(1.)
A.
B.
2.
(3.) The percentage of Allison’s egg cells that she produces during her lifetime that will carry the recessive allele and the X chromosome is seventy five percent. For the fact, twenty five percent of the cell will carry homozygous recessive allele and X chromosome, the other fifty percentage will carry heterozygous recessive allele and X chromosome.
(4.)
A. Twenty Five Percent
B. No.
C.
(X)R (X)r
(X)R (X)R(X)R - Female homozygous dominant (X)R(X)r - Female heterozygous (Y)r (X)R(Y)r – Male heterozygous (X)r(Y)r – Male homozygous recessive
Since the row for the mother and the column for the father both have heterozygous for Tay Sachs, it indicates they are Rr. The father has XX and …show more content…
For the fact, the mother is heterozygous and the father is homozygous recessive. As a result of the mother being heterozygous and the father being homozygous, there is a 50% chance that the child will have homozygous recessive. The disease is autosomal dominant indicating there is a chance just with the mother being heterozygous herself. However, if she is homozygous recessive that means that her child will have homozygous recessive. As a result, the child will not develop the …show more content…
B. The chromosomes fail to separate when a nondisjunction occurs. This process usually occurs during meiosis I or Meiosis II. As a result, they remain adjunct as they should after meiosis. An extra chromosome or a missing chromosome in the gamete is cause by the nondisjunction.
C. Meiosis has several aspect of the process to ensure success. The spindle checkpoint is an important aspect of the process to look at. The spindle checkpoint could have not perform its duty and could be responsible for this aneuploidy condition. Therefore, the spindle checkpoint of the process could most definitely be responsible. For the fact, failure of the spindle checkpoint working correctly can result in a nondisjunction.
(7.)
A. They have a fifty percent chance that they will have a boy. If the child is a boy, the chance it will be affected by SCID will be fifty percent. Therefore, the chance of the boy not having the SCID is twenty-five percent.
B.
(X)R (X)r
(X)R (X)R(X)R –Female homozygous dominant (X)R(X)r – Female homozygous
(Y) (X)R(Y) – Male dominant (X)r(Y) – Male
A.
B.
2.
(3.) The percentage of Allison’s egg cells that she produces during her lifetime that will carry the recessive allele and the X chromosome is seventy five percent. For the fact, twenty five percent of the cell will carry homozygous recessive allele and X chromosome, the other fifty percentage will carry heterozygous recessive allele and X chromosome.
(4.)
A. Twenty Five Percent
B. No.
C.
(X)R (X)r
(X)R (X)R(X)R - Female homozygous dominant (X)R(X)r - Female heterozygous (Y)r (X)R(Y)r – Male heterozygous (X)r(Y)r – Male homozygous recessive
Since the row for the mother and the column for the father both have heterozygous for Tay Sachs, it indicates they are Rr. The father has XX and …show more content…
For the fact, the mother is heterozygous and the father is homozygous recessive. As a result of the mother being heterozygous and the father being homozygous, there is a 50% chance that the child will have homozygous recessive. The disease is autosomal dominant indicating there is a chance just with the mother being heterozygous herself. However, if she is homozygous recessive that means that her child will have homozygous recessive. As a result, the child will not develop the …show more content…
B. The chromosomes fail to separate when a nondisjunction occurs. This process usually occurs during meiosis I or Meiosis II. As a result, they remain adjunct as they should after meiosis. An extra chromosome or a missing chromosome in the gamete is cause by the nondisjunction.
C. Meiosis has several aspect of the process to ensure success. The spindle checkpoint is an important aspect of the process to look at. The spindle checkpoint could have not perform its duty and could be responsible for this aneuploidy condition. Therefore, the spindle checkpoint of the process could most definitely be responsible. For the fact, failure of the spindle checkpoint working correctly can result in a nondisjunction.
(7.)
A. They have a fifty percent chance that they will have a boy. If the child is a boy, the chance it will be affected by SCID will be fifty percent. Therefore, the chance of the boy not having the SCID is twenty-five percent.
B.
(X)R (X)r
(X)R (X)R(X)R –Female homozygous dominant (X)R(X)r – Female homozygous
(Y) (X)R(Y) – Male dominant (X)r(Y) – Male