Purpose: To see if we can predict the possible results of genetic crosses
Hypothesis:
1) I think they’re all going to be Bb.
2) I think there will be a ratio of 1:2 BB, and 1:2 Bb.
3) I think a ratio of 1:4 BB, 1:2 Bb, and 1:4 bb.
Procedures:
Written procedures: Take 2 brown bags, and label them as follows: “Bag 1, Female Parent”, and “Bag 2, Male Parent”. Then make a table that has: Number of chart at the very top; A column labeled “Trials” numbered 1-10; an “Allele from bag 1, Female parent” column (to be filled with allele letters); an “Allele from bag 2, Male parent” column (also to be filled with allele letters); and an “offspring” column. Make three of these tables.
Part 1.
Now place 2 blue marbles in bag1, and mark the first data table “1”. Use the letter B for the blue ,or dominant allele. Place 2 white marbles in bag 2. Use the letter b to represent the recessive allele for white color.
(*)For trial1 take out a marble from each bag, and don’t look inside the bag. Record result in the data table. Return the marble. Do the same to bag 2. In the column labeled offspring, write BB if you took out 2 blue marbles, bb if 2 white ones, and Bb if you took out 1 white marble and 1 blue marble. Repeat this procedure 9 more times. (end *)
Part 2.
Place 2 B marbles in bag 1. Place marbles B and b in bag 2. Repeat (*) 10 times in data table # 2.
Part 3.
Place marbles B and b in both bags. Repeat (*) 10 times in data table # 3.
Materials list:
2 Small Paper Bags
Marking Pen
3 Blue Marbles
3 White Marbles
Observations/Data:
data table number one | |||
trials | Female | Male | Offspring |
| 1 | b | B | Bb |
| 2 | b | B | Bb |
| 3 | b | B | Bb |
| 4 | b | B | Bb |
| 5 | b | B | Bb |
| 6 | b | B | Bb |
| 7 | b | B | Bb |
| 8 | b | B | Bb |
| 9 | b | B | Bb |
| 10 | b | B | Bb |
Data Table Number One | |||
Trials | Female | Male | Offspring |
| 1 | b | B | Bb |
| 2 | B | B | BB |
| 3 | B | B | BB |
| 4 | b | B | Bb |
| 5 | b | B | Bb |
| 6 | B | B | BB |
| 7 | B | B | BB |
| 8 | B | B | BB |
| 9 | b | B | Bb |
| 10 | B | B | BB |
Data Table Number One | |||
Trials | Female | Male | Offspring |
| 1 | b | B | Bb |
| 2 | B | b | Bb |
| 3 | b | B | Bb |
| 4 | b | b | bb |
| 5 | B | b | Bb |
| 6 | B | b | Bb |
| 7 | B | b | Bb |
| 8 | B | b | Bb |
| 9 | B | B | BB |
| 10 | b | b | Bb |
Communication:
Graph: See next page.
Conclusion and Analysis:
Question 2
According to my results in Part 1, only 1 kind of offspring is possible when the homozygous parents (BB and bb) are crossed. The results I obtained using the marble model do agree with the results shown by a Punnett square.
Question 3
According to my results in Part 2, about 50% of offspring are likely to be homozygous when a homozygous parent (BB) and a heterozygous parent (Bb) are crossed. 50% are also likely to be heterozygous. No, my model does not entirely agree with the results shown by a Punnett square. The Punnett square shows 50-50, or 5 and 5, but my model is 40-60, or 4 and 6. But it is close enough.
Question 4
According to my results in Part 3, 3 different kinds of offspring are possible when two heterozygous parents (Bb x Bb) are crossed (BB, Bb, and bb). 25% of BB, 50% of Bb, and 25% of bb are likely to be produced. My model does not agree with the results of a Punnett square. My results were: 10% BB, 80% Bb, and 10% bb.
Question 5
For part 3, if I did 100 trials instead of 10, my results would be closer to the results shown in a Punnett square: pretend you are flipping a coin to see the percentage you are going to get for heads and tails. You flip the coin only once. It lands on heads. You decide that the percentage is always 100% heads, 0% tails. This is, of course, wrong. So, the farther you get away from this extreme, the better the chance of becoming equal with the probability amount.
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