If negligible mutation rates exist, it signifies that the heritability of this trait follows mendelian laws and would give rise to distinct ratios.
The data is given for 100 turtles with a constant ratio of
genotype
Using unitary method we can determine the Genotype distribution of
162 turtles.
For example,
For AA we have 5 turtles showing this genotype in every 100
So in 162 turtles we will have: (5/100) X 162 = 8.1
So approximately 8 turtles show this genotype.
Similar calculations can be done for all genotypes
Genotype | True Value | Rounded off Value |
AA | 8.1 | 8 |
AB | 24.3 | 24 |
AC | 51.84 | 52 |
BB | 4.86 | 5 |
BC | 32.4 | 32 |
CC | 40.5 | 41 |
The total number for the rounded up values comes out to be 162
Table 1. Observed number of Bog Turtles of each genotype in 2017 Genotype Observed number in...
Use the information below to answer questions 1-2. Leather back turtles (Dermochelys coriacea) lay 50 eggs per clutch on soft sand beaches. Let us assume that the first and second cleavage within the developing eggs is controlled by the presence of a single pair of codominant alleles, CA and C Full cleavage and development to maturity will occur if an offspring bears alleles in the heterozygous (CAC) condition. However, there is only 18% cleavage penetrance if C Cappears in the...
Part 2: Data Tables Table 1: Parent Genotypes: Monohybrid Crosses Generation Genotype of Individual #1 Genotype of Individual #2 P Yy SS P1 Sy Sy P2 yS yy P3 Sy Sy P4 Sy Sy Table 2: Generation Data Produced by Monohybrid Crosses Generation Possible Offspring Genotypes Possible Offspring Phenotypes Genotype Ratio Phenotype Ratio P YY Yy yy Yellow or Blue 1:2:1 3:1 P1 YY Yy yellow 1:1 1 P2 Yy yy Yellow or Blue 1:1 1:1 P3 Yy yy Yellow...