{"id":119,"date":"2015-07-18T04:03:34","date_gmt":"2015-07-18T04:03:34","guid":{"rendered":"https:\/\/courses.candelalearning.com\/bio2labsxmaster2\/?post_type=chapter&p=119"},"modified":"2016-01-08T22:21:04","modified_gmt":"2016-01-08T22:21:04","slug":"natal-bean-discrimination-by-bean-beetles","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-bio2labs\/chapter\/natal-bean-discrimination-by-bean-beetles\/","title":{"raw":"Natal Bean Discrimination by Bean Beetles","rendered":"Natal Bean Discrimination by Bean Beetles"},"content":{"raw":"

Adapted from C. Beck and L. Blumer by Staci Forgey, TCC Biology Faculty<\/span><\/em><\/p>\r\n\r\n

\r\n

Objective<\/b><\/span><\/h2>\r\n

Design and perform a set of experiments to evaluate whether female bean beetles (Callosobruchus maculatus<\/i>) discriminate between suitable species of beans.<\/span><\/p>\r\n\r\n<\/div>\r\n

Introduction<\/b><\/span><\/h2>\r\n

Bean beetles, Callosobruchus maculatus<\/i>, are herbivorous pest insects that are found in Africa and Asia.\u00a0<\/span>Females lay their eggs on the surface of beans.\u00a0<\/span>Eggs are layed singly, and hatch into larvae (maggots) several days later.\u00a0<\/span>The larva then burrows into the bean and will form a pupa 21\u201330 days after the egg was deposited.\u00a0<\/span>They mature 24\u201336 hours after emergence from the pupa and do not need to feed.<\/span><\/p>\r\n

Adults typically live for 1\u20132 weeks.\u00a0<\/span>Mating and oviposition occur during this time period.\u00a0 <\/span>Females will choose the best substrate (bean) to lay their offspring on, since the larvae cannot move.\u00a0<\/span>By choosing a substrate for oviposition, the female chooses the food resource available to her offspring (Brown and Downhower 1988).\u00a0<\/span>This is a critical choice for the female, as it influences the growth, survival and future reproductive success of her offspring (Mitchell, 1975, Wasserman and Futuyma, 1981).\u00a0<\/span>Females can lay eggs on a wide range of bean species, but very few bean species will result in normal development and the successful emergence of adults.\u00a0<\/span>Some species of beans are toxic to larvae (Janzen 1977).\u00a0<\/span><\/p>\r\n\r\n

Materials<\/b><\/span><\/h2>\r\n

In class, you will be provided with live cultures of bean beetles containing adults that have been raised on mung beans (Vigna radiata<\/i>). <\/span><\/p>\r\n

Female beetles are easily identified in the live cultures because they have two dark stripes on the posterior of the abdomen, whereas the posterior abdomen of males is uniformly light in color. <\/span><\/p>\r\n

You will also have access to petri dishes and several types of beans.\u00a0<\/span><\/p>\r\n\r\n

Experimental Design<\/b><\/span><\/h2>\r\n

Since the oviposition choices of females influences the survival and future success of their offspring, females may be very sensitive to the species and condition of the beans on which they are depositing eggs. <\/span><\/p>\r\n

Each group should design a set of experiments to address whether female bean beetles discriminate between suitable species of beans. We will re-visit the experiment next week and collect our data set.\u00a0 <\/span>We will choose one experiment to perform as a class.\u00a0<\/span><\/p>\r\n

As a group, list several ways we might examine female bean choice for oviposition.\u00a0 <\/span>We will pick one experiment from the class to perform.\u00a0 <\/span>Write your ideas below.<\/span><\/p>\r\n

**Stop here.\u00a0<\/span>We will go over ideas for our experiment as a class and choose one to perform**<\/b><\/span><\/p>\r\n

Outline the agreed on experiment using the criteria\u00a0below.\u00a0<\/span><\/p>\r\n\r\n

    \r\n\t
  1. Formulate a hypothesis for this week\u2019s experiment.\u00a0<\/span>Be specific!<\/span><\/li>\r\n\t
  2. We will re-visit the experiment in a week to collect data on the number of eggs laid.\u00a0 <\/span>Formulate a hypothesis for this experiment. \u00a0 \u00a0<\/span><\/span><\/li>\r\n\t
  3. Identify the independent variable(s).<\/span><\/li>\r\n\t
  4. Identify the dependent variable(s).<\/span><\/li>\r\n\t
  5. What variables will you keep standard?<\/span><\/li>\r\n\t
  6. What is your control?<\/span><\/li>\r\n\t
  7. Design data collection tables for both of your experiments on a separate sheet of paper.<\/span><\/li>\r\n<\/ol>\r\nFor this data, it is most useful to perform a Chi Square analysis. Chi Square statistical test evaluate whether this is a significant difference between groups of data. The null hypothesis (H0<\/sub>) is the hypothesis that states that there is no difference between the groups of data. The alternative hypothesis, (Ha<\/sub>) is the hypothesis you outlined above.\r\n
      \r\n\t
    1. Restate your H0<\/sub> hypothesis.<\/span><\/li>\r\n\t
    2. Ha<\/sub> hypothesis.<\/span><\/li>\r\n\t
    3. If the null hypothesis is correct, what would we expect to see?<\/span><\/li>\r\n\t
    4. If our alternative hypothesis is correct, what would we expect to see?<\/span><\/li>\r\n<\/ol>\r\n

      To do this statistical test, we need to calculate observed and expected values for the bean species and for our control.\u00a0 <\/span>Our expected values are based on our null hypothesis (that there is no difference).<\/span><\/p>\r\n

      For example, if we had 40 eggs and four different treatment groups, we would expect there to be 10 eggs on each group.\u00a0<\/span><\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      Treatment<\/th>\r\nObserved number of eggs<\/th>\r\nExpected number of eggs<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      Mung beans<\/td>\r\n5<\/td>\r\n10<\/td>\r\n<\/tr>\r\n
      Pinto beans<\/td>\r\n25<\/td>\r\n10<\/td>\r\n<\/tr>\r\n
      Chick peas<\/td>\r\n10<\/td>\r\n10<\/td>\r\n<\/tr>\r\n
      Control<\/td>\r\n0<\/td>\r\n10<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

      Fill in the data table below with our observed (what we actually found) and expected values.\u00a0<\/b><\/span><\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      Treatment<\/th>\r\nObserved number of eggs<\/th>\r\nExpected number of eggs<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      <\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n
      <\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n
      <\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n
      <\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n
      <\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

      To calculate the chi-square value, or \u03c7 <\/span>2 <\/sup><\/span>, we simply add the square differences, divided by the expected, of all the observed and expected.\u00a0 In mathematical terms:<\/span><\/p>\r\n

      [latex]\\displaystyle\\chi^2=\\Sigma\\frac{(O-E)^2}{E}[\/latex]<\/p>\r\n

      So for our example from the previous sample table, the first [latex]\\displaystyle\\frac{(O-E)^2}{E}[\/latex]<\/span>\u00a0<\/span>would be [latex]\\displaystyle\\frac{(5-10)^2}{10}=2.5[\/latex]\u00a0<\/span><\/p>\r\n

      We would then add to this value all of the other[latex]\\displaystyle\\frac{(O-E)^2}{E}[\/latex]<\/span>\u00a0in the table and then add to get the[latex]\\displaystyle\\chi^2[\/latex]<\/span>\u00a0\u00a0<\/span><\/sup><\/span>value.\u00a0<\/span><\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      Treatment<\/th>\r\nObserved number of eggs<\/th>\r\nExpected number of eggs<\/th>\r\nObserved-Expected<\/th>\r\n(Observed-Expected)2<\/sup><\/th>\r\n(Observed-Expected)2<\/sup>\/Expected<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      <\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n
      <\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n
      <\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n
      <\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n
      <\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nCompute the \u03c72<\/sup> value for your data by adding all of the (Observed-Expected)2\/Expected values.\r\n\r\nIn order to find something to compare this number with, we need to calculate the degrees of freedom, or the number of different comparisons that can be made within the table. The degrees of freedom are the number of columns (M) minus one times the number of rows (N) minus one.\r\n\r\nDegrees of freedom = (M \u2212 1)(N \u2212 1)\r\n\r\n\r\n\r\n\r\n
      A<\/td>\r\nB<\/td>\r\n<\/tr>\r\n
      C<\/td>\r\nD\u00a0<\/b><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

      How many ways can this two by two table be broken into individual comparisons?\u00a0 <\/span>Hint:\u00a0 <\/span>Use the formula above.\u00a0<\/span><\/p>\r\n

      Calculate the degrees of freedom in your experiment.\u00a0<\/b><\/span><\/p>\r\n

      Now we can compare against the Chi distribution for the likelihood that our data is generated by chance.\u00a0<\/span>Remember, we are looking at the column labeled 0.05 for our p value.\u00a0 <\/span>This means that at this level, we are 95% sure our results are real.\u00a0<\/span>The Chi distribution table gives us a critical value to compare our test value to.\u00a0<\/span><\/p>\r\n\r\n