Final Exam Review - Conceptual & Math Flashcards
Scurs on a cow are said to be sex-influenced. If the P is dominant in males and A is dominant in. females, what would the ratios be the corss between SC^P SC^A and SC^P SC^A (sPsP, sPsA, sPsA, sAsA)
75% of the males would have scurs
25% of females having scurs
All males with P will have scurs and All females with recessive genes will have scurs
Hemophilia A is an x-linked disorder. If a mother who is a carrier and a father who is normal have children, what is the probability that the daughter could have the disease?
X Y x Xa
0% of the girls will have hemophilia
50% of the boys will have hemophilia
Given a population of butterflies, yellow wings are recessive (b) to the wild-type blue wings (B). You find 20 yellow butterflies, 100 heterozygote butterflies, and 70 homozygous blue butterflies. What is the frequency of the alleles in this population?
bb=20
Bb=100
BB=70
Total = 190
of specific / 380 (190 * 2)
40 + 100 / 380
0.368
36.8% of the population will have yellow
140 + 100 / 380
0.631
63.1% of the population will have blue wings
What is the correct order of the genes
Abc - 100
aBc - 90
abc - 30
ABc - 500
aBC - 100
abC - 500
ABC - 30
AbC - 90
1) ID the highest number and the lowest number
(ABc and abC are parental)
(abc and ABC are double cross over)
2) Find the odd man out
Compare the differences and the similarities, the C’s are the only differences (middle gene)
3) Correct order will not matter ACB or BCA as long as C is in the middle
4) Compare the parental ones and see their order meaning that B is the odd man out
5) double crossover > cross w/B > cross w/ A > parental gene
Total population is 1440
Interferences from 3 point cross
B - C - A
200 + 60 / 1440 = gene distance of 0.1805
distance between C and A
180 + 60 / 1440 = gene distance of 0.166 repeating
1 - double cross over (60) / (0.1805)(0.166)(1440)
-0.3906 Interference
A snapping turtle has 30 chromosomes and its triploid. The snapping turtle therefore has _____ sets of chromosomes containing ______ chromosomes each
30 / 3 = 10 Sets of 3 chromosomes
Calculate the narrow-sense heritability of body weight between grandparents and their grandchildren, if observed phenotypic correlation coefficient is 0.5
h^2 = observed / expected
0.05 / 0.25 = 0.2
Farmer frank decides to increase the weight of his pigs. The average weight of his pigs is 75 pounds. He knows that the heritability of pig weight is 0.6. He selects and breeds his pigs that weigh 100 pounds. The expected average weight of the resulting pigs should be:
h2 = R / S where R is offspring of population and S is teh parents of the population
0.6 = x - 75 / 100 - 75
90 is the realized heritability
