complete set of organelles. During G2, the
cell goes through a control checkpoint to see if it is ready for mitosis (so that it doesn’t
create mutations or mess up the cell structure).
The four phases of mitosis include Prophase, Metaphase, Anaphase, and Telophase.
a. Prophase
i.
Chromosomes become shorter and fatter by coiling and supercoiling
ii.
Nucleolus (nuclear envelope) breaks down
iii.
Microtubules grow to form spindle between two cell poles
iv.
At the end of prophase
b. Metaphase
i.
Microtubules continue to grow and attach to centromeres of
chromosomes
ii.
Two attachment points on opposite sides of each centromere allow the
chromatids of a chromosome to attach to microtubules from different
poles
iii.
Microtubules put under tension to see if attachment is correct shortening the microtubules at the centromere
c. Anaphase
i.
Each centromere divides, allowing sister chromatids to separate
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
ii.
iii.
Spindle microtubules pull them to the poles
Mitosis produces two genetically identical nuclei because sister
chromatids are pulled to opposite poles
d. Telophase
i.
Chromatids reached the poles - now called chromosomes
ii.
At least pole, chromosomes are pulled into a tight group
iii.
Nuclear membrane reforms
iv.
Chromosomes uncoil and nucleolus is formed
v.
Starting to divide, daughter cells will enter interphase again
Cytokinesis & control
3. Outline how cyclins are involved in the control of the cell cycle.
Cyclins are a class of proteins discovered in 1982 by Timothy Hunt, that are involved
in the control of the cell cycle and are responsible for the control of specific events
such as microtubule formation and chromatid alignment. Cyclins work by binding to
enzymes called cyclin-dependent kinases (CDKs) that become active and attach
phosphate groups to the other proteins in the cell. That attachment triggers the other
proteins to become active and carry specific functions depending on the phase of the
cell cycle.
4. Outline the role of serendipity in scientific discovery using cyclins as an
example
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
During his research in sea urchins Timothy Hunt discovered that a protein increased
in concentration when fertilisation occurred and decreased afterwards in a pattern that
convinced with the phases of the cell cycle. This protein came to be known as cyclin
and led to Hunt winning the Nobel Prize in Physiology or Medicine along with Lee
Hartwell and Paul Nurse who also contributed to the discovery.
Diagnosing cancer
5. Skill: Identification of phases of mitosis in cells viewed with a microscope or in
a micrograph
Interphase
(not part of mitosis)
Prophase
Metaphase
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
Anaphase
Telophase
6.
from a micrograph
Skill: Determination of a mitotic index
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
7. Outline the use for the mitotic index
When studying cells under the microscope, the ratio of the number of cells undergoing
mitosis to the total number of cells in view is called the mitotic index.
The mitotic index in an important prognostic tool used in predicting the response of
cancer cells to chemotherapy – a low mitotic index indicates a longer survival time,
and suggests that the treatment is working. It can be less accurate when used with
elderly patients, whose cells divide more slowly. In such patients, a low mitotic index
may not indicate that a treatment is working. In the laboratory, it is possible to work
out the mitotic index of growing and dividing cells from an electron micrograph as
shown in the image below
8. Mutagens, oncogenes and metastasis are involved in the development of
primary and secondary tumours
After mitosis occurs and tissues develop, normal cells undergo a programmed form of
death known as apoptosis. However sometimes mitosis does not proceed normally.
Cell division may continue unchecked and produce an excess of cells, which clump
together.This growth is called a tumour. Tumours can be either benign, which means
they are restricted to that tissue or organ, or malignant which means cancerous.
Cancer occurs when cells from a primary tumour migrate to other tissues and form
new secondary tumours in a process known as metastasis. It is caused by damage to
genetic material, producing cells that undergo uncontrolled, abnormal mitosis. DNA
damage that leads to cancer can be caused by a range of factors. These factors
include mutations caused by mutagens which are physical, chemical and biological
agents that modify the genetic material of the cell through mutations. Examples of
mutagens include X-rays, gamma rays and ultraviolet light – and also chemical
compounds, such as those found in tobacco smoke and aflatoxins produced by certain
fungi. Another factor that leads to DNA damage is mistakes in DNA replication or
genetic predisposition as a result of inheritance. Genes with the potential to cause
cancer are known as oncogenes and are active by an alteration of the gene or
expression of the gene at abnormally high levels. Activated oncogenes can cause cells
that should die during apoptosis to survive and divide instead which can lead to cancer.
9. Application: The correlation between smoking and incidence of cancers
Reminder: These answers are detailed to aid with understanding. You will not be required to
answer IB questions in this much detail.
Smoking is widely known to be one of the major causes of several types of cancer.
Evidence to show that high smoking rates positively collates to an increased risk of
cancer of the bladder, cervix, kidney, larynx and stomach. Although the death rate due
to other cancers is not significantly different in smokers and non smokers, studies show
that smokers are several times more likely to die of all cancers compared to nonsmokers mainly due to the increased exposure to the mutagenic chemicals in
cigarettes. For example, the risk of contracting lung cancer increases with the number
of cigarettes that a person smokes and the number of years that they continue to
smoke. If a person gives up smoking, their risk of developing cancer decreases