The Inheritance And Mutation Mechanisms Biology Essay


A pedigree diagram showing the family members that are affected by one of the following metabolic disorders and express the trait: Diabetes mellitus; red, impaired glucose tolerance (IGT); blue or impaired fasting glucose (IFG); green and also the ones who are not affected (Progeny, 2013).

By observing Figure 1 it can be deduced that the apparent inheritance pattern for this trait is mitochondrial. It can be seen that in the first generation shown on the pedigree only the mother is affected by and expresses the trait. It can then be observed that all members of the second generation (her offspring; consisting of both sexes) have inherited the trait via inheriting their mother's mitochondrial DNA (mtDNA), which contains the trait's mutation. Also, it can be noticed that in the third generation all of the family members that have inherited and express the trait have mothers, not fathers, that also have the trait; which supports the theory that they inherited the trait from their mother's mtDNA. However, once the fourth generation is reached it is evident that all of the daughters in this generation have not inherited the trait; due to them not being able to inherit their affected father's mtDNA, but rather inheriting their unaffected mother's mtDNA. These observations, in combination with the statement that both male and female children inherit their mtDNA from their mother and not their father (which is typical of mitochondrial disorders) suggest that the pedigree represents a mitochondrial inheritance pattern (Bradley et al., 2006). Taking into consideration that three different metabolic disorders were represented in the pedigree (Figure 1), it can be assumed that the symptoms expressed by each trait would vary. Family members: I.2, II.1, II.4, II.6 all suffer from diabetes mellitus (DM) and can be considered to suffer from harsher symptoms (expressed at a higher level) in comparison to II.8, III2, III.3, III.4, III.6, III.7, III.8, III.10, III.12 and III.14 who suffer from either impaired glucose tolerance (IGT) or impaired fasting glucose (IFG). Taking into consideration that IGT and IFG are considered to be 'pre-diabetes' it can be assumed that they will suffer less severe forms of diabetic symptoms which include: frequent urination, constant thirst (linked to having a dry mouth), weight loss and an increase in fatigue (Diabetes, 2013).

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Another trait which follows a mitochondrial inheritance pattern is myoclonic epilepsy associated with ragged-red fibres (MERRF). MERRF can be caused by mutations in many genes, including: MTTF, MTTH, MTTK and MTND5 (OMIM, 2010). In ~85% of cases, sufferers of MERRF have an A→G conversion at the 8344 position of the mtDNA. The point mutations related to MERRF are localised to the tRNALys gene (Klug et al., 2012).

Assuming that the gene locus responsible for the 'glucose metabolism' character in the pedigree (Figure 1) can be occupied by two alternative alleles, then two symbols can be assigned to represent theses alleles: Symbol for mutant allele = A Symbols for wild type allele = a The genotype of the proband would be AA and the genotype of the her husband would be aa. If the pedigree was different and II.2 expressed the trait and was heterozygous (Aa) then her daughter (III.1) would express the trait because she has inherited the mutated mtDNA from her mother and mitochondrial traits are genotypically and phenotypically transmitted from the mother to all of her children.

By observing Figure 1 it can be determined that the risk of the proband's next child expressing the trait would be 100% (unless a new mutation was to occur) because all of her existing children have expressed the trait due to it being mitochondrial. Whether the proband's next child is a boy or a girl does not affect their chances of expressing the trait because mitochondrial diseases are not sex-linked (Jorde et al., 2010).

By observing Figure 1 it can be determined that the risk of the next child of II.1 expressing the trait would be 0% because the affected father does not copy and transmit his mtDNA to any of his offspring during zygote formation (Bradley et al. 2006).

The daughter of III.12 does not show any symptoms related to the trait because although her father expresses the trait he does not transmit the trait to his daughter. This is because sperm cells do not contribute mitochondria to developing embryos (Jorde et al., 2010) so IV.4 has not inherited the trait from her father and will not show any symptoms. It is not because she is too young.

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Taking into consideration that the affected members of the family carried a G→A substitution, the type of base change that has occurred is considered to be a transition. This is because G, a purine, has been substituted for A, which is also a purine. The original GGG codon changed to AGG, which resulted in the amino acid changing from glycine to arginine (Read and Donnai, 2011) via a point mutation. The difference in chemical nature and properties of the glycine side chain (-H) and the arginine side chain (-CH2-CH2-CH2-NH-C=N+H2-NH2) is so influential that the substitution can be deemed non-conservative. Glycine is a nonpolar amino acid whereas arginine is a basic polar amino acid. Nonpolar amino acids are hydrophobic and polar amino acids are hydrophilic, meaning that the substitution of the base has led to a missense mutation which results in changes of the function of the protein (McKee and McKee, 2009). It is this change/loss of function of the protein that has caused the 'glucose metabolism' mutation to occur.

Deamination is an endogenous (spontaneous) process whereby an amino group is removed from cytosine (C), with the use of sodium bisulfite, converting it into uracil (U). This results in the amino group being converted into a keto group. In DNA, this spontaneous conversion of C to U can occur naturally and the spontaneous deamination of 5-methylcytosine can also occur. However, in the latter deamination the 5-methylcytosine becomes thymine (T) not U (Read and Donnai, 2011). Adenine (A) can be deaminated to become hypoxanthine and guanine (G) can be deaminated to become xanthine. Our cells contain a uracil-DNA glycosylase enzyme which recognises the change of base and replaces the mutant U with C; repairing the damaged DNA strands via the use of the base excision repair (BER) mechanism. This repair mechanism mainly works for the C→U mutation and not the 5-methylcytosine→T mutation because U is not a normal base that appears in DNA, it appears in RNA, so uracil-DNA glycosylase recognises this obvious base change and repairs it. The short-path BER mechanism uses AP endonucleases to create an incision/nick at the AP site where uracil-DNA glycosylase removed U. Once the incision has been made by the AP endonuclease, DNA polymerase I can place a C nucleotide into the correct position on the DNA strand and DNA ligase can then complete the BER procedure by ligating together the new C nucleotide with the nicked DNA strand (Robertson et al., 2009). It is faults in the BER mechanism that can lead to mutations not being recognised and would result in the substitution that would occur in the next round of DNA synthesis: which is the original G-C base pair being substituted by an A-U base pair in the first replication cycle, and then the A-U base pair becoming an A-T base pair in the second replication cycle. A tautomeric mutation occurs when a proton is transferred between a nucleotides changing structure. The spontaneous change of the nucleotide structure causes a change in the conformation of the base (it becomes a base analogue isomer and can exist in two forms) which can lead to incorrect base pairing and eventually substitutions in the next round of DNA synthesis. The movement of the proton causes hydrogen bonding to occur incorrectly; the tautomeric form of the nucleotide bases end up pairing with their usually uncomplimentary base pair. The tautomeric enol form of T will base pair with the keto form of G, whereas the imino form of C will base pair with the amino form of A. During DNA replication, the enol form of the 5-Bromouracil (5-BU) base analogue can hydrogen bond with G. In the next replication cycle, G will bond with C and 5-BU will bond with A. The following replication cycle results in A bonding with T and 5-BU once again bonding with A; if it is still present within the DNA molecule. This mechanism can therefore cause the G→A base pair substitution after undergoing DNA synthesis (Klug et al., 2012). A DNA repair mechanism would not be needed in this case because during each replication cycle, the 'normal' base on the template strand will complementary base pair with its respected base; A=T and CG.

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It is possible that if the mutation had been a gene deletion instead of a point mutation then it could have occurred during meiosis in the mother. This is because deletions are a type of mutation that can happen during crossing-over and chromosome rearrangement in meiosis.

Question 2A. Linkage and positional cloning

By observing Pedigree 1 (provided in the assignment) it can be deduced that the apparent inheritance pattern for the hearing impairment trait is autosomal dominant (AD).

By observing Figure 2 it can be seen that the possible genotypes of the gametes produced by II.1 and II.2 are; Bb (heterozygous) or bb (homozygous recessive).

Figure 2. A Punnett square showing the genotypes of the gametes produced by II.1 and II.2.

The expected frequency of the heterozygote and homozygote children of II.2 would be 50% (see Figure 2).

If II.3 is unaffected by the trait, it can be assumed that his wild type allele has a higher penetrance in comparison to the mutant allele, which results in II.3 not phenotypically expressing the mutant allele which gives rise to the trait, but he instead expresses the wild type allele. However, the father (II.3) still possesses the mutant allele within his genotype, which is then transmitted to his children; who inherit the mutant allele. This means that although the children have the same genotype as their father, their phenotype will differ because they will express the trait (taking into consideration that the rest of the pedigree hasn't changed) due to them having an increased penetrance of the mutant allele.

There are four classic types of Mendelian inheritance patterns, namely: autosomal recessive (AR), autosomal dominant (AD), X-linked recessive and X-linked dominant. By observing Pedigree 1, it can be seen that the trait appears to be typically AD Mendelian because: both sexes are equally affected by the trait, the trait is apparent in each generation and its history can be tracked (Bradley et al., 2006). Because the trait appears to be AD (a typical Mendelian inheritance pattern) it can be assumed that a single mutation on a single gene has given rise to the trait and that two alleles (one from each parent) have been inherited. These types of observation suggest that the trait follows Mendelian law.

Question 2B. Linkage and positional cloning

By observing Pedigree 2 (provided in the assignment) it can be seen that II.2 inherited the [212122] alleles from her mother, I.1.

By observing Pedigree 2 it can be seen that II.2 inherited the deafness trait from her mother, I.1.

It can be deduced from i. and ii. that the markers are closely located to the affected/mutant allele of the trait, so it can be suggested that they are tightly linked.

Considering II.3 and his children, the markers that seem to segregate most closely with the trait are: D3S3544, D3S1569 and D3S1744. These markers are considered to be most closely linked to the trait.

Taking into account that III.5 and III.9 don't express the trait, it can be assumed that the D3S1744 marker is not directly linked to the allele associated with the trait. It can then be deduced that the D3S3544 and D3S1569 markers segregate most closely with the trait.

Looking at the haplotype of II.2, it can be observed that in 3 of her children recombination between the trait locus and the alleles of D3S3544 and D3S1569 has occurred.

Assuming complete linkage; = 0.

(1/2)8 = 3.91 x 10-3.

Where 2 = number of possible gametes and 8 = number of children.

The probability of the children of II.2 having the observed genotype = (1/2)8.

Assuming the 3 loci are independent, there are 8 possible gamete outcomes, so:

(1/8)8 = 5.96 x 10-8.

The probability of the children of II.2 having the observed genotype = (1/8)8.

Calculations of the LOD score:

(1/2)8 ÷ (1/8)8 = 65536

log (65536) = 4.82

LOD score = 4.82

The LOD score for the linkage of the trait with the polymorphic markers is 4.82. Taking into consideration that the LOD score is higher than 3.6, it can be suggested that the correlation between the marker and the allele was due to real linkage and not chance (Bradley et al., 2006).

Taking into consideration that the LOD score is 4.82, it can be observed that the closest number to this figure (on the table provided in the assignment) is 3.46. 3.46 falls under the recombination fraction of when = 0.05, which shows that the marker related to this number (3.46) and recombination fraction is best linked to the trait. A recombination fraction of 0.05 shows almost complete linkage to the trait (complete linkage occurs when = 0). So it can then be deduced that the marker best linked to the trait locus is marker D3S1569. Assuming the that the recombination fraction of is equal to 0.05, a recombination rate can be calculated:

Recombination rate = 0.05 x 100 = 5%.

A 5% recombination rate is equivalent to ~5 centimorgan (cM), which in turn is estimated to be ~5 Mb. Therefore, the 'map distance' between the best linked marker (D3S1569) and the trait locus is 5 Mb. Crossing-over between the marker and the allele is unlikely.

Given that the average length of a gene is 5.3 Kb and the average inter-gene distance is 50 Kb, it can be expected that ~110 genes will reside in the 5 Mb interval.

Stereo cilia resemble hair cells that line the inner ear and sit on neurons. Mutations involved in the cilia formation can cause deafness, therefore, if the authors had had sequenced the PAQR9 gene candidate they may have noticed the defected cilia which led to the hearing impairment trait.

A more rapid alternative method to linkage analysis for identifying candidate genes from a small number of subjects would be genome-wide association studies (GWAS). The use of high-density micro arrays and SNPs in GWAS allows for specific participants who have the same mutant allele as those in the studies (rather than investigate their entire family) to help deduce whether a specific SNP is linked to a trait.

A phenocopy is a phenotype that mimics a genetic phenotype, but has come about through non-genetic means (Read and Donnai, 2011). The haplotype of III.2 is problematic because the phenotypic appearance of the trait for this member of the family came about via the effect of environmental factors on his genes and was not genetically inherited. So the trait appearing in III.2 is a mimic of the genotypically inherited traits that are apparent in the other effected family members; it is a phenocopy. Although the haplotype of III.2 is known, it is unknown which of his inherited alleles are specifically affected by environmental factors, which is why his haplotype is problematic.

'Linkage disequilibrium' is the phenomenon which describes the major difference of the frequency of an allele within the general population and the frequency of the allele within a family who have the trait linked to that allele. It occurs when the allele of a marker is linked more frequently with the trait, taking into consideration the population frequency. This means that the marker allele could contribute to the trait because a person who has the single nucleotide polymorphism (SNP) that could lead to the trait will have both the affected marker allele and the SNP.

Question 3. Chromosome abnormalities

De novo translocations are often referred to as 'new mutations' that can arise during fertilisation (Genetics Home Reference, 2013). Two features of meiosis that favour the formation of these translocations are: crossing over/recombination of chromosomes and allelic mutations in daughter cells.

If a balanced chromosome translocation was to take place then a carrier would not necessarily show a related phenotype to the translocation because no genetic information is lost or gained in balanced translocations; so there would be no phenotypic difference. Balanced translocations tend not to affect a cell's ability to undergo mitosis so a carrier may grow up to be phenotypically unaffected (Read and Donnai, 2011).

Robertsonian translocation is a type of rearrangement where two acrocentric chromosomes (chromosomes that have their centromere located closer to one end rather than the middle of their structure) join together at their centromeres and lose the short arms containing DNA. (Read and Donnai, 2011). Although Robertsonian translocations are considered to be a form of balanced translocation, it differs from other balanced translocations in the sense that when the two long arms of each chromosome join together, the two short arms that contain DNA are lost; this does not happen in other balanced translocations. However, the genetic information located on the chromosomes that are contained within the short arms of the acrocentric chromosomes is similar to the genetic information in the other arms, so when the short arms are lost no phenotypic difference will be observed.

By observing Figure 3 it can be seen that the karyotypes of the six embryos are: 46 XY, 45 XY, 47 XY, 45 XY, 47 XY and 45 XY, respectively. The embryo which is labelled 'normal' is done so because it has two copies of chromosome 13 and 14 and therefore contains the correct amount of genetic information that they code for; it can therefore also be assumed that this embryo is likely to reach full term of the pregnancy. The embryo labelled 'balanced carrier' has one chromosome 13 and one chromosome 14 which are independent of each other and although it also has the recombined chromosome 13/14, no genetic information has been lost or gained in this embryo; which implies that it is also likely to reach full term of the pregnancy. The embryo labelled 'trisomy 13' is deemed to be an aneuploidic trisomy because it has three copies of chromosome 13 and is not balanced. Although there is an excess of genetic information in this embryo, it is still possible for it to reach full term of the pregnancy but the child is likely to suffer from Patau syndrome, which occurs on average in 1 out of 10,000 live births as a result of this mutation. If the embryo does reach full term of the pregnancy, there is a 95% chance that the child will die within the first year of its life whilst suffering from deformities such as: cleft lip, closely spaced eyes, heart abnormalities and a malfunctioning central nervous system (Jorde et al., 2010). The embryo labelled 'monosomy 13' only has once copy of chromosome 13; this embryo is unlikely to reach full term of the pregnancy because monosomies are not compatible with pregnancy due to a severe lack of genetic information. The embryo labelled 'trisomy 14' contains three copies of chromosome 14, instead of the usual two. Although trisomic embryos are more likely to survive than monosomic embryos, trisomy 14 disorders are so rare that a sufficient amount of information related to this mutation hasn't been collected thus far, so it can be deduced that the embryo is not likely to reach full term of the pregnancy (Jorde et al., 2010). The embryo labelled 'monosomy 14' only contains a single copy of chromosome 14. The loss of such a vast amount of genetic material means the embryo is unlikely to reach full term of the pregnancy.


Figure 3. A diagram showing the possible gamete outcomes after meiosis and the zygotes formed after fertilization. The karyotype of all the six embryos is also shown.

Statement: 'For any single meiosis II, both gametes of each pair shown in the diagram will be present in sperm.' In oocytes, meiosis II can only begin once the gametes have undergone fertilization via a sperm cell. It can be deduced that after fertilization the gametes would have become zygotes (in meiosis II), meaning the statement is not true for oocytes (Jorde et al., 2010).

The authors' Sankaran et al. (2011) used partial trisomy cases to map the human trisomy 13 syndrome (Patau syndrome) and found that the trait's gene locus was 13q14. They discovered that the two main microRNA genes involved in this persistence of foetal haemoglobin (HbF) condition were miR-15a and miR-16-1; elevated levels of HbF were a result of increased expression of these two microRNAs. Sankaran et al. (2011) also found that the specific miR-15a and miR-16-1 target, MYB, is vital in silencing the HbF gene and also suggested that miR-15a, miR-16-1 and MYB studies may be advantageous to us to help aid sufferers of sickle cell diseases and β-thalassemia (Sankaran et al., 2011).

Two cytological detection methods that would show the presence of the chromosome translocation t(13;14) are fluorescence in situ hybridization (FISH) and spectral karyotyping (SKY). FISH allows for the detection of a specific chromosome via the use of fluorescently labelled DNA or RNA probes (Read and Donnai, 2011) whilst SKY uses 5 basic dyes combined together to create 31 different paints which provide a unique colour to each chromosome. The images produced by these methods enable translocated chromosomes to be detected. A suitable source of genetic material for these methods to be used on is amniotic fluid. The amniotic fluid contains amniotic cells which can be separated from waste urea cells (which are also present in amniotic fluid) by being spun (centrifuged). The remaining amniotic cells can then be stained to produce a karyotype using Giemsa staining.

Question 4. Investigation of a named congenital disorder

Disorder: Sickle cell disease; sickle cell anaemia (SCA).


Sickle cell anaemia (SCA) is a common form of sickle cell disease that affects millions of people globally. It is most apparent in people who have African or Mediterranean heritage; which provides a link as to why suffers of sickle cell anaemia are less prone to contracting malaria and show more resistance to the disease. On average, 1 out of 400 live births in America of African American babies will result in them having SCA (Marieb and Hoehn, 2010). The high levels of frequency of SCA sufferers within the world's population has allowed for many studies and research to be conducted regarding the disease and has contributed to developing methods of dealing with its affects.


SCA is an autosomal recessive Mendelian disease that requires two homozygous mutant alleles in order for the trait to become apparent; although heterozygotes can be deemed to have a 'sickling trait' but are usually not severely affected by it and can be considered to be carriers of the trait (Read and Donnai, 2012). SCA is the result of defected haemoglobin (Hb) molecules distorting the shape of erythrocytes and affecting their oxygen-transporting properties. This type of cell morphology can induce the effect of SCA. The structure of Hb consists of two α-globin subunits, two β-globin subunits and four heme groups. The abnormal Hb shape comes about via a single change of an amino acid at position 6 on one of the β-globin chains; a glutamic acid (E) group is replaced by valine (V) as a result of the adenine (A) base in E mutating into thymine (T) in the nucleotide sequence, which leads to the nature of the original nucleotide point mutation becoming a missense, non-conservative mutation overall. It can therefore be established that the SCA mutation can be donated as E6V. The authors Lonergan et al. (2001) showed that the gene locus for the E6V mutation is represented on chromosome 11p6 (Lonergan et al., 2001). The mutation is non-conservative because the polar E side chain has become a non-polar V side chain which causes the haemoglobin to become abnormal haemoglobin; haemoglobin S (HbS), as a result of a HBB gene mutation (Genetics Home Reference, 2012). The mutation of the amino acid allows the β-globin chains to connect together when oxygen levels are low and form rigid rod-like structures, causing the HbS to become spiked, which subsequently results in the erythrocytes becoming sickle shaped when transporting and uploading oxygen or when blood oxygen levels are low (Marieb and Hoehn, 2010). It is estimated that 0.1 of every 1000 live births in the UK will be of a child who suffers from SCA. General symptoms include: increase in fatigue, loss of breath and delayed development (Genetic Home Reference, 2012). If the child is unfortunate enough to suffer from a sickle cell crisis (where vital organs are deprived of oxygen) their kidney, lung and brain tissues could be severely affected by the attack, along with increasing joint pains (Klug et al., 2012). These sufferers are also anaemic because the abnormally sickle shaped cells are rapidly destroyed and removed from the body in comparison to the destruction of regular erythrocytes (Lonergan et al., 2001). There are certain treatments which can be utilised in order to stop the sickling of erythrocytes. A useful functional property of foetal haemoglobin (HbF) is that it does not sickle; so drugs such as Hydroxyurea have been designed in order to switch the HbF gene on to reduce the severe pain felt by many sufferers. Stem cells and gene therapy are also undergoing testing to assess how well they could potentially treat SCA (Marieb and Hoehn, 2010).

For more information regarding the intron/exon structure for the nucleotide sequence of the HBB gene related to SCA, click on the following links:;r=11:5246694-5250625