Question

I need help with number 12 giving the information in the graph

[12] 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.

Procedure I - Part C - Baby bugs when parents are bb and Bb

Data Table - Enter your Baby Bug Counts from each data run

Data Run	BB Baby Bug Count	Bb Baby Bug Count	bb Baby Bug Count
1	0	7	3
2	0	3	7
3	0	7	3
4	0	4	6
5	0	6	4
6	0	3	7
7	0	5	5
8	0	3	7
9	0	3	7
10	0	9	1

Data Averages Table - Enter your average Baby Bug Counts

Tip: BB Baby Bug Count Average = Sum of BB Baby Bug Counts / Number of Data Runs

BB Baby Bug Count Average	Bb Baby Bug Count Average	bb Baby Bug Count Average
0	50	50

Percentage Tables - Enter the Baby Bug percentages

Tip: Baby Bug Percent = 100% ´ (Baby Bug Count Average) / (Total Number of Baby Bugs)

BB Baby Bug Percentage	Bb Baby Bug Percentage	bb Baby Bug Percentage
0%	50%	50%

Tip: Blue Rimmed Baby Bug Percentage = BB Baby Bug Percent + Bb Baby Bug Percent

Blue Rimmed Baby Bug Percentage	Yellow Rimmed Baby Bug Percentage
50%	50%

Observations and Questions

[10] Complete the Punnett square below when the parents are bb and Bb.

Punnett Square	Male
Female	Alleles/Genes	B	b
	b	Bb	bb
	b	Bb	bb

Answer 1

12) Yes, it is possible to draw conclusions about the genotype counts for examining the phenotypes. Because the given phenotypes of baby bug count said that it is a testcross between hetemozygous Bb and homozygous recessive bb baby bug counts. So heterozygous dominant must contain one necessive allele b in their parental genotype and recessive genotype bb mut be from parental genotype. So, we can easily determine that heterozygous dominant Bb allele must be contain ballele from the parent and Bb is the parental genotype. Hence, we can easily determine the genotype coont after examining given baby bug phenotypes.