SEMESTER 1 FINAL Syllabus Statements

Outline the procedure for DNA electrophoresis.

1. Prepare the agrose gel

2. Place gel into electrophoresis chamber

3. Load dna samples

4. Apply electric current

5. Dna fragments are separated by size

6. Stain the gel

7. Visualize bands

Describe how and why DNA fragments separate during electrophoresis.

• Dna is negatively charged, so it moves towards the positive electrode

• Small fragments move quickly and travel farther

• Big fragments move slowly and travel slowly

Outline the process of DNA profiling.

1. Collect dna samples

2. Extract dna

3. Amplify str regions using pcr

4. Cut dna

5. Separate fragments using gel electrophoresis

6. Compare band lengths

List applications of DNA profiling.

• Forensic investigations

• Paternity tests

• Disease gene tracking

List example sources of DNA that can be used in DNA profiling.

• Blood 

• Saliva

• Semen

• Skin cells

• Hair

• Urine

• Sweat

The units for measuring genome size are Mb (mega bases). 1 Mb = 1,000,000 bp (base pairs). Use a database to compare genome sizes to determine if there is a relationship between the number of genes in a species and the species complexity in structure, physiology and/or behavior.

Genome size does not consistently correlate with complexity. This known as c-value paradox. The regulation of genes is what drives complexity, not the number of genes.

Chromosome number is a distinguishing characteristic of a species. Explain why the typical number of chromosomes in a diploid cell is an even number.

A dipoloid has chromosomes in homologous pairs. One chromosomes comes from the mother, and the other comes from the father. Because the chromosomes are inherited in pairs, the total number must be even. Having an even number ensures the proper pairing during meiosis, correct formation of gametes, and equal separation of chromosomes.

Humans have 46 chromosomes and chimpanzees have 48. Evaluate the evidence for the hypothesis that chromosome 2 in humans arose from the fusion of chromosomes 12 and 13 with a shared primate ancestor.

1. Telomeres normally occurs only at the ends of the chromosomes. Humans chromosomes 2 contains internal degenerated telomere sequences in the middle of the chromosomes

2. One nomal chromosome has one centromere. A human chromosome 2 shows one active centromere, and a second inactive centromere

3. The top half of human chromosome 2 matches one chimp chromosome. The bottom half matches the other

Define karyotype and karyogram.

A karyotype is the number and types of chromosomes in an organism. A karyogram is a photograph or arrangement of chromosomes in homologous pairs.

List the characteristics by which chromosomes are paired and arranged on the karyogram.

• Length

• Centromere position (metacentric - centric, submetacentric - off center, acrocentric - near one end)

• Banding patters after staining

• Prescient of homologous pairs

Describe the process of creating a karyogram.

1. Inhibition of mitosis with drugs

2. Separate the chromosomes

3. Staining 

4. Photography 

5. Karyotype preparation

Define autosomes and sex chromosomes.

Autosomes 

• Chromosomes 1-22 in humans

• Do not determine sex

• Homologous pairs with matching size and banding

Sex chromosomes

• The 23 pair

• Determines sex

• X is large and metacentric, y is small and acrocentric

Define “DNA barcode” and “environmental DNA.”

Dna barcode - a short dna sequence that is unique to each species and can be used to identify organisms Environmental dna - genetic material collegected from soil, water, air, or surfaces which include cells shed by organism, secretions, saliva, waste, etc.

Outline the use of “DNA barcoding” to identify species from environmental DNA.

1. College enviormental sample

2. Extract dna

3. Amplify barcode regions using pcr

4. Sequence the pcr 

5. Compare sequences to BOLD

6. Identify species in the enviorment

List applications of DNA barcoding using environmental DNA.

• Biodiversity 

• Invasive species

• Detections

• Endangered animals monitoring

• Food chain and tropic studies

• Water quality and ecosystem health

Outline the mechanism by which the presence of lactose regulates the expression of genes related to digestion and use of lactose in E. coli.

1. The lac operon enables e. Coli to digest lactose only when lactose is present, and glucose is low.

When lactose is absent:

1. Repressor protein binds to operator region

2. Rna polymerase is blocked

3. Genes for lactose metabolism is off

When lactose is present:

1. Lactose is converted to allolactose

2. Which then binds to the lac repressor, causing it to change shape

3. The repressor detaches from the operator

4. Rna polymerase binds the promotor

5. The lac genes are transcribed

Outline the mechanism by which the presence of oestrogen in a cell’s environment regulates the expression of genes related to endometrium development and maintenance during the uterine cycle.

1. Ostergeon enters the cell

2. It binds to the oestrogen receptor 

3. The receptor changes shape and formas a hormone-receptor complex

4. This moves in the nucleus

5. It binds to specific dna sequences (oestrogen response elements)

6. Binding recruits transcription factors and rna polymerase

7. The target genes involved endometrical growth, blood vessel formation, and nurtirent secreation are activated

Increases transcription of genes

Define phenotypic plasticity

The ability for a single genotype to produce different phenotypes in response to environmental conditions, due to changes in patterns of gene expressions. These are reversible within the organisms lifetime.

Outline an example of phenotypic plasticity.

Himalayan rabbit:

1. Temperature sensitive gene for melanin production

2. Cold areas activate the enzyme and calls for darker fur

3. Warm areas deactivate, with calls for while fur

Describe the inheritance of epigenetic tags in differentiated cells of a multicellular organism.

Differentiated cells have specific patters of gene expression controlled by dna methylation and histone modifications. They are inherited during dna replication, maintenance methyltransferase, and histone modifications. This results in the daughter cells inheriting the same epigenetic patter, maintain the same cell identity, and allow tissues to keep their specialized function throughout life

No change in dna sequence, only in expression.

Discuss the consequences of reprogramming and imprinting of epigenetic tags in haploid gametes.

Reprogramming - before gametes form, most epigenetic tags are erased; this is to reset the genome so that the next embryo can develop normally. 

Imprinting - some genes retain methylation after reprogramming; the consequence of this is that only one allele is expressed while the other is silences (either maternal or paternal); this affects growth, metabolism, and development

Explain the reason why monozygotic twin studies are often used to measure the impact of the environment on gene expression.

Monozygotic twins comes from a single fertilized egg, have identical dna sequences and starts with identical epigenetics patterns.

This leads scientists to hold genetics constant so the differences myst be due the environment; compares the lifestyle, nutrition, stress, etc.

All in all, identical twins are the best experiment for isolatiing environmental influences on gene expression.

Outline the impact of air pollution on the epigenetic regulation of genes associated with the immune response.

• Increased dna methylation on promotor of genes involved in immune signaling, inflammation, and antioxidant.

• Decreased expression of these genes; reduces ability to fight infections, higher inflammation is respiratiory tract

• Hypomethylation of inflammatory genes leads to higher risk of autoimmune disorders, and overactivation of immune responses.

Discuss how imprinting of epigenetic tags impacts gene expression in a diploid cell.

Imprinting - one allele is silenced by methylation and the other is active

The cells behave as if it only has one functional copy of the imprinted genes. These genes regulate growth and development. If imprinting is abnormal (silencing the wrong allele), disease occurs.

Outline the epigenetic origins in the difference in size between tigons and ligers (lion–tiger hybrids).

Liger = male lion and female tiger

Tigon = male tiger and female lion

• Paternally expressed genes = growth (ligers - growth promototing genes, so they grow extremely large)

• Maternally expressed genees = suppresses growth (tigons - limit growth so they are smaller than either parent species)

Outline evidence that viruses evolved after the origin of cells.Outline evidence that viruses evolved after the origin of cells.

• Viruses depend on host cells for replication that cannot reproduce → suggests that they originated after the cell already exists

• Viral genomes contain genes that are derived from host cells indiciating that they evolved through reduction

• No viruses have ribosomes or metabolic pathyways, which are evolved early in life. So the virus must have diverged later

• The genetic code shared by viruses mirror the host organisms, showing they adopted the code from cells rather than evolving it independently.

Outline the progressive and regressive hypotheses for the origin of viruses.

Progressive (escape) = viruses evolved from mobile genetic elements. These elements escaped cells and gained the ability to move between cells. This allows them to be infectious particles

Regressive (reduction) = viruses originated from once-free-living cells that became parasitic. Over time, they lost genes that was unnecessary for the life, and eventually lost all metabolic and structural genes.

Outline how convergent evolution could result in the commonality of some structures shared by viruses.

• Different viral linages evolved independently but faced similar selective pressures like enteing host cells, protect genomes, and replicate efficiently

• The could have lead to the evolution of similar structural features

Outline three reasons for rapid evolution in viruses.

1 - High replication rate/short generation time - produces billions of virions per day in the affected person

2 - Lacks proofreading - introduces many mutations into the DNA

3 - Short generation time - leads to the formation of many different genetic variants within an affected individual

4 - small genome

Discuss the consequences of rapid virus evolution on treating diseases caused by viruses (influenza and HIV).

Influenza

• Antigenic drift (constant small mutations) alters surface proteins (HA and NA) which leads to the immune system no longer recognizing the virus

• This leads to producing highly infectious strands which could be pandemics. So antiviral drugs have to become not effective and vaccines must be updated annually.

Hiv 

• Extremely high mutation rate which leads to evolution of drug-resistant strains

• Requires communication therapy to reduce resistance

• Rapid evlolition could allow hib to evade immune detention and contribute to infection.

State that gametogenesis involves mitosis, cell growth, two divisions of meiosis and differentiation.

1. Mitosis - to create many parent cells

2. Cell growth - the growth of these parent cells

3. Two divisions of meiosis - producing haploid cells

4. Differentiation - turning the haploid cells into sperm or eggs

Define oogenesis and spermatogenesis.

1. Oogenesis - the production of female gametes (eggs) in the ovaries

2. Spermatogensis - the production of male gametes (sperm) in the testes

Compare the processes of spermatogenesis and oogenesis, including the number of gametes, size of games, the timing of formation and release of gametes.

Spermatogensis - in the testes, begins at puberty and continues throughout life, it has 4 functional sperm per spermatocyte, it divides equally to small cells, it is a continout production, and they are flagellum

Oogenesis - in the ovaries, it beings before birth and stops at menopause, 1 functional ovum and 2-3 polar bodies prodicts per parent cell, it divides unequally into 1 large egg, it releases 1 egg per menstrual cycle

Outline unequal cytokinesis during human oogenesis.

In oogensis, cytosis during meiosis 1 and 2 is unequal. Almost all of the cytoplasm goes to one daughter cell which then produces the primary oocyte in meiosis 1, and the secondary oocyte in meiosis 2. The other cells are called polar bodies, and they recieve almost no cytoplasm and eventually break down due to no structural support. This ensures that one large egg with enough nutrients and organelles are there to support early embryo development.

Compare motility, size, energy reserves and production rate of male and female gametes of animals.

Sperm - high motile (it swims using a flagellum), it is very small in size, little to none energy reserves as it depends on the surrounding fluids, produces millions per day, and its main role is to deliver DNA quickly to the egg.

Egg - non motile (stays in reproductive tract), it is much larger in size, has large energy reserves in the cytoplasm, produces typically one per month during menopause, and its job is to provide DNA and cytoplasm to support the embryo

Define “carrier” as related to genetic diseases.

A carrier is a person who has one recessive allele for a genetic disease but doesn’t show symptoms of that disease. They can pass along the allele to their offspring.

Explain why genetic diseases usually appear unexpectedly in a population.

Recessive alleles are usually hidden in carriers. The disease appears when a child inherites two recessive alleles, where one comes from each parent. Parents may not know that they carry the allele, which is why the disease seems to appear out of no where in the population.

Outline the genetic cause of phenylketonuria.

It is caused by a recessive mutation in the gene for phenyalaine hydroxylase. Which is an enzyme that converts phenyalanine to tyrosine. Without this enzyme, phenylalanine builds up to harmful levels.

List consequences of phenylketonuria if untreated.

• Brain damage 

• Seizures

• Behavioral problems

• Light skin/hair

State how phenylketonuria is treated.

Having a strict low-phenyalanine die, medical prescriptions, and regular blood tests to monitor phenyalanine levels are used to treat phenylketonuria.

Compare continuous to discrete variation.

1. Continuous variation - traits vary along a range with many possible values

2. Discrete variation - traits fit into categories with no intermediates

State that a normal distribution of variation is often the result of polygenic inheritance.

Polygenic inheritance, where many genes control one trait, often produces a normal distrubution of phenotypes (bell shaped curve)

Explain polygenic inheritance using an example of a two gene cross displaying incomplete dominance.

Capital letters = more skin pigment 

Lowercase letters = less skin pigment

Each of the dominant alleles contributes additively

With incomplete dominance:

AA BB is the darkest skin color

Aa Bb is the intermediate skin color

aa bb is the lightest skin color

Because so many of these combinations exist, the phenotypes fall along a gradient, like a continuous variation.

State example human characteristics that are associated with polygenic inheritance.

• Skin color 

• Height

• Eyecolor

• Body mass

Outline two example environmental factors that can influence phenotypes.

1. Sun exposure - increases the melanin production which affects skin color

2. Nutrition - influences height, weight, and cognitive development

State the direction of movement of gases exchanged in leaves.

Carbon dioxide is from the atmosphere to the leaf into the spongy mesophyll cells

The oxygen goes from the lead to the atmosphere out the leaf

Epidermis

thin and transparent, and allows light to reach mesophyll and gases to diffuse efficiently

Waxy cuticle

waterproof layer, reduces excessive water loss while still allowing gas exchange through the stomata

Stomata

pores mainly on the lower epidermis, allow diffusion of co2 into the leaf and o2 out of the leaf

Guard cells

controls opening and closing of the stomata, regulate gas exchange and water loss

Air spaces

large intercellular spaces inside the leaf, increase surface area for gas diffusion

Spongy mesophyll

loosely packed cells, short diffusion distance between air spaces and photosynthesizing cells

Veins

xylem supplies water needed for photosynthesis, phloem removes sugars produced during photosynthesis

A plan diagram shows the distribution of tissues, but not individual cells. Draw and label a plan diagram to show the distribution of tissues in a transverse section of a leaf.  Include upper and lower epidermis, palisade and spongy mesophyll, xylem and phloem.

Calculate stomatal density from a leaf cast or micrograph.

Stomatal density = number of stomata / area of lead surface (mm^2)

Interpret micrographs or cast of leaf surfaces to compare stomatal density on different leaf surfaces.

Lower epidermis - higher stomatal density, reduces water loss while maintain gas exchange

Upper epidermis - lower stomatal density, limits direct water loss from sunlight exposure

Stomata are usually more numerous on the lower surface of leaves to balance efficient gas exchange with reduced transpiration

Draw a plan diagram to show the distribution of tissues in a stem, including vascular bundles, xylem, phloem, cambium, cortex, pith and epidermis.

Outline the function of tissues in a stem, including vascular bundles, xylem, phloem, cambium, cortex, pith and epidermis.

Epidermis - protects stem and reduces water loss

Cortex - storage of carbohydrates, structural support

Vascular bundles - transport system of the plant

Xylem - transports water and mineral ions upward, provides mechanical support

Phloem - transport sugars and organic nutrients (translocation)

Cambium - produces new xylem and phloem, allows secondary growth

Pith - storage of nutrients, help maintain stem rigidity

State two ways xylem and phloem can be differentiated in cross sections of stem.

Position - xylem on the inside, phloem on the outside of vascular bundles

Structure - xylem has thick and lingified walls, phloem has thin non lingnified walls

Draw a plan diagram to show the distribution of tissues in a root, including vascular bundles, xylem, phloem, cortex and epidermis.

Outline the function of tissues in a root, including vascular bundles, xylem, phloem, cortex and epidermis.

Epidermis - absorbs water and mineral ions, often has root hairs

Cortex - stores carbohydrates, allows movement of water to xylem

Xylem - transports water and mineral ions upward

Phloem - transports sugars to root tissues

Vascular bundle - conducts materials throughout the plant

Describe how the structure of xylem vessels are adapted for the transport of water under low pressure.

No cytoplasm = minimal resistance to flow

No or incomplete end walls = continuous water column

Narrow diameter = maintains cohesion of water

Pits allow movement of water between vessels

Outline how xylem is able to maintain rigidity even under low pressure or mechanical disturbance.

Lignified walls strengthened vessels

Prevent collapse under tension

Provide resistance to mechanical stress

List conditions in which a plant may generate root pressure to transport water.

• High humidity

• Nighttime

• Early spring

• Low transportation rates

Outline the mechanism by which roots maintain a positive pressure potential when evaporation from leaves is insufficient to move water through a plant.

1. Mineral ions actively transported into xylem

2. Water enters xylem by osmosis

3. Positive pressure develops in xylem

4. Water is pushed upward through the stem

Define translocation, phloem sap, source and sink.

Translocation - movement of organic nutrients through phloem

Phloem sap - sucrose rich solution

Source - tissue that releases sugars

Sink - tissues that use or stores sugar

List example source and sink tissues.

Sources - leaves, storage organs releasing sugars

Sinks - roots, fruits, seeds, growing shoots

Phloem transport is bidirectional. Outline the stages of phloem translocation including loading of carbohydrates at a source, transport of carbohydrates through the plant, and unloading of carbohydrates at a sink.

1. Loading - sucrose actively loaded into sieve tubes at source

2. Transport - water enters by osmosis which leads to a mass flow

3. Unloading - sugars removed at sink for respiration or storage

Outline the structure and function of sieve tube cells, with specific mention of the rigid cell wall, reduced cytoplasm and organelles, no nucleus and sieve plates.

Rigid cell walls → support

Reduced cytoplasm and organelles → low resistance 

No nucleus → more space for sap

Sieve plates → allow flow between cells

Outline the structure and function of companion cells, with specific mention of mitochondria and plasmodesmata.

Many mitochondria –> Atp for active transport

Plasmodesmata → direct transfer of sugars to sieve tubes

Contrast positive and negative tropism.

Positive tropism - growth toward a stimulus

Negative tropism - growth away from a stimulus

Outline phototropism and gravitropism in roots and stems.

Phototropism 

• Shoots = positive photoropism (grow toward light)

• Roots = negative phototropism (grow away from light)

Gravitropism 

• Roots = positive gravitropism (grow downward, toward gravity)

• Shorts = negative gravitropism (grow upward, away from gravity)

Outline the cause and consequence of positive phototropism in a plant shoot.

Causes - light detected by the shoot tip, auxin redistrubtes to the shaded side of the shoot

Consequences - cells on the shaded side elongated more, shoot bends toward the light source, maximizes light absorption for photosynthesis

Phytohormones are signalling chemicals that control growth, development, and response to stimuli in plants. List examples of chemicals that function as phytohormones.

• Auxin

• Cytokinin

• Gibberellin

• Ethylene

• Abscisic acid

Outline role of phytohormones in plant growth, development and response to stimuli.

• Controls cell division and elongation

• Regulate development (roots, shoots, flowers)

• Coordinate responses to environmental stimuli

• Integrate growth across different tissues

State two roles of the hormone auxin.

1. Prompts cell elongation in shoots

2. Controls directional growth (phototropism and gravitropism)

Describe the mechanism of movement of auxin into and between plant cells.

• Auxin diffuses into plant cells freely

• Auzin cannot diffuse out

• Auxin efflux carriers actively pimp auxin out on one side of the cell

• Coordinated placement of carriers creates directional transport

• Results in auxin concentration gradients within tissues

Explain how auxin concentrations allow for phototropism.

• Auxin accumulates on the shaded side of the shoot

• Higher auxin concentrations causes faster cell elongation

• Unequal growth rates cause bending toward light

Describe the mechanism of action of auxin in the phototropic response, including the role of H+ ions and cellulose crosslinks.

1. Auxin simulates proton pumps in the cell membrane

2. Hydrogen ions enter the apoplast

3. Cell wall ph decreases

4. Acidic conditions loosen cellulose cross-links

5. Cell wall becomes more extensible

6. Cell elongates under turgor pressure

Outline the source and transport of auxin and cytokinin in plants.

Auxin - produced in shoot tips, transported downward to roots

Cytokinin - produced in root tips, transported upward to shoots

Explain how root and shoot growth are regulated by the interaction of auxin and cytokinin.

High auxin : cytokinin ration = promotes root growth

High cytokinin : auxin ration = promotes shoot growth

Interaction ensures balanced development 

Coordinates growth between roots nd shoots

Outline the lifecycle stages of flowering plants.

• Seed dispersal

• Germination

• Growth of seedling into mature plant

• Flower formation

• Pollination

• Fertilization

• Seed and fruit development

Identify the location of gametogenesis in flowers.

Male gametes - formed in pollen grains in the anthers

Female gametes - formed in ovules inside the ovary

Contrast pollination and fertilization.

Pollination - transfer of pollen from anther to stigma

Fertilization - fusion of male and female gametes to form a zygote (embryo)

Draw and label an insect pollinated flower, including: petals, sepals, stamen, anthers, filaments, pollen, carpel, stigma, style, ovary and ovule.

State the function of the different parts of the animal-pollinated flower.

Petals - attract insects

Sepals - protect flower bud

Anther - produces pollen

Filament - holds anther in position

Pollen - contains male gametes

Stigma - receives pollen

Style - supports stigma and guides pollen tubes

Ovary - contains ovules

Ovule - contains female gametes, becomes seeds after fertilization

Outline structures of insect-pollinated flowers  that aid in the attraction of insects and transfer of pollen between flowers.

• Brightly colored petals

• Scent

• Nectar

• Sticky or spiky pollen

• Stigma positioned to brush insects

Outline the reason for homologous structures in flowers.

Flowers share a common evolutionary ancestor

Same basic structures adopted for different pollination methods

Cross-pollination is the transfer of pollen from the anther (male part) of a flower on one plant to the stigma (female part) of a flower on a different plant of the same species. Outline the benefits of cross-pollination and self-incompatibility in flowering plants

• Increases genetic varuation

• Produces more vigorous offspring 

• Reduces risk of inherited defects

• Improves adaptability to environmental change

List methods for promoting cross-pollination in flowering plants

• Different maturation times of pollen and stigma

• Separate male and female flowers

• Male and female plants

• Wind pollination

• Animal pollination

Outline why self-pollination is generally avoided, even in hermaphroditic plants.

• Leads to inbreeding

• Reduces genetic diversity

• Decreases survival and vigor

Outline the mechanism that promotes self-incompatibility in flowering plants.

• Stigma recognizes pollen from the same plant

• Chemical recognition prevents pollen tube growth

• Fertilization only occurs with pollen from a different plant

Distinguish seed dispersal from pollination.

Pollination - movement of pollen

Seed dispersal - movement of seed away from parent plant

The primary function of a seed is to propagate and disperse a plant species to new locations and ensure its survival through periods unfavorable for growth. List mechanisms of seed dispersal.

• Wind 

• Water

• Animals (inernal and external)

• explosion

Germination is the process by which a plant grows from a seed or spore after a period of dormancy. Outline why water, oxygen and warmth are required for germination

Water - activates enzymes and metabolism

Oxygen - for aerobic respiration

Warmth - optimum enzyme activity

Outline the role of gibberellin during germination.

• Released by embryo

• Stimulates aleurone layer to produce amylase

• Amylase breaks starch into sugars

• Sugars fuel embryo growth

The primary functions of fruits are to protect the developing seeds and to facilitate their dispersal away from the parent plant.  List changes that occur to a fruit as it ripens.

• Fruit softens

• Starch converted to sugars

• Color changes

• Aroma develops

Describe the positive feedback mechanism of fruit ripening.

• Ethylene simulates ripening

• Ripening increases ethylene production

• Increasing ethylene accelerates ripening further

Outline why fruit ripening has evolved to be rapid and synchronized.

• Ensures efficient seed dispersal

• Attracts animals at the same time

• Reduces risk of seed loss

What properties of RNA suggest it was the first genetic material?

RNA can store information, self-replicate, and catalyze reactions (ribozymes). It is versatile and central in modern cells (mRNA, tRNA, rRNA), supporting the “RNA world” hypothesis.