Passive Absorption — Some General Concepts
For a herbicide to cause plant injury or death, it must first enter the plant cell. To do this, most herbicides are thought to move across plant membranes via passive diffusion. Passive diffusion can be defined as solute movement down a concentration gradient (i.e. from high to low concentration), and thus it requires no energy inputs (Figure 3).
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Figure 3. Diagram illustrating the concept of diffusion, with herbicide molecules passively moving from an area of high concentration on the right, to a lower concentration on the left and into the plant root system. (Image by Dusti Duffy, Tracy Sterling, Scott Nissen, and Deana Namuth)
The most important physiochemical properties of a herbicide molecule in terms of passive absorption by plant cells are lipophilicity and acidity. These are discussed next.
Lipophilicity is measured by the octanol:water partition coefficient (log Kow). The log Kow is the ratio of herbicide that is soluble in octanol (organic or non-polar solvent) as compared to water (a polar solvent). Log Kow is a good indicator of a herbicide’s lipophilic or hydrophilic nature. A herbicide with a low log Kow value such as 0.6 is more water soluble (hydrophilic or polar), while a herbicide with a high log Kow value, such as 2.6 is more lipid soluble (lipophilic). Molecules with a charge, or water-loving functional group (i.e. –OH), are more hydrophilic, while molecules with many hydrocarbons (i.e. -CH2CH2CH3) are more lipophilic. Some examples of herbicides with low log Kow values include glyphosate and paraquat. Examples of herbicides with high log Kow values include atrazine and oxyflurofen. By following the link for “cellular absorption” in this animation, Herbicide Uptake by Leaves and Cells, you can test the different absorption rates of molecules with four different log Kow values. Figures 4a and 4b show the screen where you can find this animation.
Figure 4a. This image shows a screen capture of an animation. You can view the animation. If you are inside the interactive animation, click and access the cellular absorption activity, indicated with a red box in this image. (Image by Dusti Duffy, Tracy Sterling, Scott Nissen, and Deana Namuth)
Figure 4b. A screen shot of the animation view. You can view the animation depicting different effects of Kow. If you are using the interactive animation, click on the leaf and then click past the to find “cellular absorption” with different Kow values. (Image by Dusti Duffy, Tracy Sterling, Scott Nissen, and Deana Namuth)
A note on Kow You will normally see ‘Kow’ expressed as ‘log Kow’. ‘Log Kow’ is the more useful expression because values for Kow can range over 7 orders of magnitude. For example, Kow = 10,000 means that there is 10,000 times more herbicide partitioned into the octanol phase compared to the water phase. In scientific notation, 10,000 = 1 x 10 to the 5th power. The log of 1x10 to the 5th is ‘5’. So, the log Kow for this herbicide is 5. Another example: Kow = 0.01 means that there are 100 times more herbicide molecules partitioned into the water phase compared to the octanol phase. In scientific notation, 0.01 = 1 x 10 to the minus 2nd power. The log of 1 x 10 to the minus 2nd power is -2. So, log Kow for this herbicide is -2.
Thinking Question:
What is the log Kow for a herbicide that is equally soluble in octanol and water?
ANSWER: If the herbicide is equally soluble in water and octanol, the ratio would be 1:1 and the log of 1 is zero. So the log Kow for this would be 0.
In general, those herbicides that are more lipophilic in nature will have an easier time moving across membranes. If they are extremely lipophilic (high log Kow), these lipophilic molecules may actually partition into the membrane and remain there. In contrast, hydrophilic herbicides (low log Kow) will move more slowly across lipophilic membranes, and if the log Kow is small enough, these hydrophilic molecules may not be able to enter the cell at all. Further details on the importance of lipophilicity in herbicide transport throughout the plant are discussed in the Foliar Absorption and Phloem Translocation lesson.
Acidity is the second important physiochemical property in herbicide absorption. Acidity, or the dissociation constant (pKa), determines the relative proportion of an acid and its conjugate base present at a particular pH. The pKa value is that pH at which the herbicide is present in a 1:1 ratio of ionized (charged, hydrophilic) to non-ionized (uncharged, hydrophobic) form. If the concepts of ionization, pH, and pKa are unclear for you, view this animation:
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Figure 5. A screen shot of the animation. You can view the animation on Ionization. (Image by Jeremy Steele, Vishal Talpallikar, and Alex Martin)
The relative amounts of ionized versus non-ionized forms of a herbicide can affect how readily it crosses the plasma membrane. For example, ionized forms will move across membranes more slowly than non-ionized forms. For ionizable herbicides (i.e. bentazon), it is the pH of the solution and the pKa of the molecule which determines the degree to which the herbicide is present in an ionized or non-ionized form. At physiological pH in the cytoplasm (i.e. pH 7), bentazon with its pKa of 2.45 would be predominately ionized and thus less likely to traverse a lipophilic membrane. However, if bentazon were in the cell wall free space where the pH is acidic, more of it would be in its acid form, or be protonated, thus without a charge and lipophilic, readily able to traverse a lipophilic membrane. Follow the “what is COO-?” link in the Herbicide Uptake by Leaves and Cells animation for more information on this concept (Figure 6a). In Figure 6b, a screen capture of this animation, you will notice that the herbicide depicted is in its acid form or protonated on the left side of the figure, and ionized or charged on the right side; the proportions of these two molecules will change depending on the solution pH. Let's follow an example with a weak acid herbicide such as 2,4-D, which has a pKa = 2.2. If 2,4-D were in an acidic solution, the equilibrium would shift toward the left such that more of the herbicide would be in its acid, lipophilic form. In contrast, if 2,4-D were in a solution with an alkaline pH, the equilibrium would shift toward the right, with more of the herbicide present as a charged, hydrophilic molecule. Thus, the pH in which the herbicide resides will dictate how much of it is able to traverse a membrane.
Figure 6a. This image shows a screen capture of an animation. You can view the animation. If you are in the interactive animation, click and access the “What is COO-?” activity, indicated with a red box in this image. (Image by Dusti Duffy, Tracy Sterling, Scott Nissen, and Deana Namuth)
Figure 6b. Animation screen capture indicating the equilibrium balance of acidic and basic molecules of a herbicide. This would be an example of the herbicide, 2,4-D. (Image by Dusti Duffy, Tracy Sterling, Scott Nissen, and Deana Namuth)
In general, those herbicides that have ionizable groups vary in their ability to cross lipophilic membranes depending on the pH of the solution. Those herbicides that are uncharged are more lipophilic and therefore more readily able to cross a lipid bilayer. Those that are charged are more hydrophilic and therefore will traverse a lipid bilayer more slowly. This concept is explained further in the next section of this lesson. Further details on the importance of acidity in herbicide transport throughout the plant are discussed in the Foliar Absorption and Phloem Translocation lesson.
Herbicides that are absorbed via passive diffusion can be separated into two classes, depending on the physiochemical characteristics of the herbicide molecule: 1) a lipophilic, neutral molecule or 2) a lipophilic molecule with a pH-sensitive functional group which can dissociate into a less lipophilic ion. This will be our next topic of discussion.