Procedure I - Part B - Baby bugs when parents are BB and Bb Data Table - Enter your Baby Bug Counts from each data run Data BB Baby Bug Run Bb Baby
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Procedure I - Part B - Baby bugs when parents are BB and Bb
Data Table - Enter your Baby Bug Counts from each data run
Data
BB Baby Bug
Run
Bb Baby Bug
bb Baby Bug
Count
Count
Count
7
3
0
2
4
6
0
3
3
7
0
4
4
6
0
5
0
6
O
7
7
0
8
5
5
0
9
4
6
0
10
5
5
0

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Data Averages Table - Enter your average Baby Bug Counts Tip: BB Baby Bug Count Average = Sum of BB Baby Bug Counts I Number of Data Runs BB Baby Bug Bb Baby Bug bb Baby Bug
Count Average Count Average Count Average Percentage Tables - Enter the Baby Bug percentages Tip: Baby Bug Percent = 100% x (Baby Bug Count Average) I (Total Number of Baby Bugs) BB Baby Bug Bb Baby Bug bb Baby Bug
Percentage Percentage Percentage Tip: Blue Rimmed Baby Bug Percentage = BB Baby Bug Percent + Bb Baby Bug Percent Blue Rimmed Baby Bug Yellow Rimmed Baby Bug
Percentage Percentage

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[6] Complete the Punnett square below when the parents are BB and Bb. [7] Using your Punnett Square, calculate the expected percentage of Blue Rimmed Baby
Bugs and Yellow Rimmed Baby Bugs. Show your work. How do your percentage table
results compare with the Punnett Square calculations? (higher, lower, similar) Explain
your answer. 4 mm Science — All Rights Reserved [8] Why do we use multiple data runs for this procedure? Explain your answer. [9] For this set of parents, is it possible to draw conclusions about the genotype counts
from examining the phenotypes? Why or why not? Use counts from one of your BB vs
Bb data runs as part of your discussion.

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Procedure 1 - Pawl B when parents or BB and Bb
The baby bug
bb
count will be gero (o ).
D's
in
parents cron below
BB X Bb
from
com
mo
baby bug count.
Hence
data
run
The Table shown
In
bb
counts ....

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