Biology Chromosome
1.
2.
3. All of Allison’s eggs will carry the X chromosome and 50% of Allison’s egg cells will carry the recessive allele (hexa).
4.a. There is a 25% chance that Allison and Tim will have a baby boy who is heterozygous for Tay-Sachs.
b. No, the baby boy will not have Tay-Sachs he will be a carrier for the disease. The boy would only have the disease if he was homozygous recessive.
| X | Y | X | XX | XY | X | XX | XY |
1:2 1:2 | T | t | T | TT | Tt | t | Tt | tt |
1x1=1
2X2=4 1:4= 25%
5. a. Yes, there is a 50% chance that the young woman will develop FFI disease.
B.
| f | f | F | Ff | Ff | f | ff | ff | This Punnett square shows that there is a 50% chance that the young woman will have the dominant allele and become affected by FFI and a 50% chance that the young woman will be homozygous recessive and be unaffected by FFI.
6.
a.
b. The egg that has non-disjunction and therefore two X chromosomes in it after going through the entire process of meiosis. If that egg is then fertilized by a sperm with an X chromosome from the father this would result in the XXX Triple-X syndrome.
c. This syndrome can only be held “responsible” to both the parents chromosome contributions. The mother could achieve nondisjunction anywhere throughout meiosis to achieve the double XX chromosome in one of the final egg gametes. The father if had nondisjunction in the final division could end up with two XX chromosomes in the final sperm gamete at the end of meiosis.
7. a. The probability that their boy will be will not have SCID is 0% because there is a 100% chance he will have it.
This Punnett square shows that if a baby girl was born she would not have SCID but she would be a carrier for it. If a baby boy was born he would have 100% chance of having SCID. | X-r | X-r | X-R | XRXr | XRXr | Y | XrY | XrY |
b.
The Punnett square shows that if this couple had a female child, they would not have SCID but would be
2.
3. All of Allison’s eggs will carry the X chromosome and 50% of Allison’s egg cells will carry the recessive allele (hexa).
4.a. There is a 25% chance that Allison and Tim will have a baby boy who is heterozygous for Tay-Sachs.
b. No, the baby boy will not have Tay-Sachs he will be a carrier for the disease. The boy would only have the disease if he was homozygous recessive.
| X | Y | X | XX | XY | X | XX | XY |
1:2 1:2 | T | t | T | TT | Tt | t | Tt | tt |
1x1=1
2X2=4 1:4= 25%
5. a. Yes, there is a 50% chance that the young woman will develop FFI disease.
B.
| f | f | F | Ff | Ff | f | ff | ff | This Punnett square shows that there is a 50% chance that the young woman will have the dominant allele and become affected by FFI and a 50% chance that the young woman will be homozygous recessive and be unaffected by FFI.
6.
a.
b. The egg that has non-disjunction and therefore two X chromosomes in it after going through the entire process of meiosis. If that egg is then fertilized by a sperm with an X chromosome from the father this would result in the XXX Triple-X syndrome.
c. This syndrome can only be held “responsible” to both the parents chromosome contributions. The mother could achieve nondisjunction anywhere throughout meiosis to achieve the double XX chromosome in one of the final egg gametes. The father if had nondisjunction in the final division could end up with two XX chromosomes in the final sperm gamete at the end of meiosis.
7. a. The probability that their boy will be will not have SCID is 0% because there is a 100% chance he will have it.
This Punnett square shows that if a baby girl was born she would not have SCID but she would be a carrier for it. If a baby boy was born he would have 100% chance of having SCID. | X-r | X-r | X-R | XRXr | XRXr | Y | XrY | XrY |
b.
The Punnett square shows that if this couple had a female child, they would not have SCID but would be