MSGC Assignment

Question 1

Presentation: congenital ventricular septal defect, cleft palate, tubular nose, asymmetric crying facies , speech and gross motor skill developmental delay

Answer 1

DiGeorge Syndrome

heart defect + cleft palate = DiGeorge/22q11.2 Deletion Syndrome

Question 2

What is the gene for DiGeorge Syndrome and what testing should be ordered to look for this condition?

Answer 2

22q11.2 deletion
Chromosomal Microarray (CMA) for CNV analysis
targeted FISH would also be an option, but not the best choice
NOT a panel

Question 3

DiGeorge Syndrome is considered a microdeletion syndrome. What size are microdeletions?

Answer 3

<5 million base pairs (5Mb)
22q11.2 encompasses 30 genes and 1.5 - 3 million base pairs

Question 4

What testing should be ordered if Turner Syndrome is suspected?

Answer 4

Karyotype and CMA
Can’t do just karyotype b/c only about half of Turner Syndrome patients are truly Monosomy X, the rest are mosaic. Need CMA so that mosaicism can be detected.
FISH is not very applicable here.

emailed Dr. Tallis about this to be sure

don’t delete anything on this slide for now please :)

Question 5

Presentation: childhood obesity, increasing appetite, floppy as a baby, very small in first year, mild ID

Answer 5

Prader Willi Syndrome

Question 6

What is the best testing to order if Prader Willi Syndrome is suspected?

Answer 6

CMA/SNP analysis and methylation testing
NO SEQUENCING TESTING

SNP: CMA technique that can detect triploidy

Question 7

What are the causes of Prader Willi Syndrome?

Answer 7

• 15q deletion = 60-70%
• UPD 15 = about 30%
• Imprinting center deletion or defect = <5%

Question 8

What is the recurrence risk for the child of a parent with a cleft?

Answer 8

3-6%

Question 9

What is the most common craniofacial cleft: isolated cleft lip, cleft lip and palate, or isolated cleft palate?

Answer 9

Cleft lip and palate
isolated cleft lip = 25%
isolated cleft palate = 30%
cleft lip and palate = 45%

Question 10

What is the recurrence risk for a pregnancy of a cleft when the parent and one child both have had a cleft?

Answer 10

15-16%

Question 11

What is the recurrence risk for a child who has a sibling with a cleft, but whose parents are unaffected?

Answer 11

2.5-4%

Question 12

If considering an X-linked inheritance pattern for intellectual disability, WES/CMA or WGS are appropriate, especially if you’re not sure that it is X-linked, but what other testing/condition must be considered/included?

Answer 12

Fragile X
Needs it’s own testing b/c WES/CMA and WGS do not check for triplet nucleotide repeat disorders.

Question 13

What does this nomenclature mean?
arr Yp11.31q11.223(2650278-24445033)x0~1

Answer 13

Microarray
- This result shows mosaicism of the Y chromosome from the location on the p arm to the location on the q arm, so includes the centromere. Some cells have 1 copy of this region, while other cells have a deletion of this region.

Question 14

arr Xp22.33q28(409876-155237808)x1
What does this nomenclature mean?

Answer 14

That there is one X chromosome (these are all of the markers for X chromosome)

Question 15

What does this nomenclautre mean?
ish Xcen(DXZ1x1),Ycen(DYZ3x0)[79]/Xcen(DXZ1x1),Ycen(DYZ3x1)[21]

Answer 15

• FISH
• 79 cells were missing the Y centromere while 21 cells had the Y centromere, indicating mosaicism for the Y chromosome
• All cells have 1 X chromosome, no cells with 2 X chromosomes

Question 16

What are the unique risks to mosaic Turner (XO/XY)?

Answer 16

• Ovarian gonadoblastoma in patients with Y chromosome cell lines require removal of streak gonads after puberty
• 45,X/46,XY. Increased risk (up to 20%) of gonadoblastoma if the gonadal tissue remains in the abdomen

Question 17

A couple are both carriers for an autosomal recessive disorder. They have four children. What is the probability that at least one child is affected?

Answer 17

¾ x ¾ x ¾ x ¾ = 81/256
1 – 81/256 = 175/256

Question 18

Hardy-Weinberg equation

Answer 18

p2 + 2pq + q2 = 1

Question 19

Assume you discover a new X-linked platelet antigen that can be detected by a monoclonal
antibody. Heterozygous affected individuals as well as homozygous positive females are antigen positive (i.e. can be detected by the antibody). You find 20% of males to be antigen positive and 80% to be antigen negative. Assuming Hardy-Weinberg equilibrium, what percent of females would you predict to be antigen positive?

Answer 19

Males:
p + q = 1
p = 0.80
q = 0.20
Females:
p2 + 2pq + q2 = 1
q2 = homozygous females
2pq = heterozygous females (female carriers)
p = unaffected
q = affected
2pq = 2(0.80)(0.20) = 0.32
q2 = (0.20)2 = 0.04
0.32 + 0.04 = 0.36
I would predict 36% of females to be antigen positive

Question 20

What does this nomenclature mean?
arr 15q11q13(22,876,632-28,557,186)x1 pat
What condition is this likely to be?

Answer 20

• Microarray
• Paternally inherited (deletion on paternally inherited chromosome 15)
• Deletion on chromosome 15 location q11 – q13 spanning base pairs 2,876,632-
  28,557,186
• Prader Willi Syndrome

Question 21

What is the equation for rate of new mutation?

Answer 21

Question 22

A disorder is inherited as autosomal dominant with complete penetrance, reproductive fitness of 0.8, and disease frequency of 1/50,000. What is the rate of new mutations?

Answer 22

Question 23

Explain in your own words what are the assumptions a population must meet in order to be in Hardy-Weinberg equilibrium

Answer 23

Hardy-Weinberg equilibrium assumes that the genotype frequency proportions in a population are always constant. This is true when:
- Mating is random.
- Large enough population that the frequency of alleles is unlikely to change drastically.
- A low enough rate of mutation that allele frequency isn’t changed.
- Capacity of passing on genes is equal, maintaining equality between each allele
frequency and its chance of inheritance.
- Populations have not undergone significant migration which would lead to altered allele
frequencies.

Question 24

Given the following genotype frequencies, determine for EACH of the populations below if it meets Hardy-Weinberg equilibrium?
- A/A 0.49, A/a 0.42, a/a 0.09
- A/A 0.01, A/a 0.18, a/a 0.81
- A/A 0.64, A/a 0.04, a/a 0.32

Answer 24

1. solve for p and q by taking the sq rt of each homozygous frequency
2. solve for 2pq to see if this matches the heterozygous frequency
3. if they match = yes, if they don’t match = no
   - A/A 0.49, A/a 0.42, a/a 0.09 = yes
   - A/A 0.01, A/a 0.18, a/a 0.81 = yes
   - A/A 0.64, A/a 0.04, a/a 0.32 = no

Question 25

What is the risk that a person with autosomal recessive phenylketonuria (PKU) would have a
child with PKU, assuming their partner is not related but is of the same ethnicity? Assume a
population frequency of 1/10,000.

Answer 25

First, you need to go from disease frequency to carrier frequency.