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Family engagement factor 2 is not a two-way communication is…
Family engagement factor 2 is not a two-way communication is facilitated through multiple forms and is responsive to the linguistic preference of the family.
Children will not benefit from seeing their families represe…
Children will not benefit from seeing their families represented on a daily basis in your setting.
Most families want to be involved in their children’s lives…
Most families want to be involved in their children’s lives in the program. Family involvement and support can be an important resource for the program.
Given the relationship covariance chart constructed, calcula…
Given the relationship covariance chart constructed, calculate the relationship coefficient between Cricket McCue and Jack McCue. Report values within the equation as fractions. Report the final solution rounded to the 4th decimal place. R[BLANK-1] = COV[BLANK-2] / (COV[BLANK-3])(COV[BLANK-4]) = [BLANK-5] / ([BLANK-6])([BLANK-7]) = [BLANK-8]
The probability of being homozygous dominant given a clean t…
The probability of being homozygous dominant given a clean test – P(HD|CT) – can be found as: P(HD|CT) = P0 / (1 – (LOC x (1 – P0)) Therefore, we can update the probability of Bull A with a clean test to be: P(HD|CT)male = [BLANK-1] Report solution of the equation to the 4th decimal place. You should use the solution (4 decimal places) reported for Po and LOC in prior problems to solve the equation.
Animal Order (i.e., column/row order): J, B, E, A, F, C Co…
Animal Order (i.e., column/row order): J, B, E, A, F, C Column 3, rows 3 – 6: Row 3: COV[BLANK-1] = 1 + F[BLANK-2] = 1 + 1/2([BLANK-3]) = [BLANK-4] Row 4: COV[BLANK-5] = 1/2(COV[BLANK-6] + COV[BLANK-7]) = 1/2([BLANK-8] + [BLANK-9]) = [BLANK-10] Row 5: COV[BLANK-11] = 1/2(COV[BLANK-12] + COV[BLANK-13]) = 1/2([BLANK-14] + [BLANK-15]) = [BLANK-16] Row 6: COV[BLANK-17] = 1/2(COV[BLANK-18] + COV[BLANK-19]) = 1/2([BLANK-20] + [BLANK-21]) = [BLANK-22]
Given the relationship covariance chart constructed, report…
Given the relationship covariance chart constructed, report the inbreeding coefficient of Cricket McCue. Report values within the equation as fractions. F[BLANK-1] = 1/2 * COV[BLANK-2] = 1/2 * [BLANK-3] = [BLANK-4]
Of the ancestors identified in Question 10, identify the anc…
Of the ancestors identified in Question 10, identify the ancestor(s) that is(are) inbred by clicking on the hot spot(s) (dashed boxes). You can click on more than one option. If none are inbred, click on the blank space to indicate none.
Instructions: Use letters from the pedigree to label equatio…
Instructions: Use letters from the pedigree to label equations (e.g., A, MO, M_, etc.). Report values as whole numbers (e.g., 0, 1, 2, etc.) or fractions (1/2, 5/4). The relationship coefficient between Antonio and Teresa can be shown as: R[BLANK-1] = COV[BLANK-2] / (1 + F[BLANK-3])(1 + F[BLANK-4]) Based on the pedigree and solutions from using the path method, the covariance between Antonio and Teresa is based on: [BLANK-5] ancestor(s) (a number), [BLANK-6] path(s) (a number), and is equal to: [BLANK-7] (a whole number or fraction). Therefore, the relationship coefficient between Antonio and Teresa is equal to: [BLANK-8] (round to 4 decimals).