Corn Lab
Purpose
To examine genetic crosses using corn.
Objectives
- Solve monohybrid and dihybrid cross problems.
- Use sampling to determine phenotypic ratios of a visible trait in the corn.
- Form hypotheses about genotypic and phenotypic ratios in the F2 generation of corn crosses.
- Use a chi square test to determine whether observed results are consistent with expected results.
Let's Get Started!
Procedure: Develop Null Hypothesis - Monohybrid Corn A
- Create a monohybrid cross with the corncob labeled A.
- Draw 2 punnett squares for the possible crosses to represent the null hypothesis.
- Label each punnett square as either Null Hypothesis #1 or Null Hypothesis #2.
Null Hypothesis #1:
Look at the punnett square and asses the information about the F2 generation.
Possible Gametes of F1:
Look at the punnett square and asses the information about the F2 generation.
Possible Gametes of F1:
- Parent #1 - P and p
- Parent #2 - p and p
- Pp : pp (2 heterozygous and 2 homozygous recessive)
- 1 : 1 (2 purple to 2 yellow)
- Expected proportion of kernels on your cob to be purple is 50%.
- Expected proportion of kernels on your cob to be yellow is 50%.
Null Hypothesis #2:
Look at the punnett square and asses the information about the F2 generation.
Possible Gametes of F1:
Look at the punnett square and asses the information about the F2 generation.
Possible Gametes of F1:
- Parent #1: P and p
- Parent #2: P and p
- PP : Pp : pp (1 homozygous dominant, 2 heterozygous, 1 homozygous recessive)
- 3 : 1 (3 purple to 1 yellow)
- Expected proportion of kernels on your cob to be purple is 75%.
- Expected proportion of kernels on your cob to be yellow is 25%.
Count 100 kernels of the corn on your cob. Record the observations:
- 83 Purple
- 17 Yellow
- 100 Total
Chi Square formula: X^2 = E (observed - expected)^2 / Expected
Use the data obtained by counting the 100 kernels and calculate the chi-square value.
Conclusions:
The null hypothesis #2 (3 : 1) for the chi square probability is correct because it is 3.4133 which is less than 3.84 and likely to happen because the observed deviation is due to chance alone. The chance is greater than 5% so it is accepted.
Conclusions:
- Degree of Freedom = 1
- Results agree with proportions expected after completing the chi square test.
- Chi square probability value for null hypothesis #1: 43.56 p<0.001.
- Chi square probability value for null hypothesis #2: 3.4133 p>0.05.
The null hypothesis #2 (3 : 1) for the chi square probability is correct because it is 3.4133 which is less than 3.84 and likely to happen because the observed deviation is due to chance alone. The chance is greater than 5% so it is accepted.
Dihybrid Cross: P1 cross=PPSS x ???? F2 Generation is CornB
- Obtain corncob labeled B.
- There are four grain phenotypes as shown in the picture below. A) Purple & Starchy (smooth) B) Purple & Sweet (wrinkled) C) Yellow & Starchy D) Yellow & Sweet.
- Phenotypes are produced by two pairs of heterozygous genes, each on a different chromosome.
- Law of Independent Assortment can easily be demonstrated through F1 x F1 crossing to equal F2.
- Punnett squares for the P1 x P1 cross and for the F1 x F1 cross are shown below.
From the F1 x F1 cross record the F2 generation results:
Observed...
- Gametes: PS, Ps, pS, ps (Both Parents)
- Genotypes: PPSS (1/16), PPSs (2/16), PpSS (2/16), PpSs (4/16), PPss (1/16), Ppss (2/16), ppSs (2/16), ppss (1/16), ppSS (1/16).
- Phenotypes: 9:3:3:1 (9 PS, 3 Ps, 3 pS, 1 ps)
- PPSS x ppss
- Purple & Smooth = 9/16
- Purple & Wrinkled = 3/16
- Yellow & Smooth = 3/16
- Yellow & Wrinkled = 1/16
Observed...
- Purple & Smooth = 54
- Purple & Wrinkled = 19
- Yellow & Smooth = 20
- Yellow & Wrinkled = 7
- Total = 100
- Degree of Freedom = 3
- Chi square probability value = 0.2963 p>0.9
Dihybrid Cross: F1 cross = P?S? x ???? F2 Generation is CornC
- Obtain corncob labeled C.
- There are four different genes and grain types once again - the same letters (A-D) apply to the corresponding phenotypes.
- Phenotypes are produced by two pairs of genes located on two pairs of homologous chromosomes.
- The original P1 was purple & smooth and the other P1 was yellow & wrinkled.
- Observe the corncob and create a null hypothesis.
- Punnett square of possible F1 x F1 cross is shown below.
From the F1 x F1 cross record the results:
Observed...
- Gametes: PS, Ps, pS, ps (Parent 1) and ps (Parent 2)
- Genotypes: PpSs, Ppss, ppSs, ppss
- Phenotypes: 1 : 1 : 1 : 1 (1 A, 1 B, 1 C, 1 D)
- PpSs x ppss
- Purple & Smooth = 1/4
- Purple & Wrinkled = 1/4
- Yellow & Smooth = 1/4
- Yellow & Wrinkled = 1/4
Observed...
- Purple & Smooth = 27
- Purple & Wrinkled = 25
- Yellow & Smooth = 23
- Yellow & Wrinkled = 25
- Total = 100
- Degree of Freedom = 3
- Chi square probability value = 0.32 p>0.9