Corn Lab Report

Introduction

The lab that my group preformed last Friday was based on counting the phenotypes for corn. There were Mono Hybrid and Di-hybrid crosses, and we had to find the dominant and recessive alleles.

Mono Hybrid

Di-hybrid    

Purpose: The purpose of this lab was to find out how phenotypes work and how common the different alleles were in corn. 

Hypothesis: If there are more dominant genes in corn, then there will be less recessive genes left.

Materials: 2 Mono hybrid corn cobs, 1 Di hybrid corn cob. Ipad, and Mr. Wong

Procedures: 1. We were to first count the amount of each phenotype per corn, on the mono hybrid crosses.
2. After counting the corn and putting the data in, we counted the dihybrid crossed corn and put the data in too.
3. We then created a punnet square:

4. We then found out the phenotype ratios for each seed.

Purple and Smooth: 9/16

Purple and Shrunken: 3/16

Yellow and Smooth: 3/16

Yellow and Shrunken: 1/16

5. We then found the Chi Squared value for our corn. It was a good fit. 

Data 

MonoHybrid Cross:

Table 1	Number of Kernals	Kernal Percentage
Purple	206	81.1%
Yellow	48	18.9%
Total	254	100%

Table 2	Number of Kernals	Kernal Percentage
Smooth	166	77.8%
Shrunken	49	22.2%
Total	215	100%

Dyhibrid Cross

Table 3	Expected Number	Observed Number	[Observed - Expected]2/expected
Smooth & Purple	Total; x 9/16 = 137.25	137	0.0046
Shrunken & Purple	Total x 3/16 = 45.75	47	0.03410
Smooth & Yellow	Total x 3/16 = 45.75	43	0.167350
Shrunken & Yellow	Total x 1/16 = 15.25	17	0.20080
		CHI SQUARE VALUE =============>	0.400728

Conclusions:

I can conclude that my hypothesis was correct, because there was a greater ratio of dominant genes than there were recessive. Our chi square value of 0.400 is very close to the good fit value of 0.58 This makes me believe that we had a good chi square fit. 

Analysis(Answers)\

1. The dominant phenotypes appear to be purple and smooth. 

2. The probable parents could be Pp and Pp because there is almost a 3 to 1 ratio of dominant to recessive genes.

3. The probable parents could also be Pp and Pp because again, there is almost a 3 to 1 ratio of dominant to recessive genes.

4. The expected percent is 75 purple to 25 yellow. 

5. My data is very close to what is expected, but falls short on the yellow.

PROBLEMS:

1. Problem:  A large ear of corn has a total of 433 grains, including 271 Purple & starchy, 73 Purple & sweet, 63 Yellow & starchy, and 26 Yellow & sweet.

Hypothesis: This ear of corn was produced by a dihybrid cross

(PpSs x PpSs) involving two pairs of heterozygous genes resulting in a theoretical (expected) ratio of 9:3:3:1.

Objective: Test Hypothesis using chi square and probability value.

 Total Kernals = 840    
 Expected                                        
 Purple and Smooth 9/16 of 433 = 243.6
Purple and Wrinkled 3/16 of 433 = 81.2
Yellow and Smooth 3/16 of 433 = 81.2
Yellow and Wrinkled 1/16 of 433 = 27.1
(Math using Calculator and paper)

3.08 + .82 + 4.07 + .04 = 8.01

This is a poor fit

2. Problem:  In a certain reptile, eyes can be either black or yellow. Two black eyed lizards are crossed, and the result is 72 black eyed lizards, and 28 yellow-eyed lizards.
Hypothesis: The black eyed parents were Bb x Bb.

Objective: Test your hypothesis using chi square analysis. In this set, because only two values (traits) are examined, the degrees of freedom (df) is

Expected
Punnet Square of Bb x Bb, I get an expected percentage of 25% and 75% or 1/4 and 3/4
Total number of lizards is 100 so the ratio works.

Expected ....................... Observed

Black Eyes = 75 ................ 72
Yellow Eyes = 25 ............. 28

.12 + .36 = .48

This is a good fit

3. Problem: A sample of mice (all from the same parents) shows
58 Black hair, black eyes 16 Black hair, red eyes
19 White hair, black eyes 7 White hair, red eyes
Your tentative hypothesis: (what are the parents?)

Objective: Use a chi square analysis to support your hypothesis:

 9/3/3/1 ratio, so the parents are HhEe x HhEe

Total number of offspring = 100

Expected

9/16 of 100 = 56.25
3/16 of 100 = 18.75
3/16 of 100 = 18.76
1/16 of 100 = 6.25

.05 + .40 + .003 + .09 = .54 = GOOD FIT

Meiosis the Movie Lab report

Synopsis: During the meiosis movie, the focus of the lab was to make a video showing the stages of meiosis 1 and 2. We used IMovie to edit our video. Here is a link of the assignment:
https://bcp.instructure.com/courses/1384/files/201663/download?wrap=1

youtube video: https://m.youtube.com/watch?feature=youtube_gdata_player&v=MCOlOMVeYrs

• What is the function of meiosis?

There are many functions to mitosis, but one of the main functions is genetic diversity. People need to be different and adapt to avoid deadly illnesses, weaknesses, and to have an advantage over other organisms. Another function is to repair a damaged genetic line. For example, if Generation A was dying to a virus, Generation B or C could develop an immunity and let the species live longer.

• What events promote genetic variation during meiosis?

One event that promotes genetic variation is crossing over during metaphase. The chromosomes share genetic material which gets passed on and changes genes. Mutation also causes genetic variation and changes the chromosomes. 

• What causes non-disjunction?

Non-disjunction is the failure of separation of homologous chromosomes during cell division. This happens when going into anaphase, the chromosomes never split up. This can end up causing down syndrome and other harmful genetic disorders. Normally there is a checkpoint making sure that all of the chromosomes are in working order before going to the next stage, but failure of this system can cause non-disjunction.

• Panda Bears have 42 chromosomes compared to the 74 chromosomes found in most bears. How could this occur? Explain in terms of non-disjunction.

This could have occurred because of a mutation, or because the Panda Bears require less genetic material. This could also mean that way back in the past, there was a massive checkpoint failure that allowed massive non-disjunction, leaving the bears with less chromosomes. Another explanation is that somewhere along the line there was a non-disjunction that actually improved the bears.

• This lesson could be improved by including more about non-disjunction in the lesson. The response talks all about it, but it was barely mentioned in the actually activity.