IB Biology A1.1: Water worked solutions

C earns [1]. Oxygen attracts the shared electrons more strongly. The resulting charges are partial; the molecule has no overall charge.

Identify an H already covalently bonded to O in molecule 1 and an O in molecule 2 [1]. An O–O line or a bond within one molecule does not represent this interaction.

(a) Adhesion acts between different materials [1]. (b) Cohesion acts between water molecules [1]. Cellulose can attract water without dissolving.

Surface tension resists increasing or disrupting the interface [1]. Cohesion arising from hydrogen bonds between water molecules produces this effect [1].

Neighbouring water molecules attract through hydrogen bonds [1]. This cohesion resists separation, transmitting tension along the continuous column [1]. Cohesion does not itself create the pull.

Water is attracted to polar or charged groups in the wall [1]. Adhesion wets the surface and draws water into the narrow space, while cohesion allows adjacent water to follow [1].

Oxygen faces \mathrm{K^+} [1]. Opposite charges attract; the oxygen end of water is partially negative [1]. The ion retains its positive charge when hydrated.

Water earns [1]. With equal Q and m, \Delta T=Q/(mc), so doubling c halves the rise. These are supplied rounded model values.

0.75\times80=60\,\mathrm{g} [1]. The animal initially accelerates upwards [1] because the displaced-water weight (84 g equivalent) exceeds its own weight (80 g equivalent) [1]. Internal water content alone does not determine buoyancy.

Impacts of water-bearing asteroids are a proposed source [1]. Sufficient planetary gravity limits loss to space, or cooling permits vapour to condense [1]. Delivery and retention are different claims.

Draw each molecule with one O bonded to two H atoms in a bent arrangement [1]. Label oxygen \delta- and both hydrogens \delta+ [1]. Show solid O–H bonds within each molecule [1]. Draw and label a dashed link from H on one molecule to O on the other [1]. No numerical bond angle is required.

Water orients its opposite partial charge towards each ion, stabilising separated ions [1]. Glucose has polar hydroxyl groups that form hydrogen bonds with water [1]. The mainly non-polar triglyceride lacks sufficient favourable interactions to disperse in water [1]. In animals, triglycerides can travel in lipoprotein particles whose outer surface interacts with water [1]. Do not claim that every protein is soluble.

(a) Use Q=mc\Delta T [1]; 25\times4.2\times4.0=420\,\mathrm{J} [1]. (b) Vaporisation separates molecules by overcoming intermolecular hydrogen bonding [1]. The covalent O–H bonds remain, so the vapour consists of water molecules [1].

\Delta m=0.20\,\mathrm{g} [1], so \Delta Q=0.20\times2400=480\,\mathrm{J} [1]. Energy is needed to overcome intermolecular attractions as water escapes [1]. Transfer of this energy from the pad and nearby material can lower its temperature [1]. This alone does not determine final temperatures because heat can enter from the surroundings.

Under the stated equal conditions the rate is proportional to conductivity [1]. 0.60/0.025=24, giving water:air 24:1 [1]. A warm seal loses heat down its temperature gradient to cold water [1]. Blubber provides a relatively poorly conducting, thick layer that reduces this transfer [1]. The ratio is not a measurement of total heat loss from a swimming seal.

Water provides greater upthrust for a given displaced volume because it is denser [1]. Its greater viscosity contributes to resistance to movement [1]. A streamlined body reduces drag [1]. Webbed feet provide a surface for propulsion [1]. A pointed bill helps seize aquatic prey [1]. The bird still has to breathe air; buoyancy does not supply oxygen.

The orbital region permits temperatures compatible with surface liquid water under suitable conditions [1]. An appropriate atmosphere and surface pressure are needed [1]. There must be water available and retained on the planet [1]. Possible liquid water indicates an opportunity for life, not evidence of organisms or metabolism [1].

Water transmits useful visible light, allowing submerged producers to receive light for photosynthesis [1]. Dissolved reactants can move and encounter enzymes in aqueous cell contents [1]. Water acts as a reaction medium, and is also a reactant in processes such as hydrolysis [1]. Ionic or polar nutrients can dissolve by interacting with water [1]. The dissolved particles can then be carried in fluids such as xylem sap or blood plasma [1]. Transparency does not guarantee sufficient light at every depth.

Use internal radius as the independent variable and measure rise height at fixed times, or time to a fixed height [1]. Keep water composition and starting temperature constant [1]. Keep tube material and cleanliness constant [1]. Keep tube orientation and immersion depth constant [1]. Repeat each radius and compare mean curves to account for variation [1]. A smaller radius can increase equilibrium rise but also changes flow resistance; maximum height does not show that movement is faster, and soil geometry is more complex than a straight tube [1].

For the closed copy use \Delta T=2520/(120\times4.2)=5.0\,\mathrm{K} [2: equation and result]. Evaporation takes 0.50\times2400=1200\,\mathrm{J} [1]. Energy available for warming is 2520-1200=1320\,\mathrm{J} [1]. The open rise is 1320/504=2.62\,\mathrm{K} [1]. Real skin exchanges heat by radiation, convection and conduction and has changing blood flow; choose one reason why the isolated energy balance is incomplete [1]. The fixed-mass approximation is supplied, not a biological claim.

The inlet drop is 11.0-8.6=2.4\,\mathrm{mg\,L^{-1}} [1], or 100\times2.4/11.0=21.8\% [1]. Extraction is inlet minus outlet: 1.8, 2.4 and 3.6\,\mathrm{mg\,L^{-1}} [1]. At equal flow this is consistent with increasing oxygen uptake rate [1]. The measured outlet concentrations and differences are observations; faster aerobic respiration is a possible explanation [1]. Inlet solubility and uptake both matter; temperature correlation alone does not establish the mechanism, so measure uptake independently or control oxygen supply [1].

Asteroid delivery could supply water to both planets [1]. Weak gravity makes atmospheric loss easier for R [1]. Strong heating favours vaporisation and can contribute to loss, while suitable pressure and cooler conditions on S permit condensation/liquid water [1]. S is therefore more favourable, but the amount retained also depends on its history and water supply [1]. Isotope agreement is consistent with a shared source [1]. It does not identify a unique source or show that all water arrived that way; mixing and later processing can alter the comparison [1].

Hourly loss is 1.8\times60/15=7.2\,\mathrm{g\,h^{-1}} [1]. Evaporation from wet mesophyll surfaces removes water and lowers water potential locally [1]. Water drawn from xylem contributes to tension [1]. Hydrogen-bond cohesion maintains continuity of the water column [1]. Adhesion keeps water in contact with polar cell-wall surfaces [1]. Stomatal closure usually reduces water-vapour loss and hence transpiration, subject to the stated conditions [1].

The substrate dissolves through favourable interactions with water [1]. Diffusion permits encounters with the enzyme active site [1]. Polar phospholipid heads face the internal and external water [1]. Non-polar tails associate away from water, helping form a stable bilayer [1]. A sodium ion is charged and surrounded by interacting water molecules [1]. Entering the hydrophobic centre is energetically unfavourable, so passage generally requires an appropriate transport protein [1].

Water conducts heat away more readily than air under matched conditions [1]. Blubber or a suitable trapped-air layer limits heat loss [1]. Water resists movement and exerts drag [1]. A streamlined shape or propulsive limbs helps movement [1]. Dissolved oxygen is available only if the organism has a suitable exchange surface; an air-breathing seal must surface [1]. Buoyancy provides upward force and supplies neither oxygen nor metabolic energy [1].