Question

Two normal parents have a daughter with hemophilia (an X-linked recessive

disorder) that also has Turner Syndrome.

a. Write out the cross as described above illustrating the genotypes and

phenotypes of the parents and child.

b. Did the non-disjunction event occur in the mother, the father, or could

have occurred in either of the two parents?

Did the nondisjunction event occur during meiosis I, meiosis II, or could it

have occurred in either division during meiosis.

Answer 1

a.

• Factor VIII hemophilia is inherited as X-linked recessive state.

Representation:

Since the affected condition is caused by recessive state, representing Factor VIII affected condition as h, corresponding normal state as H.

Thus, genotypes may be represented as:

· Affected Male: XhY (hemizygous, recessive)

· Normal Male: XHY (hemizygous, dominhnt)

· Affected Female: XhXh (homozygous, recessive)

· Carrier female: XHXh (heterozygous, dominant)

· Normal female: XHXH (Homozygous dominant)

• Turner's syndrome is resulted due to non-disjunction causing one less X chromosomein females.

In the given scenario, the offspring has hemophilia and Turne's syndrome.

Thus, it has one X chromosome and genotype is Xh

The parents are normal. thus, genotype

Mother= XHXh

Father = XHY - non-disjunction= X missing

XHXh x Y

Offspring:

XHY and

Xh (female offspring with Hemophilia ns Turner's syndrome.

b.

Non-disjunction may occur due to- unable to pair, synapse, or cross-over during prophase of meiosis, failing to align during metaphase, or improper or failure to separate or dis-join during anaphase and telophase.

In meiosis-I, non-disjunction will be occurring in Anaphase-I, when pairs of homologous chromosomes fail to separate.

For meiosis –II in anaphase-II, the sister chromatids may not separate.

In meiosis 1- two monosomic may be resulted.

In meiosis II- one monosomic may be resulted.

Thus, Turner’s syndrome is more probably caused in meiosis I non-disjunction.