This is not so much a test as an exercise to work through at intervals. It is designed to consolidate (and in some cases to extend) what we have covered in class so far, and to help you realise the connections between the topics we’re covering. You probably know more than you realise ! Some sections are more challenging than others. Very few of the questions are about recall of facts’, but will require you to apply genetic concepts. This means that you should not worry if you cannot simply state the answer’ immediately in response to the question. Graduates are expected to have skills in deduction, interpretation, reasoning, conceptualisation: some of my exercises are designed to encourage these faculties.
I will be very surprised , and delighted, if anyone gets 100%, so do not worry if you are not sure of your answers.
You may use any sources to answer the questions, and any one of the reference books mentioned the module handbook will be useful, but some of the exercises require thought and cannot be tackled by reading or surfing. (Draw diagrams when you get stuck). You will need a table of the genetic code and perhaps a chart of aminoacid side chains for one or two problems. You may type your answers into the page in track-change mode, and upload it via Turnitin. It is not assessed, but you will get useful feedback if you attempt the exercises. You will also acquire more insights which will be helpful in assessment and exams. Deadline: 30 November if you want feedback by Christmas.
A. The following two mendelian diseases both have the same mode of inheritance (this might not appear obvious, but SAQ9 is more typical of its kind!)
1. Name the inheritance pattern
2. In the second pedigree, give the genotypes of individuals IV.1, III.2 , II.2, and II.1 (two possibilities for this last one). (Can you see how IV.1 came to inherit the trait ? )
3. What is the risk that the next child of the proband III.8 will express the trait?
4. Can you name two common mendelian traits with the same inheritance pattern? What is their approximate frequency in the UK population?
B. Draw the pedigree described in the next paragraph, and answer the questions. Remember to organise male-female couples with the male on the left. Number the individuals within each generation; for instance, the woman called Audrey is referred to as II.9 . You do not need to upload your chart (but if you do want to draw a pedigree using a simple tool, see ** at the end of the file).
Jim is 5 years old. Small, benign bone tumours have been detected in the long bones and pelvis. His parents are surprised to be referred to the genetics clinic, since neither of them has the condition. They are planning another child, and are anxious to know the risks of transmission. The clinician asks about family history of illness. Jim’s father Max is 35. His mother Mabel is 44. Max has four brothers and two sisters, all older than him, who are all well. Mabel has a twin sister Audrey who is well, and had no issue (ie, she is childless). Max’s father is 72, his mother is 75, and both are well. Mabel’s mother Elizabeth suffered from swellings that she called arthritis, but is now dead and cannot be asked any more about this. The clinician will inquire into Elizabeth’s medical records. Mabel’s father is 80, and has no symptoms.
a. Two individuals in this pedigree might express the suspected trait. How are these two referred to (using the same notation as for Audrey, II.9)?
b. Considering Jim alone, if no other family member had any symptoms, what would be the most likely mode of inheritance? (Choose from AD, AR, X-linked). Give a reason for this.
c. Given the mode of inheritance you suggested in b), what are the likely genotypes of III.7 and III.8 ?
d. You receive new information, based on a molecular test of Jim’s blood cell DNA. This reveals a mutation in one allele at locus 11p11 converting the first nucleotide of intron 6 from G to T. Is this a transition or a transversion?
e. The identity of the gene in d) tells the molecular geneticist that this is in fact a common autosomal dominant trait. Looking at the pedigree you have drawn, why is this surprising? The gene is EXT2, and the trait is hereditary multiple exostoses, one of the common traits I showed in lecture 1.
f. This trait shows incomplete penetrance’/IP in females; women carrying the mutant allele sometimes have mild or no symptoms even though the trait is AD. In this case, what is the likely genotype of I.3 and II.8? [Note that incomplete penetrance often complicates pedigrees]. [The basis for IP in this trait is not known, but might be due to gender differences in bone development. If you want to read about the condition, which is one of the top 10′ from the table in the first lecture, see http://www.ncbi.nlm.nih.gov/pubmed/12417417 If you can’t obtain the pdf, let me know].
g. In the comparison below, the upper sequence is the normal allele, and the lower is the mutant allele (these are not from the same person they are not shown as 5′ and 3′)
i) Why are no codons shown after Gln 391 ?
ii) In the mutant allele, what is shown at the first base in intron 6?
iii) In the mutant allele, a new splice site has been created. This leads to the removal of the whole of exon 6, which contains 94 nucleotides. Is 94 a multiple of three? What do you suppose will happen to the aminoacid sequence when exons 5 and 7 are spliced together?
iv) Here is a partial sequence of the gene, showing exons 5,6,7,8 in alternating colours. (There are 13 exons in all). Note the underlined arginine codon, AGA, which crosses exons 5 and 6. Delete the red exon 6. What happens to the arginine codon, and which aminoacid does it now encode? What is the next codon after that? Can you see what happens to the valine codon GTG at the junction of exons 7 and 8? Does this help to explain the origin of the mutant allele ?
C. Are these statements about inheritance true or false?
1. Fathers transmit their X chromosome to all their sons
2. Fathers transmit their Y chromosome to all their sons
3. Fathers cannot transmit an X-linked trait to their daughters
4. Fathers cannot transmit an X-linked trait to their sons
5. Most X-linked traits normally appear to be inherited in a dominant mode and the associated allele is denoted as A in pedigrees.
6. Females have two X chromosomes, and one of them (maternally or paternally inherited) is inactivated at random in every tissue so that the X-chromosome gene dosage is the same in males and females.
D. These questions all relate to screening of new borns in the UK.
1. Name the 5 congenital traits screened for in UK newborns
2. Assign the following inheritance patterns to these 5 traits:
a. Autosomal recessive
b. Autosomal codominant
E. Individuals homozygous for the allele causing sickle cell disease produce an altered form of beta-globin. This mutation exposes a hydrophobic pocket in the protein, causing beta globin chains to aggregate through hydrophobic interactions. This process disorts the shape of red blood cells, which disrupts their membrane and leads to cell death and anemia.
a. The mutation is GAG->GTG. Is this a transition or a transversion?
b. What would be the effect of the mutation GAA -> GTA ? is this a transition or a transversion?
c. What does E6V mean ?
d. Would you classify the mutation as a substitution, insertion or deletion? (Tick v all that apply)
e. Is the mutation sense, missense or nonsense? (Tick all that apply).
f. Is the mutation synonymous, conservative or nonconservative? Why?
g. Does it cause a frameshift, a splicing error or a stop codon?
h. In homozygous sickle cell patients, why does anemia not lead to rapid death?
i. Individuals heterozygous for the E6V mutation at the beta globin locus are resistant to malaria. Is resistance to malaria a recessive or a dominant trait? Is this trait conferred by the wild type or the mutant allele? Is this allele dominant or recessive?
j. Individuals heterozygous for the E6V mutation at the beta globin locus do not develop anemia but still produce sickle cells. Is sickle cell trait recessive or dominant? Is this trait conferred by the wild type or the mutant allele? Is this allele dominant or recessive?
k. For dominant traits, we usually say that the heterozygote and homozygote have the same phenotype. Does this cause you to reconsider your answers to i and j above?
F. The following three traits share a common molecular basis.
Fragile X syndrome ; Huntingdon’s Disease; Friedreich’s Ataxia.
I) What is this common basis called?
II) Can you assign these three traits to their respective inheritance patterns?
III) Which of these three traits involves A) the coding region of the gene? B) the 5′ untranslated region? C) An intron?
G. Several of these statements about inheritance are true: which ones?
a. Sporadic traits are never congenital
b. Sporadic traits may arise in either the oocyte or spermatocyte of the carrier parents
c. Sporadic traits normally affect the carrier parent.
d. Mendelian traits are always caused by a unique mutant allele (on either or both chromosomes)
e. Mendelian traits are always caused by one or more mutations in the same gene
f. In carriers of Huntingdon’s Disease, the mutant allele is dominant
g. The expression of a dominant trait can be modified in some inheritors, a situation called incomplete penetrance
h. About 2500 genes are known to cause mendelian disorders
i. The genetic basis of most mendelian disorders is known
H. Which of these statements about genome organisation are true?
a. The human genome contains ~ 3 X109 bases (3000 megabases) of DNA
b. The largest human chromosome is ~ 250 X 106 bases (250 megabases) of DNA
b. Each human chromosome contains on average 2000 genes
c. The average distance between human genes is ~ 10 kilobases of DNA
d. The average length of a gene is 5.3 kilobases of DNA, including introns and exons
e. Genes are arranged along chromosomes without gaps between them.
f. Deletions of DNA totalling 50 Mb across the genome would necessarily lead to a chromosome disorder like DiGeorge syndrome
g. The X chromosome contains about 800 genes, and the Y-chromosome 80 or fewer
h. Because of their size disparity, the X and Y chromosomes cannot pair up and cross over in meiosis
i. The average length of genes does not vary greatly across the genome .
I. Which of these statements about human genetic variation are true?
a. A single nucleotide polymorphism/SNP is a variation in the DNA sequence that occurs in > 1% of the population
b. There are about 3X103 known SNPs in the human genome
c. The pattern of SNPS is often different in diseased and control individuals
d. Diseases can normally be attributed to SNPs when these are detected
e. Deleterious gene mutations arise through the same mechanisms as SNPs
f. Deleterious gene mutations are commoner than SNPs in the genome
g. Most humans carry 50-100 mutations predisposing to autosomal recessive traits
J. Consider this passage about dominant and recessive traits.
Normally, both alleles at a gene locus are expressed (biallelic expression’). A mutation in one allele may terminate translation, or produce a protein without normal function. The wild-type allele will be expressed normally, and produce sufficient protein on its own to mask the effect of the mutation in the other allele. If on the other hand, the mutant allele is present in two copies, the absence of normal protein encoded on the wild type allele, produces effects observable as a recessive trait.
When one allele is inactivated by a mutation, or otherwise produces abnormal protein, and the other, wild type allele cannot produce enough protein by itself to overcome the inactivation, the effect of the mutation appears as a dominant trait. The wild type allele is said to be haplo-insufficient when it is expressed monoallelically.
Based on this passage, and information about inheritance mode in the list of common traits you were given after Lecture 12, which of the following deductions are reasonable?
a. Cells expressing the Huntingtin protein require only one wild type allele for sufficient protein to be produced, thereby escaping Huntingdon’s disease.
b. Cells expressing CFTR protein require only one wild type allele for sufficient protein to be produced, thereby escaping cystic fibrosis.
c. Cells expressing the FRM1 protein require both copies of the wild type allele for sufficient protein to be produced, thereby escaping fragile X mental retardation syndrome.
d. In individuals expressing the cystic fibrosis trait, the wild type allele must be haplo-insufficient.
e. For some individuals expressing the familial hypercholestrolemia trait (FAH), mutations in the low density lipoprotein receptor gene prevent normal transport of the receptor to the membrane. The mutant allele is haplo-sufficient.
f. If the expression of the mutant allele is weak, a dominant trait might skip a generation altogether.
K. Consider the following comparison of a wild type and a frameshifted DNA sequence.
a. In the upper sequence, deletion of a C caused a change of amino acid from Histidine to Isoleucine (Ile). What would happen to the histidine codon in the upper sequence if a C was inserted at position 3, after C A?
b. What would happen to the serine codon after that ?
c. What would happen to the Tyrosine (Tyr) in the lower, frame shifted sequence if an A was inserted at position 3 after T A?
d. Following the change in c) above, what would the following amino acid be instead of Serine (Ser)?
L. Which of these processes is relevant to spontaneous ( as opposed to induced) genetic lesions in DNA?
a. Endogenous alkylation eg methylation by SAM
b. Oxidation of guanine by reactive oxygen species (ROS) to produce 8-oxo dG
c. Deamination of cytosine to produce uracil
d. UV irradiation leading to ROS production and eventual DNA damage
e. Hydrolysis of a base-sugar linkage -> Depurination
f. Incorporation of chemical base analogues such as Acridine Orange
g. Tautomerisation of keto-enol forms followed by false pairing during DNA replication
h. False pairing of 8-oxo dG to other bases (principally T, G)
i. UV irradiation of DNA, causing dimers between adjacent thymines on the same chain
M. Which of these statements about DNA repair is true?
a. Cytosine can deaminate spontaneously to uracil, generating G:U pairs which can be detected by uracil-DNA glycosylase/UDG
b. Methylcytosine can deaminate spontaneously to uracil, generating G:U pairs which can be detected by uracil-DNA glycosylase/UDG
c. Methylcytosine is a common modification of cytosine, particularly near the 5′ untranslated region of genes
d. Base excision repair (BER) removes a sequence of bases to restore a misincorporated base
e. Nucleotide excision repair (NER) removes a sequence of bases to restore a misincorporated base
f. NER preferentially repairs distortions of the DNA helix produced by exogenous damage, for example by UV giving rise to thymine dimers.
g. Defects in the BER mechanism are associated with an inherited disease, Xeroderma pigmentosum
h. Depurination of DNA is a common occurrence, leading to loss of a C or T on one strand of the DNA about 100,000 times per cell per day
i. Depurination is repaired by BER
j. Deamination is repaired by BER
k. Unrepaired DNA lesions can result in point mutations at that site
l. Unrepaired UV damage to DNA in skin cells will be transmitted to the next generation
m. Missing bases are preferentially replaced by C residues, whatever the nature of the missing base
N. Notice the salient points in the text that follows. When you were asked after Lecture 1 to research a mendelian trait, you should have covered the same kind of ground: named trait, inheritance pattern, genetic basis, nature of mutation(s), mechanism of mutation where known (here, recombination with pseudogene), function of the wild-type protein, possible effects of mutation on either transcription or protein function, biochemical consequence of altered protein linked to development or physiology, protein structure information where known, symptoms, treatments, any other genetically interesting facts about the trait. In your research, make a record of your sources; very important if you want to go back and check. The best places to start are OMIM and GeneTests; you can type diseases, genes, tissues or any key words into the search boxes. For primary and review data, use Pubmed.
Congenital adrenal hyperplasia is one of the commonest autosomal recessive traits. CAH is caused by one of several mutations in the gene for CYP21A2. These mutations appear to arise through recombination (during meiotic crossing-over) of the CYP21A2 gene with a highly homologous pseudogene located nearby on the same chromosome. This pseudogene probably arose through duplication of CYP21A2, followed by mutations that inactivated it. These inactivating mutations were not de-selected (by impairing the reproductive success of the carrier) since an intact gene remained. Because recombination with the pseudogene is rarely precisely aligned, the parents usually carry different recessive alleles; the child is then said not to be homozygous for the mutation, but a compound heterozygote (having two different mutant alleles). One of the many mutant alleles contains the substitution P31L, which is predicted from protein crystal studies to disrupt the association of strands in the protein. The aminoacid Proline introduces a turn in a protein chain, which would be disrupted by replacement with Leucine. CYP21A2 encodes an enzyme localised within microsome membranes that converts progesterone to a precursor of glucocorticoids and mineralocorticoids. These classes of steroid hormone affect homeostatic responses to stress and salt imbalance, respectively. In CAH, the absence of cortisol leads to increase of ACTH hormone (which would normally be antagonised by cortisol) and consequent hyperplasia, or overgrowth of the adrenal. Moreover, the progesterone substrate of CYP21A2 is inefficiently converted, and is diverted instead to the production of excess sex steroid. Symptoms in children may include ambiguous genitalia in girls and early onset of puberty in boys (both due to perturbations of sex steroid synthesis) and vomiting and extreme dehydration resulting from excessive sodium elimination in urine (salt wasting), resulting from inadequate synthesis of mineralocorticoids. Some symptoms can be treated by administration of glucocorticoids. The precise appearance of the disorder depends on the extend of recombination with the pseudogene; some mutant alleles confer adult onset and mild reduction in fertility.
OMIM. http://omim.org/entry/201910. ADRENAL HYPERPLASIA, CONGENITAL, DUE TO 21-HYDROXYLASE DEFICIENCY
GeneReviews. http://www.ncbi.nlm.nih.gov/books/NBK1171/ 21-Hydroxylase-Deficient Congenital Adrenal Hyperplasia
Trakakis E, et al (2010). An update to 21-hydroxylase deficient congenital adrenal hyperplasia. Gynecol Endocrinol. 2010 Jan;26(1):63-71. Review.
Comprehension based on this passage: Based on the passage above, would you say that:
a. This form of CAH is always caused by the substitution P31L
b. The allele causing CAH is duplicated
c. In CAH, progesterone is converted to excessive cortisol
d. Pseudogenes can arise through back-conversion of RNA to DNA, followed by reinsertion of the new DNA, now lacking introns, into the genome.
e. It is a reasonable deduction that parents of a child diagnosed with CAH would not themselves experience any symptoms
f. Glucocorticoids affect homeostatic responses to salt imbalance.
g. A compound heterozygote for a recessive trait has two mutant alleles, neither of them identical.
**A drawing tool for pedigrees
This is so simple, even I can use it (I am not a gifted user of buttons and keypads).
Open the tool.
Left- click anywhere on the screen to AID an Individual.
Right click in the box or circle to AID a relative; youy do this by clicking on the + buttons.
Left click in the box or circle to AID subtext (eg, 5 years’), to change status (deceased’), to apply symbols (I personally like symbol 10, to denote a person expressing a trait), or to show properties (proband’, no issue’ etc). You may find other useful features as you play with it. When you’re satsified with the pedigree, you can right click on the screen to save the image as a JPEG. This can be copy-pasted into a Word document.
For an example, notice in the pedigree below, that individual III.3 has been coded for three separate traits using the apply symbols button. The image comes from a paper that has a neat description of gene mapping using microsatellite markers (rather like genetic fingerprinting at crime scenes), and we’ll cover this later in the module: see