. Blue eyes and inability to roll the tongue are recessive traits in humans.
Brown eyes and tongue rolling are dominant traits.
Illustrate the F2 generation of two people: one who are heterozygous for both traits.
Let B stand for brown eyes and b stand for blue eyes, R for the ability to roll the tongue and r for the
inability to roll the tongue. Determine the phenotypic ratio in their offspring.
Parent Genotypes _BbRr x __________
BR Br bR br
BR
Br
bR
br
phenotypes: ______ tongue rollers
______ brown/nonrollers
______ blue/rollers
______ blue eyes
8a. _____ Marfan syndrome is an autosomal dominant disorder. It causes a defect in an elastic
connective tissue called fibrillin. James is a 15-year old male. At the end of his high school basketball
game he was pronounced dead due to a bursting of the wall of the aorta.
Which of the following could represent his parents’ genotypes if one of them was normal?
A. MM x MM B. Mm x Mm C. mm x mm D. Mm x mm
8b. What percentage of the other children in the family described in question #8a could also be
expected to have the same condition?
_______________% Marfan Syndrome
1)
BR | Br | bR | br | |
BR | BBRR( brown roll) | BBRr (brown roll) | BbRR (brown roll) |
BbRr (brown roll) |
Br | BBRr (brown roll) | BBrr (brown non roll) | BbRr (brown roll) |
Bbrr (brown non roll) |
bR | BbRR ( brown roll) | BbRr ( brown roll) | bbRR ( blue roll) |
bbRr ( blue roll) |
br | BbRr (brown roll) | Bbrr ( brown non roll) | bbRr ( blue roll) |
bbrr (blue non roll) |
phenotypes -
tongue rollers - 12 are rollers out of the 16 , as mentioned above
brown non rollers - 3 are brown non roller as mentioned in the table above
blue rollers - 3 are blue rollers out of 16
blue eyes - 4 are blue eyes out of 16 as mentioned above
8 a) CORRECT OPTION IS D
as it is mentioned that the one of the parent are normal and this disease is an autosomal dominant disease therefore, out of the two parents one need to be heterozygous, therefore, the genotypes of parents would be 'mm' , the one who is normal and 'Mm' for the one who is abnormal, therefore, the cross would be Mm*mm
b) CORRECT ANSWER IS 50percent
as when the cross is carried out with the above genotypes then there will be four outcomes as follows
m | m | |
M | Mm | Mm |
m | mm | mm |
as we can see in the above table, the one who are heterozygous would be affected , and out of 4 , there are only 2 are heterozygous dominant, therefore, correct option would be 50 percent
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