Problems in Genetics and Their Solutions

Problems. Please calculate and/or hypothesize answers to the following questions or problems.

1. Duchenne muscular dystrophy is caused by a relatively rare X-linked recessive allele. It causes progressive muscular wasting, and usually leads to death before age 20.

a. What is the probability that the first son of a woman whose brother is affected will be affected? (3 points)

Answer:

Probability of mother inheriting the gene =

Probability of a son inheriting the gene =

Therefore, probability of the first son inheriting the gene = x

=1/4

b. What is the probability that the second son of a woman whose brother is affected will be affected if her first son was affected? (3 points)

Answer: Probability of first son inheriting the gene =

Probability of a son inheriting the gene =

Probability of second son inheriting the gene = x

Probability = 1/8

c. What is the probability that a child of an unaffected man whose brother is affected will be affected? (3 points)

Answer:

Zero. This is because, the X-linked gene cannot be transferred to the son by the father

2. Familial incontinentia pigmenti (IP) is an X-linked dominant disorder that causes a disturbance of skin pigmentation and can be associated with malformations of the eyes, teeth, and skeleton. The disorder is most often lethal prenatally in males.

a. If a woman who exhibits IP marries a man with no family history of the disease, what is the probability that they will have a child with the disease? Please explain how you determine this probability (3 points)

Answer:

P: Dominant

p: Recessive

Parents

Female Male

3810001352553333741676400034290016763900 P p p p Pp Pp pp ppRatio: 2Pp: 2pp

Chance: 50%

b. Continuing with the previous problem, the couples first child is a girl who does not exhibit this disorder. Given that the mother has the disorder, the couple is quite informed about the disorder. After the ultrasound of their second child, they are told that they are expecting a boy. The couple is quite concerned that the woman will not be able to carry the baby to term. However, she is able to give birth to a baby boy. After a week, the couple starts to notice symptoms of IP on the skin of their baby boy. The couple is concerned that their son might die soon. This does not happen. The child is currently two years of age and does not suffer from the disorder. How is this possible? Please explain. (3 points)

Answer: There could be a case of genetic mosaism where by the mutant genes are distributed unequally in the body cells. In mosaism, both normal and mutant genes are presentbut due to differential distribution, the normal gene masks the mutant gene. Also,there could be a case of hypomorphic mutations which could lead to a less potent IP that may not present with symptoms of IP, or, a case of Klinnefelters syndrome, where the boy has XXY genotype, where the mutant X is inactivated to leave behind a normal X-chromosome.

3. What is the most likely mode of inheritance of this trait in this family? (2 points)

Answer: Autosomal dominant

Please provide four reasons why you selected this mode of inheritance. (4 points)

Unaffected parents could not transmit the condition to their children

The condition is distributed equally among along all sexes

In the case of an affected parent, at least one of the children acquired the condition

There is an affected child or children through all the generations

4. Here is the inheritance of another trait in the same family. What is the most likely inheritance for this trait? (2 points)

Answer: X-Linked recessive

Please provide four reasons why you selected this mode of inheritance. (4 points)

i) All affected mothers transmit the disorder to at least one of their sons

ii) The disorder cannot be transmitted from an affected male to his sons

iii) This disorder has appeared most frequently in males

iv). Daughters to affected fathers are carriers who give birth to affected sons

5. In tomatoes, yellow flowers (Wf) are dominant over white flowers (wf), and yellow fruit (r) are recessive to red fruit (R). A cross was made between true-breeding plants with yellow flowers and yellow fruit, and plants with white flowers and red fruit. The F1 generation plants were then crossed to plants with white flowers and yellow fruit. The following results were obtained:

225 white flowers, red fruit

75 white flowers, yellow fruit

37 yellow flowers, red fruit

256 yellow flowers, yellow fruit

a. What is the genotype of the F1 plant? Please write your answer is such a way that you show which alleles are present on each homologous chromosome. (3 points)

Answer: Wf R

4953005143500

wf r

b. How many plants with yellow flowers and yellow fruit should we have expect to observe from these results? (3 points)

WfwfwfwfRow total

Rr225 37 262

Rr75 256 331

Collumn total 300 293 593

Grand total

Observed Expected=Row total x Collumn totalGrand totalWfwfRr225 132.54

Wfwfrr75 167.45

wfwfRr37 129.45

wfwfrr293 163.55

Answer: Yellow flower and yellow fruit will be expected to be 163

c. In testing to determine if these two traits are linked, what is the chi-square statistic associated with the test of independence? (3 points)

X2=(Observed-expected)2expected

(225-132.54 )2 132.54+ (75- 167.45 )2 167.45 + (37- 129.45)2 129.45 + (293-163.55)2163.55

= 64.50+ 51.04+ 66.02 + 102.46

= 284.02

d. What are the degrees of freedom and probability associated with this statistic? Are these two traits linked? (3 points)

Degree of freedom = (n-1) x (n-1)

= (2-1) x (2-1)

= 1

Since X2= 284.02 then Probability score=0.00001

The value is significant. Meaning that the genes are linked

e. What is the distance in terms of map units between the two genes that control these traits? (3 points)

Number of recombinant genestotal number of offsprings = (37 +75)/593

= 18.89%

=18.89 LMU

6. In fruit flies, the mutations, brown eyes (bw) are recessive to the wild type eyes (+), curved wings (cv) are recessive to wild type wings (+), and ebony body color (e) is recessive to the wild type body color (+). Fruit flies that exhibit all three wild type characteristics and are heterozygous are crossed to flies that exhibit all three mutant traits. The following number of progeny were obtained from these crosses. If not stated, the progeny class exhibits the wild type phenotype for that trait.

All mutant338

All wild type346

Brown eyes107

Brown eyes, curved wings24

Brown eyes, ebony189

Curved wings188

Curved wings, ebony106

Ebony34

Determine if brown eyes are linked to curved wings, curved wings linked to ebony body color, and ebony body color linked to brown eyes. Please show your work. Explain how the genetic map would look for the genes that control these traits. What are the map distances between the three genes? Please show your work. Calculate the coefficient of coincidence and interference. Please show your work. (18 points).

Answer:

bwcve 338

b+c+e+ 346

bwc+e+ 107

bwcve+ 24

bwc+e 189

b+cw e+ 188

b+cwe 106

b+c+e 34

Total 1332

bwcw Recombinantstotal progeniesx 100

107+189+188+1061332 x 100

(592/1332) x 100

= 44.44%

Not linked

Gene distance bw-cw= 44.44 LMU

cwe 24+ 189+ 188+ 34 1332 x 100

32.66%

Not linked

Gene distance cw-e= 32.66 LMU

ebw 107+24+34+1061332 x 100

=20.35%

Not linked

Gene distance e-bw = 20.35 LMU

The parental classes tend to maintain the gene order seen in the parent genes. These will be seen to have the highest frequencies among the progenies (besides the offspring that maintain the parental genotypeswild and mutant).

Therefore: bwc+e 189

b+cw e+ 188 are the parental classes. c is the central gene as seen from the double cross shown above.

The gene order, thus, is: bwcwe

bw cw e

5715030480

44.44LMU32.66LMU

Coefficient of coincidence and interference.

Probability of cross bw/cw =0.44

cw/e =0.33

Probability of a double cross =0.44x0.33

=0.15

Expected double crosses =0.15 x 1332

= 199.8 crosses

Coefficient of coincidence=observed crossesExpected double crosses

Double crosses are observe least frequently. Therefore:

bwcve+ 24

b+c+e 34 are involved.

(24+34)/199.8

=0.29

Interference = 1- Coefficient of coincidence

1-0.29

=0.71