Corn Lab CORN LAB: CORN LAB: Looking at Monohybrid Crosses and Chi Square Tests

Lesson Objectives  Minds ON  Reminders  Yesterday’s Recap  Corn Lab – Count first, then wait…  Ratios and Chi Square Lesson

Minds ON  COOL GENETICS – This clip brings it all together!!

Reminders  Corn Lab: due Friday, March 21 st  Quiz: Thursday, March 20 th (all lessons from Thursday before March Break to Wednesday, March 19 th )  Homework (monohybrids sheets): Wednesday, March 19 th ** I will be starting to do random homework checks to make sure we are all on task!

Yesterday’s Recap  Gregor Mendel: “Father” of Genetics ( ): Chose to study pea plants to determine if there was a way to predict genotypic and phenotypic traits  Punnett Squares allow us to predict the probable genotypes and phenotypes of offspring.  Monohybrid Cross: the crossing of two F2 heterozygous dominant offspring.  F1 Generation: (PP x pp); F2 Generation: (Pp x Pp) G:0:4:0G: 1:2:1 P: 4:0P: 3:1  Test Cross: Cross a homozygous recessive individual with an individual of unknown genotype; resulting offspring will determine parent genotype.

Corn Lab  Corn reproduces sexually.  Male and female flowers are separate. Each kernel of corn represents a separate fertilization of an ovum by a sperm.  Each cob of corn holds the results of hundreds of separate monohybrid genetic crosses.  Purple is dominant, yellow is recessive

Corn Lab 1.Fill out the data table in the lab handout. 2.Just simply count the corn of each colour. 3.STOP when you have done this for the next set of instructions.

Some Minor Details:  Null hypothesis: An assumption that the observed results and the expected results are not related; result of chance and not statistically significant.

Figuring Out Ratios Results for genetic crosses are often recorded as ratios. Calculate the ratio of purple to yellow. To do this, use their totals. The smaller total you call 1, and write it in the appropriate space. Then divide the larger one by the smaller to two decimal places. Record this number in the appropriate space. For monohybrid crosses, we know that the phenotypic ratio should be 3:1

Figuring Out Ratios # of purple seeds# of yellow seeds Actual number TOTAL = 625 Actual ratio2.57 (450/175) (purple / yellow) 1 Expected number  469 (625 x 0.75)  156 (625 x 0.25) Expected ratio3.01  (469/156)1 Since we are looking at a monohybrid cross, the phenotypic ratios are ¾ for purple (dominant) and ¼ for yellow (recessive), which is 0.75 and 0.25 respectively

Chi Square  Tool that allows you to determine if your data is statistically valid

Chi Square in Table Format Phenotyp e Observed (O) Expected (e) d = o – e Purple Yellow Total

Chi Square Chart ^2= 3.05  3.05 is between 0.10 and 0.05 in degree of freedom category 1, thus its range of probability is about 0.07, which technically renders is not statistically significant. Therefore, I have to accept the null hypothesis.