Chapter 2. Soil Genesis

Answer Key

  1. The properties (e.g. texture, mineralogy) of the parent materials that soils form in are inherited from the rocks that were pulverized by the glaciers.
  2. (From table 2.1): Till, Glacio-fluvial gravels, glacio-fluvial sands, glacio-lacustrine silts and clays.
    (The answer will depend perhaps on the region the student is in)
    Least suitable: glacio-fluvial gravels – poor medium for seeding and for water retention (and hard on equipment!)
    Glacio-Fluvial sands: Poor water retention and prone to droughts and wind erosion
    Till: Loam textures are good for seed beds and water retention but gravel complicates field operations
    Glacio-lacustrine silts and clays: good water holding capacity, flat landscapes, few or no stones.
  3. Unglaciated landscapes like the Cypress Hills or the unglaciated region in Yukon.
  4. Additions, losses, transfers, and transformations
    (Alternative answers are possible)
    Figure 2.12: Addition of organic matter
    Figure 2.14B: Transformation (to form red horizons)
    Figure 2.16: Transfers (of clay to create contrasting structures)
    Figure 2.18: Transformation (to create gray and reddish colours)
    Figure 2.19: Addition (of organic matter to form peat)
    Figure 2.20: Transfer (of material through cryoturbation)
    Figure 2.22: Addition (of organic matter)
    Figure 2.24: Addition (of organic matter) or Transfer (of clay to form contrasting structures)
    Figure 2:29: Transformation (to create red soil material)
  5. Podzolization occurs on parent material derived from acidic, coarse bedrock and clay translocation (lessivage) occurs on parent materials derived from sedimentary rocks, which lead to loamy soils with carbonates. Acidic, coarse parent materials are less suitable for plant growth.
  6. The addition of organic materials through paludification builds up great reservoirs of peat in these wetland soils.
  7. Soil mammals mix the organic matter from roots through the upper part of the soil and create the organic-rich topsoil.
  8. Wind, water and tillage erosion. In drier regions like the Prairies we can expect that climate change may lead to more droughts and hence more wind erosion. In moister areas like the rest of the agricultural zone in Canada we may experience more intense precipitation events and hence more water erosion.
  9. Canadian soils are much younger (because of the glaciations) and hence are less weathered and have a higher content of primary minerals such as potassium and calcium; these are important plant nutrients.

Chapter 3. Soil Organic Matter

Chapter 4. Soil Physics

Chapter 5. Soil Chemistry

Chapter 6. Soil Biodiversity and Ecology

Chapter 7. Soil Nutrient Cycling

Answer Key

  1. Water-soluble nutrient ions in the soil pore water move towards the root surface via mass flow and diffusion processes. Growing roots make contact with exchangeable ions on organo-mineral surfaces, a process referred to as root interception. Water-soluble ions in the soil pore water enter the root hairs and diffuse through the epidermis. Once inside the root, water-soluble ions can move through the apoplastic, symplastic or transmembrane pathways until they reach the endodermis. Nutrient ions may be incorporated into structural compounds within these plant organs, or support metabolic processes in the cytoplasm. Furthermore, nutrient ions may be remobilized and translocated to other tissues or organs if they are needed in another part of the plant.
  2. Nutrient Macronutrient or Micronutrient? Source in soil? Role in plant nutrition
    N macronutrient Mineralization of organic matter, biological N2 fixation CO2 fixation, protein synthesis
    Cu micronutrient Weathering/dissolution of rocks and minerals, desorption from exchange sites, chelated forms Enzyme cofactor
    P macronutrient Mineralization of organic matter, weathering of rocks like apatite, dissolution of minerals like Ca phosphates Energy relations
    Ca macronutrient Weathering and dissolution of rocks and minerals, desorption from exchange sites Structural compounds
    Fe micronutrient Weathering/dissolution of rocks and minerals, desorption from exchange sites, chelated forms Heme proteins and cytochromes
    S macronutrient Mineralization of organic matter, weathering of rocks, dissolution of minerals like gypsum Protein synthesis
  3. Potassium fixation causes K+ tends to bind in the interlattice of 2:1 clays, and K+ is also lost from the soil pore water by adsorption in surface complexation reactions.
  4. The oxidation of NH3 to NO3 releases protons (H+) that acidify the soil pore water.
  5. Although nitrogen is a limiting factor for crop growth, nitrogen is susceptible to biological and chemical reactions that reduce its accessibility to crops (e.g., immobilization, denitrification, NH4+ fixation) as well as physical processes that leach NO3 below the root zone. Since there are limits to how much nitrogen fertilizer can be acquired by a crop, because the probability of a profitable yield response to fertilizer declines when more fertilizer is added, adding excessively high rates of nitrogen fertilizer will be costly, will not improve crop growth and will contribute to environmental pollution.
  6. Plant tissue analysis indicates whether crop nutrition is sufficient, marginal or deficient. While it can be an indication that the fertilization plan met the crop nutrient requirements this year, the diagnosis might arrive too late to take corrective action. Therefore, soil fertility testing is the most important diagnostic tool to avoid a negative economic outcome at the end of the growing season.
  7. The advantage of applying micronutrient fertilizers to crop foliage is that the micronutrients can be absorbed through pores in the leaf and enter the vascular system. This avoids the possibility that micronutrient fertilizers will be transformed into unavailable forms due to biological and chemical reactions in the soil. Foliar micronutrient fertilizers are in direct contact with the plant, whereas soil-applied micronutrient fertilizers must be transported to the root surface and absorbed by the root hairs before they can enter the vascular system.
  8. The right fertilizer source for your crop should be cost-effective. Water-soluble fertilizers with higher analysis values are preferred, since they will be a source of plant-available nutrients during the current growing season. Organic fertilizers that are bulky and wet should be obtained from a local supplier, to lower the transportation and land application costs. Controlled-release fertilizers that release of plant-available nutrients in synchrony with crop demands will be more efficient and give better return on investment, since the nutrient input will translate into higher crop yields.
  9. There are many suitable soil and water conservation practices to prevent nutrient loss through erosion. The general rule is to keep the soil covered with vegetation and residues, and to protect the downstream aquatic environments by establishing buffers with perennial crops and trees along the margins of the agricultural land.

Chapter 8. Soil Classification and Distribution

Answer Key

  1. Climate, Organisms, Parent Material, Topography, Time, Groundwater, Human Activity.
    Topography is most likely responsible for the sequence of soils in the figure. Topography
    controls the redistribution of water in the landscape as shown in Figure 8.2.
  2. Ah (grassland), O (wetland), Bh (forest).
  3. Bv (parent material – high clay glacio-lacustrine parent materials) and By, Cy (climate – caused
    by frost action).
  4. Stripping of the topsoil by wind erosion would leave a minimally developed Regosol behind.
    Human activity removed the native grassland and exposed the soil to the wind. Soils that were
    still in native grassland did not eroded despite the severe drought that occurred in some years in
    the 1930s.
  5. Salts are dissolved in the soil and then transported through the landscape by groundwater.
    Where the groundwater discharges at the soil surface the water is evaporated off and the salts
    deposited on the soil surface.
  6. The CSSC uses a threshold of a pH value of 5.5 to distinguish between the parent materials. The
    split is used in the Brunisolic order to place soils into the appropriate great group.
  7. Folisols have thick organic mats with many thick tree roots present; tree roots are very difficult
    to dig through.
  8. Climate controls the type of native vegetation found in the Prairie Provinces, which in turn
    determines the organic inputs into the soil (and the colour of the Ah horizon).
  9. Soil base status and the exchange capacity (or activity) of clay minerals. Both soil base status and
    the activity of clay is low in older, highly weathered landscapes.
  10. It is not feasible – SIL 4 is a Broad Reconnaissance survey, which would typically only have one
    observation (soil pit) per 100 to 1000 ha. A much more detailed survey at SIL 1 is required for
    pipeline planning purposes.

Chapter 9. Soils of Western Canada: British Columbia and Yukon

Chapter 10. Soils of the Prairie Provinces

Answer Key

  1. The glaciers flowed from Hudson Bay over the Prairies. When they passed over the Paleozoic limestones in central Manitoba (see Figure 10.2) they incorporated large amounts of limestone (calcium carbonate), and the amount of limestone decreased as they spread further through the landscap..
  2. Deflection of warm, moist air masses northward by the Rocky Mountains brings more temperate conditions to the Peace River area and enables the formation of Chernozemic soils.
  3. The Vertisols form on heavy clay parent materials associated with lacustrine/glacio-lacustrine parent materials.
  4. Highest: Humic Vertisol: These soils have high amounts of organic matter and are clay-textured, which allows them to store high amounts of water for crop growth.
    Black Chernozem: High amounts of soil organic matter (SOM) can release nutrients and create a good rooting medium for growth.
    Gray Luvisol: Low SOM limits nutrient availability and the surface mineral horizon (Ae) is susceptible to soil crusting and structural problems.
    Solodized Solonetz: Root-impeding Bnt horizons and salts in the C horizon both severely limit plant growth..
  5. Dystric Brunisols, Organic Cryosols, Organic Mesisols, and Gray Luvisols – there is also lots of exposed rock but rock is not a soil great group. The Gray Luvisols are out of place because they normally occur south of the Shield on till from sedimentary rocks. They occur on the Shield on a large glacio-lacustrine lake bed formed during deglaciation (see Figure 10.3).
  6. High sodium soils and saline soils. High sodium soils are classified into the Solonetzic order but salinity is only represented at the phase level (eg Orthic Black Chernozem, saline phase).
  7. Gleysolic soils are most common in Prairie wetlands. The wetlands these soils are found in are very important waterfowl habitat and are major reservoirs of organic carbon.

Chapter 11. Soils of Ontario

Chapter 12. Soils of Quebec

Chapter 13. Soils of Atlantic Canada

Chapter 14. Digital Soil Mapping

Chapter 15. Soil Health and Management

Answer Key

  1. Set a condition where pH is too low or too high and the test score is divided by 2 or set equal to zero.
  2. Visual assessments (can’t sell, need to be done in field); sulfur (difficult to quantify relative to crop need/management), with justification.
  3. Similarities: growing plants and adding carbon contrasts: apply microbes (bioremediation) or apply N+P (bioaugmentation) to enhance plant growth or breakdown organic contaminants; typically very degraded soil with reclamation/remediation.
  4. Similarities: all increase plant growth or microbial activity; all centre around Carbon; Contrasts: for crop diversity and compost the main mechanism is enhancing carbon inputs while conservation tillage minimizes C and soil loss but compaction reduction, continuous living, CCs do both.
  5. Any crop that has a limited root system or low crop residues would provide low carbon inputs. Any root crop will require significant soil disturbance to harvest which could lead to compaction and extensive tillage. Any short season crop would not assimilate C during a significant portion of the year.
  6. After or before main crop, intersown into standing crop or fallow.
  7. Short term expenses but changes to soil health are expected in the long-term, time constraints, extra management, requires change from the status quo.

Outdoor Activities

  1. DESIGNER SOIL HEALTH TEST:  With justification, test must include chemical, physical and biological indicators and must include a measure of carbon quantity or cycling.

Digital activities

  1. DECISION TOOL: Goals: compaction reduction, weed fighter, organic matter builder, perhaps another goal if well explained. The cover crops selected should be rated 3 or 4 (out of 4) for the goals selected with reasonable explanations for selecting one cover over the other.

Chapter 16. Soil Mineralogy

Chapter 17. Soil Reclamation and Remediation of Disturbed Lands

Answer Key

Problem 1

A) Calculate mass of soil to be amended

0.15 m depth x 10,000 m2 ha-1 = 1,500 m3 ha-1 soil volume

1,500 m3 ha-1 x 1.5 tonnes m-3 = 2,250 tonnes ha-1 soil mass

2,250 tonnes ha-1 x 1,000 kg tonne-1 = 2,250,000 kg ha-1 soil mass

B) Calculate mass of organic matter required

Have 1% OM; 5% – 1% = 4% OM required

Need 2,250,000 kg soil ha-1  x  4% OM = 90,000 kg OM ha-1

C) Calculate % organic matter in wheat straw

OC in straw varies; given 45.5% OC in wheat straw determine OM

OM = 56% OC

1 / 0.56 = 1.78, therefore % OM = 1.78% x OC

45.5% OC x 1.78 = 80.99% OM

D) Determine OM after mineralization

Assume bacterial respiration releases approximately 60% OM as CO2 and 40% into soil

1 kg wheat straw x 80.99% OM x 40% OM to soil = 0.324 kg OM kg-1 wheat straw after mineralization

E) Calculate number bales required to increase OM by 4%

1,200 lb bale-1 / 2.205 lbs kg-1 = 544.22 kg bale-1

90,000 kg OM as straw / 0.324 kg OM to soil = 277,777.8 kg ha-1

277,777.8 kg ha-1 straw / 544.22 kg bale-1 = 510.41 bales ha-1

Therefore need 511 bales ha-1

This was for a 4% OM increase so a 1% OM increase requires ~ 128 bales of straw

Problem 2

A) Determine plant nutrient requirements for fescue grassland

30 kg ha-1 N and 15 kg ha-1 P205 recommended

B) Determine N content of beef manure from literature

Total N = 10 kg tonne-1

Ammonium N = 2.6 kg tonne-1

Organic N = Total N – ammonium; 10 – 2.6 = 7.4 kg tonne-1

C) Determine P content of beef manure from literature

Total P205 = 2.4 kg tonne-1

D) Determine available organic N from manure for first year

25 to 30% N is released from manure in first year

7.4 kg tonne-1 in manure x 0.25 = 1.85 kg tonne-1

E) Determine available ammonium N from manure for first year

25% volatilization loss of ammonium N from literature

2.6 kg tonne-1 x 0.75 = 1.95 kg tonne-1

F) Determine total available N from manure for first year

1.85 kg tonne-1 + 1.95 kg tonne-1 = 3.8 kg N from a tonne of manure

G) Determine application rate based on fescue grassland N requirements

30 kg ha-1 / 3.8 kg tonne-1 = 7.9 tonnes ha-1

7.9 tonnes ha-1 application rate for manure

H) Determine available P2O5 if manure applied for N requirements

2.4 kg tonne-1 in manure x 50% P2O5 available first year = 1.2 kg tonne-1

7.89 tonnes ha-1 x 1.2 kg tonne-1 = 9.47 kg ha-1 P2O5 from manure

Need 15 kg ha-1 P2O5 – 9.47 kg ha-1 P2O5 = 5.53 kg ha-1 additional P2O5 for fescue grassland

I) Determine phosphorous based manure application rate

15 kg ha-1 required / 1.2 kg tonne-1 available first year = 12.5 tonnes ha-1

This application rate of manure will over supply N

Not a good use of manure

Look for alternative sources of P2O5

J) Determine phosphorous based application rate from 11-48-0 fertilizer

5.53 P2O5 / 0.48 = 11.525 kg ha-1 of 11-48-0 should be applied

This rate would apply an additional 11.525 x 0.11 = 1.27 kg ha-1 N

Problem 3

A) Amount of crude oil based C present in a hectare furrow slice (HFS)

Mg = megagram = metric tonne

Volume of HFS: 10,000 m2 x 0.15 m = 1,500 m3

HFS soil mass: volume HFS x bulk density = 1,500 m3 x 1.15 Mg m-3 = 1,725 Mg (dry mass basis)

Total crude oil per HFS: 1,725 Mg soil x 1.25 Mg crude oil 100 Mg-1 soil = 21.6 Mg crude oil

Total crude oil C per HFS: 21.6 Mg crude oil x 0.85 Mg C Mg-1 crude oil = 18.3 Mg crude oil C

B) Amount of fertilizer N needed per HFS

Required = 18.3 Mg crude oil C / 25 Mg crude oil C Mg-1 fertilizer N = 0.733 Mg fertilizer N

C) Amount of urea fertilizer needed per HFS

0.733 Mg fertilizer N / (0.450 Mg N Mg-1 urea) = 1.62 Mg or 1.62 tonnes urea per HFS

Problem 4

A) Amount of Pb to be removed from a HFS

1 Mg (megagram) = 103 kg = 106 g = 109 mg

HFS soil mass: volume HFS x bulk density = 1,500 m3 x 1.22 Mg m-3 = 1,830 Mg (dry mass basis)

Initial soil concentration Pb = 1,640 mg Pb kg-1 soil = 1,640 g Pb Mg-1 soil = 1.640 kg Pb Mg-1 soil

Initial mass Pb in HFS: = 1,830 Mg soil x 1.640 kg Pb Mg-1 soil = 3,001 kg Pb in HFS

Desired soil concentration Pb: = 300 mg Pb kg-1 soil = 300 g Pb Mg-1 soil = 0.300 kg Pb Mg-1 soil

Desired final mass Pb in HFS: = 1,830 Mg soil x 0.300 kg Pb Mg-1 soil = 549 kg Pb in HFS

Required Pb mass removal:  = 3,001 – 549 = 2,452 kg Pb per HFS

B) Amount Pb that can be removed by field grown plants per HFS per season

Dry above ground biomass = 10.0 tonnes per HFS per season (year)

Pb concentration in biomass = 1,215 mg Pb kg-1 biomass = 1,215 g Pb Mg-1 biomass = 1.215 kg Pb Mg-1 biomass

Total Pb in above ground biomass per HFS per season = 10.0 Mg biomass x 1.215 kg Pb Mg-1 biomass = 12.15 kg Pb removed per season

C) Number of seasons (years) required to remove the required amount of Pb

= [ total amount of Pb to be removed per HFS] / [amount removed per HFS per season]

= [2,452 kg Pb per HFS] / [12.15 kg Pb removed per HFS per season]

= 197 years

Thus, this remediation strategy is not very practical for this scenario.



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