Heredity Probability?

Question

Hi. I have a question about this heredity probabilty. We are doing this in my biology class and I am confused. Here is some questions:

Indicate the probability of obtaining the genotype in an offspring.
AAbb x AaBb and the offpsring is AAbb (i put 1/4)
AABbcc x aabbCC and offspring is AaBbCc (1/4)
AaBbCc x AaBbcc and off is aabbcc (1/32)

Then it gives me three traits.
Flower posiiton Axial(A) Termial(a)
Stem length Tall(T) Short(t)
Seed shape Round(R) Wrinkled (r)
if plant is heterozygous for all traits, probabilty for
a. homzygous for al dominant traits (i put 3/4)
b. homozyhous for three rec traits (1/4)
c. heterozegous for three thraits (6/36=1/6)
d. Homo for axial and tall, hetero for round (2/36 x 1/3 = 1/54)

Finally it says a disease is determined by a rec allele. if husband and woman are both carrier, prob of.

a. all three children normal (3/4)^3
b. one or more of the three children have disease (1/4+1/4+1/4=3/4)
d. at least one child will be normal (No idea)

any help?tnks!

Answer 1

Lets start with AAbb x AaBb and the offpsring is AAbb
If the first parent's A combines with the second parent's A you get AA
If the first parent's A combines with the second parent's a you get Aa
If the first parent's other A combines with the second parent's A you get AA
If the first parent's otherA combines with the second parent's a you get Aa
The A trait can make the following combinations:
AA AA AA Aa
(the odds of getting AA are 3/4
and the odds of getting Aa are 1/4
The odds of getting aa are 0)

The B trait can make the following combinations:
bB bb bB bb
Now you have to combine all possible combinations of the A trait with all possible combinations of the B trait:
AAbB AAbb* AAbB AAbb*
AAbB AAbb* AAbB AAbb*
AAbB AAbb* AAbB AAbb*
AabB Aabb AabB Aabb
the probability of getting AAbb is 6 out of 16 possible combinations or 6/16 = 3/8

The next problem is AABbcc x aabbCC and offspring is AaBbCc
All four combinations of the A trait give you Aa, making the probability of getting Aa 100%
The B trait can make the following combinations:
Bb bb Bb bb
The C trait can make the following combinations:
cC cC cC cC

All possible combinations of A and B traits
AaBb Aabb AaBb Aabb
AaBb Aabb AaBb Aabb

All possible combinations of the three traits:
AaBbcC* AaBbcC* AaBbcC* AaBbcC*
AabbcC AabbcC AabbcC AabbcC
AaBbcC* AaBbcC* AaBbcC* AaBbcC*
AabbcC AabbcC AabbcC AabbcC
AaBbcC* AaBbcC* AaBbcC* AaBbcC*
AabbcC AabbcC AabbcC AabbcC
AaBbcC* AaBbcC* AaBbcC* AaBbcC*
AabbcC AabbcC AabbcC AabbcC
The combination AaBbCc occurs in 16 of the 32 combinations
which is 16/32 = 1/2

Now try the third one by yourself

The flower traits AaTtRr and AaTtRr if all gene come from completely heterozygous plants

AATTRR# AATtRR AAtTRR AAttRR
AaTTRR AaTtRR AatTRR AattRR
aATTRR aATtRR aAtTRR aAttRR
aaTTRR aaTtRR aatTRR aattRR

AATTRr AATtRr AAtTRr AAttRr
AaTTRr AaTtRr* AatTRr* AattRr
aATTRr aATtRr* aAtTRr* aAttRr
aaTTRr aaTtRr aatTRr aattRr

AATTrR AATtrR AAtTrR AAttrR
AaTTrR AaTtrR* AatTrR* AattrR
aATTrR aATtrR* aAtTrR* aAttrR
aaTTrR aaTtrR aatTrR aattrR

AATTrr AATtrr AAtTrr AAttrr
AaTTrr AaTtrr AatTrr Aattrr
aATTrr aATtrr aAtTrr aAttrr
aaTTrr aaTtrr aatTrr aattrr~

a. homzygous for all dominant traits = 1/64#
b. homozyhous for three rec traits = 1/64~
c. heterozegous for three thraits = 8/64 = 1/8*
d. Homo for axial and tall, hetero for round (you figure it)

Recessive Disease
Dd and Dd

DD Dd dD dd

Each child has a 3/4 chance of being normal or 75%
Each child has a 1/4 chance of having the disease or 25%
The odds do not change no matter how many children you have so having three children there is a 75% chance that at least one of the children will be normal.

Whew, Well that's what it looks like to me.
Good Luck!