Materials: Corn (Zea mays) kernels Rice (Oryza sativa) grains Mongo (Vigna radiata) seeds Squash (Cucurbita maxima) seeds 1-L Container Procedure: 1. Into a container put 270 corn kernels, 90 rice grains, 90 mongo seeds and 30 squash seeds. Mix them together by stirring or shaking. 2. Supposing that we are crossing two double heterozygotic pea plants (AaBb x AaBb). A=tall; a=Dwarf; B=Normal; b=wrinkled. 3. The four types of seeds represent the four possible phenotypes of the cross: Tall-normal (Corn), Tall-wrinkled (Rice), Dwarf-normal (Mongo) and Dwarfwrinkled (Squash). Note: The materials (e.g. seeds) may be changed with whatever are available at home. 4. From the container, select randomly 96 seeds and categorize them into four phenotype classes as above. The actual number of pellets per phenotype class will constitute the observed (O) frequency or actual observation. 5. The expected (E) results are not exactly equal to the actual observation. This raises the problem of how much deviation from the expected can be accepted due purely to chance. Thus, a mathematical tool such as Chi- square(X?) is needed to determine the "goodness of fit". 6. Determining the expected or predictive ratio. The expected number of individual that are homozygous recessive for both characters will be
Materials: Corn (Zea mays) kernels Rice (Oryza sativa) grains Mongo (Vigna radiata) seeds Squash (Cucurbita maxima) seeds 1-L Container Procedure: 1. Into a container put 270 corn kernels, 90 rice grains, 90 mongo seeds and 30 squash seeds. Mix them together by stirring or shaking. 2. Supposing that we are crossing two double heterozygotic pea plants (AaBb x AaBb). A=tall; a=Dwarf; B=Normal; b=wrinkled. 3. The four types of seeds represent the four possible phenotypes of the cross: Tall-normal (Corn), Tall-wrinkled (Rice), Dwarf-normal (Mongo) and Dwarfwrinkled (Squash). Note: The materials (e.g. seeds) may be changed with whatever are available at home. 4. From the container, select randomly 96 seeds and categorize them into four phenotype classes as above. The actual number of pellets per phenotype class will constitute the observed (O) frequency or actual observation. 5. The expected (E) results are not exactly equal to the actual observation. This raises the problem of how much deviation from the expected can be accepted due purely to chance. Thus, a mathematical tool such as Chi- square(X?) is needed to determine the "goodness of fit". 6. Determining the expected or predictive ratio. The expected number of individual that are homozygous recessive for both characters will be
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
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Transcribed Image Text:Materials:
Corn (Zea mays) kernels
Rice (Oryza sativa) grains
Mongo (Vigna radiata) seeds
Squash (Cucurbita maxima) seeds
1-L Container
Procedure:
1. Into a container put 270 corn kernels, 90 rice grains, 90 mongo seeds and 30
squash seeds. Mix them together by stirring or shaking.
2. Supposing that we are crossing two double heterozygotic pea plants (AaBb x
AaBb). A=tall; a=Dwarf; B=Normal; b=wrinkled.
3. The four types of seeds represent the four possible phenotypes of the cross:
Tall-normal (Corn), Tall-wrinkled (Rice), Dwarf-normal (Mongo) and
Dwarfwrinkled (Squash).
Note: The materials (e.g. seeds) may be changed with whatever are
available at home.
4. From the container, select randomly 96 seeds and categorize them into four
phenotype classes as above. The actual number of pellets per phenotype
class will constitute the observed (0) frequency or actual observation.
5. The expected (E) results are not exactly equal to the actual observation. This
raises the problem of how much deviation from the expected can be
accepted due purely to chance. Thus, a mathematical tool such as Chi-
square(X) is needed to determine the "goodness of fit".
6. Determining the expected or predictive ratio. The expected number of
individual that are homozygous recessive for both characters will be
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