Chance in Genetic Variation, Online PDF

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We are familiar with many examples of variation within a species and even within a family. A strong element of chance enters with each reproductive event – a chance that a certain ovum and a certain sperm, each having received a random mix of chromosomes, will meet in fertilization. To simulate th...

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Frankline Olum Chance in Genetic Variation Part 1. EXPECTATIONS VS. REALITIES Making predictions is a part of the scientific problem- solving process. 1. As an example, consider the toss of a coin. If a coin is tossed 10 times, Tails: _5____ what head/tail count do you predict? Heads: 5 What is your reason for this prediction? both tail and head have equal chances of happening, so the probability is 0.5 or 50% chance of happening.

Now toss a coin 10 times and record your results: Heads: 4 Tails: 6 If the results of the trial do not match your prediction, what are possible reasons? My Prediction was based on unbiased coin, it seems that my coin is biased in favor of the tail.

When experimental results match one’s expectations perfectly, there may be cause for rejoicing. Often there is not a perfect match, however, and doubts arise about the accuracy and adequacy of the data. Would you demand a perfect result of 5 heads and 5 tails to consider the trial to be a fair one? If not, how far of would you allow the result to be before you became suspicious of a trick? How did you decide that this is acceptable? I would conduct several experiments to make a conclusion, probable 3 or four experiments. Also, my prediction cannot be perfect due to factors that might influence during coin tossing. It may occur that the toss is affected by unintentional means such as not landing on a flat surface, landing up against an object (such as a shoe) or making contact with a hand in free fall, wind, the age of the coin etcetera. I will give my experiment a margin of error or degree of freedom where I will not subject the test at 100% accuracy but rather at 99% or 95% or 90% level of significance.

Part 2. VARIATION IN FAMILIES We are familiar with many examples of variation within a species and even within a family. A strong element of chance enters with each reproductive event – a chance that a certain ovum and a certain sperm, each having received a random mix of chromosomes, will meet in fertilization. To simulate the se chance events, we will use playing cards. Monohybrid Cross Let the two colors of cards represent alleles. 1. Shuffle the deck well, then split it into two equal piles. One pile represents sperm, the other ova. Letting red cards represent dominant genes (R) and black cards represent recessive genes ( r), you have represented parents containing Rr genes. 2. What is your hypothesis for the result of 20 “fertilizations” ie. How many RR, Rr and rr would you expect? What is the basis for your prediction? (Hint: Use a Punnett square to get the predicted ratio and then use those ratios to come up with your numbers.) Rr x Rr

R

r

R

RR

Rr

r

Rr

rr

1. 25% RR = 1/4( x 20= 5 2. 25% rr = ¼ x 20 = 5 3. 50% Rr = 2/4 x 20 = 10

3. Draw one card from each pile, representing the end result of meiosis, a sperm and an ovum. Together these represent one ofspring. Keep a tally (of the genotypes from 20 fertilizations, and complete the data table to show how your results compare to your expectations. Calculate the

ratios as percents or decimals for easy comparison with the expected, predicted ratio. (See the PowerPoint for additional tips on completing this procedure.) Genotyp RR Rr rr

Tally ////// /////////// ///

Total Ratio 6 30% 11 55% 3 15%

Expected PhenotypeRatio 5 (25%) Red 85% 10 (50%) 5 (25%) Black 15%

Expected 75% 25%

Do your results match your expectations satisfactorily? Yes, acknowledging the margin of error in reshuffling card, I am satisfied with the results, it is within my degree of freedom of + or – 0.5%.

Dihybrid Cross Let the four suits represent alleles. 1. This time pile hearts and spades separately from diamonds and clubs, and shuffle both piles. Let hearts (H) = allele A, spades (S) = a, diamonds (D) = B, and clubs (C) = b. Using both parents AaBb and presuming complete dominance as suggested by letter size, predict the variation within the offspring of this mating. What is your hypothesis for the numbers of each phenotype after 32 fertilizations (HINT: Use a Punnett square to come up with the ratio and then determine how much of each if you have 32 individuals.) AaBb x AaBb AB

Ab

aB

ab

AB

AABB

AABb

AaBB

AaBb

Ab

AABb

AAbb

AaBb

Aabb

aB

AaBB

AaBb

aaBB

aaBb

ab

AaBb

Aabb

aaBb

aabb

1. 2. 3. 4. 5. 6. 7. 8. 9.

OUT AABB = AABb = AaBB = AaBb = AAbb = Aabb = aaBb = aaBb = aabb =

OF 16 1 2 2 4 1 2 1 2 1

OUT OF 32 (Fertilization) 2 4 4 8 2 4 2 4 2

2. Draw one card from each pile to form a sperm, then draw again one from each pile to form an ovum. Place these together representing a fertilization, and tally the genotype in the space below. Continue for at least 32 fertilizations, re-shuffling as needed, and complete the table. Calculate the ratios as percents or decimals for easy comparison. (See the

PowerPoint for additional tips on completing this procedure.) Hand

Genotype Tally ////

Phenotype

Total

Ratio

Expected Ratio

HHDD

AABB

20

62.5% 56.2%

HSDD

AaBB

/////

Hearts and Diamond

HHDC

AABb

////////

HSDC

AaBb

///////

HHCC

AAbb

//

9

HSCC

Aabb

///////

Hearts and Club

28.13 18.75% %

SSDD

aaBB

2

6.25% 18.75%

SSDC

aaBb

//

Spade and Diamond

SSCC

aabb

/

Spade and club

1

3.13% 6.25%

Do your results match your expectations satisfactorily? __yes, considering bias in shuffling of cards, the variance is margin of error, I wouldn’t test my hypothesis at 100% level of significance but rather at 99% or 95% or even 90%______ If they do not, what are possible reasons? Card shuffling bias....