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GENETICS MAJORS ONLY PLEASE, AND DO NOT COPY THE LAST ANSWER GIVEN ON THIS SITE AS IT WAS INCORRECT!!!!!

Question: A mother is heterozygous for an over-expressed plastin allele on her X chromosome, thus has longer than normal axons. She is also heterozygous for the SMN1 gene on chromosome 5. Her children’s father is normal for plastin and has normal length axons. He too is heterozygous for the SMN1 gene. Both genes influence spinal muscular atrophy type 1 (SMA1); two abnormal copies of the SMN1 gene are required to cause spinal muscular atrophy, but their effect is prevented by high expression of the plastin gene (Lewis 92).

a) Give the genotypes of the mother and the father; provide a legend for your symbols. For each gene, explain why the abnormal alleles are dominant or recessive. (4 marks)

b) List all the possible genotypes and proportions of their offspring for the plastin gene. (1 mark)

c) List all the possible genotypes and proportions of their offspring for the SMN1 gene. (1 mark)

D) Showing your work, determine the probability that this couple will have a daughter with SMA. (2 marks)

e) Showing your work, determine the probability that this couple will have a son with SMA. (2 marks)

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Related questions

Nondisjuction is a cell division error in which either homologous chromosomes or sister chromatids fail to separate and thus do not migrate to opposite poles. It is responsible for conditions such as monosomy (one member of a chromosome pair is missing) and trisomy (three copies of a chromosome is present rather than two).

Nondisjunction can occur during the first or second meiotic division.

Part A- Jill is heterozygous for gene A and is going to have a child with Jack, who is homozygous recessive for gene A.

Which of the following offspring genotypes could result from fertilization of a normal haploid sperm with an egg that had undergone nondisjunction in meiosis I?
Check all that apply.

1)AAa

2)Aaa

3)aaa

4)Aa

5)aa

6)a

I put AAa, Aaa, Aa, but it was not correct

Part B-

Which of the following offspring genotypes could result from fertilization of a normal haploid sperm with an oocyte that had undergone nondisjunction in meiosis II?

Check all that apply.

1)AAa

2)Aaa

3)aaa

4)Aa

5)aa

6)a

Part E-

If Jack and Jill have a child with an aaa genotype, during which meiotic division, and in which parent, could nondisjunction have occurred?

Check all that apply.

1) Meiosis I in the mother

2 Meiosis II in the mother

3) Meiosis I in the father

4) Meiosis II in the father

The answer is not Meisis I in the mother

1. If you had a fruitfly with phenotype A, what test would you make to determine if it was Aa or AA? Explain the different outcomes expected if the fly in question is AA or Aa.

2.The sex of birds, some insects, and other organisms is determined by a ZW chromosomal arrangement in which the males have like sex chromosomes (ZZ) and females are ZW (similar to XY in humans). Assume that a recessive lethal allele on the Z chromosome causes death of an embryo in birds. What sex ratio would result in the offspring if a cross were made between a male heterozygous for the lethal allele and a normal female?

3.A recessive gene for red-green color blindness is located on the X chromosome in humans. Assume that a woman with normal vision (her father is color-blind) marries a color-blind male. What is the likelihood that this couple's first son will be color-blind?

4. One form of hemophilia is caused by a sex-linked recessive gene. Assume that a man with hemophilia marries a phenotypically normal woman whose father had hemophilia. What is the probability that they will have a daughter with hemophilia? (Note: In this problem, you must include the probability of having a daughter in your computation of the final probability.)

5. One form of hemophilia is caused by a sex-linked recessive gene. Assume that a man with hemophilia marries a phenotypically normal woman whose father had hemophilia. What is the probability that their first son will have hemophilia?

According to the multiplication rule of probability, the chance of crossing two pea plants that are heterozygous for height and have none of four offspring showing the recessive short phenotype is:A. ¼ × ¼ × ¼ × ¼.

B. ¾ × ¾ × ¾ × ¾.

C. (¼ × ¼ × ¼ × ¼) + (¾ × ¾ × ¾ × ¾).

D. ¼ + ¼ + ¼ + ¼.

E. None of the above.

What characteristic is exhibited by an X-linked trait?

a) Only females have the trait.
b) Females inherit X-linked traits from their maternal parent only.
c) Females can inherit a gene for an X-linked trait from either parent.

d) Males inherit X-linked traits from their paternal parent only.

e) Males can inherit a gene for an X-linked trait from either parent but normally inherit it from their fathers.
What characteristic is exhibited by a Y-linked trait?
a) Y-linked traits appear only in male offspring of men with the trait.
b)Y-linked traits are only passed from father to son.
c) Y-linked traits can be passed from mother to son.
d) Both A and B are characteristics of Y-linked traits.
e) Both A and C are characteristics of Y-linked traits.
Which of the following codes for a phenotypic human male?
a) XO with a copy of SRY gene on an autosomal chromosome?
b) XXYY with one copy of the SRY gene deleted?
c) XX with a copy of SRY gene on an autosomal chromosome?
d) All of the above sex compliments results in a person with a sex phenotype of male.
e) XY with the SRY gene located on an autosomal chromosome?
In Drosophila, yellow body is due to an X-linked gene (Xy) that is recessive to the gene for gray body (X+). A homozygous gray female is crossed with a yellow male. What are the genotypes of the F1 progeny?
In some goats, the presence of horns is produced by an autosomal gene that is dominant in males and recessive in females. A horned female is crossed with a hornless male. The F1 offspring are intercrossed to produce the F2. What proportion of the F2 females will have horns?
a)1/2
b)3/4
c)1/4
d) All
None
a) Females are X+X+; males are X+Y.
b)Females are X+X+; males are XyXy.
c)Females are X+X+; males are XyY.
d) Females are X+Xy; males are XyY.
e) Females are X+Xy; males are X+Y.

Females inherit X-linked traits from their maternal parent only.
Females can inherit a gene for an X-linked trait from either parent.
Males inherit X-linked traits from their paternal parent only.
Males can inherit a gene for an X-linked trait from either parent but normally inherit it from their fathers.
Males can inherit a gene for an X-linked trait from either parent but normally inherit it from their fathers.

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