Mary is 20 years old and has cancer. She is concerned that her family has a diso
ID: 275987 • Letter: M
Question
Mary is 20 years old and has cancer. She is concerned that her family has a disorder that can be inherited and has come to you for advice. She has two older brothers who also had cancer in their early twenties. Mary's oldest brother has 2 children but neither of them show any symptoms of cancer (his wife has no family history of cancer). Mary's other brother has no children. Mary's mother had cancer at the age of 35. Mary's father never had cancer. Mary's mother has a younger brother who had cancer at the age of 32. This brother has four children (his wife has no family history of cancer) and they all have normal vision. Finally, Mary's mother's mother (her maternal grandmother) had cancer at the age of 50. Mary's maternal grandfather never had cancer. a. (3 pts) Given the information above, diagram a pedigree for this family. 29. in mice, genes for albinism (a), chronic inflammation (i), and reduced sense of smell (r) are linked. You initialy crossed a mouse that is homozygous dominant for all three traits (AA II RR) with a mouse that is homozygous recessive l three traits (aa i rr). You then crossed several of the resulting Fi offspring to mice that are homorygous recessive for all three traits. You observe the following genotypes and phenotypes in the offspring: Genotype Aa li Rr colored, healthy, normal smell sense aalirr albino, chronically inflamed, reduced smell sense Aa ii rr colored, chronically inflamed, reduced smell sense aa li Rr albino, healthy, normal smell sense Aa li rr colored, healthy, reduced smell sense aa ii Rr albino, chronically inflamed, normal smell sense Aa i Rr colored, chronically inflamed, normal smell sense aa li rr albino, healthy, reduced smell sense Phenotype Number 89 91 20 23 23 25 278 Total mice C. (2 pts) Diagram a map between the genes and calculate the map distances. d. (2 pts) What is the coefficient of coincidence among these genes?Explanation / Answer
Answer:
29).
c). A----------17.99cM--------R-----------19.78cM--------------I
d). Coefficient of Coincidence (COC) = 0.71
Explanation:
Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.
Hence, the parental (non-recombinant) genotypes is AIR/air.
1).
If single crossover occurs between A & I..
Normal combination: AI / ai
After crossover: Ai/aI
Ai progeny= 20+23=43
aI progeny = 23+25=48
Total this progeny = 91
The recombination frequency between A&I = (number of recombinants/Total progeny) 100
RF = (91/278)100 = 32.73%
2).
If single crossover occurs between I&R..
Normal combination: IR/ir
After crossover: Ir/iR
Ir progeny= 3+25=28
iR progeny = 4+23=27
Total this progeny = 55
The recombination frequency between I&R = (number of recombinants/Total progeny) 100
RF = (55/278)100 = 19.78%
3).
If single crossover occurs between A&R..
Normal combination: AR/ar
After crossover: Ar/aR
Ar progeny= 20+3=23
aR progeny = 23+4=27
Total this progeny = 50
The recombination frequency between A&R = (number of recombinants/Total progeny) 100
RF = (50/278)100 = 17.99%
Recombination frequency (%) = Distance between the genes (cM)
A----------17.99cM--------R-----------19.78cM--------------I
Expected double crossover frequency = (RF between A & R) * (RF between R & I)
= 17.99% * 19.78%
= 0.1799 * 0.1978 = 0.0356
The observed double crossover frequency = 3+4 / 278 = 0.0252
Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency
= 0.0252 / 0.0356
= 0.71