Non-Mendelian Inheritance

NON – MENDELIAN GENETICS

This is simply a pattern of inheritance in which the basic Mendelian ratios are modified. Examples: INCOMPLETE DOMINANCE (under gene interactions) Incomplete dominance/ Blending – Is a type of inheritance in which there occur the apparent failure of one allelic gene to dominate the other that when the two genes are together, they produce a character between them. Example A cross between a white Andalusian (fowl) and a black Andalusian produce a blue variety in the F1 generation. When the F1 members are selfed, the F2 individuals are a mixture of phenotypes i.e.: black, blue and white in the ratio of 1:2:1 Illustration: Genotype ratio:        1 BB : 2 BW : 1 WW Phenotype ratio:       1 black : 2 blue : 1 white EXAMPLESA genetist who was verifying Mendel’s first law and second law crossed 45 homozygous red flowered plants with 45 homozygous white flowered plants. The resulting F1 were 530 plants all with pink flowered plants, the seeds obtained were planted and F2 offspring with the following phenotypes were obtained. 1292 red flowered. 2570 pink flowered. 1290 white flowered.

1. Illustrate using symbols the crosses made and the results obtained in the experiment described above.
2. (b)(i) What is the name above experiment?
3. How do the above observations differ from the results of Mendelian work which     led     him to formulate his laws of inheritance?
4. Describe the genetical test you would carry out to prove weather not the appearance of the pink flower in the above experiment is true deviation from Mendel’s principles of inheritance (20marks).

Solution: Let      R – allele for red colour. W – allele for white colour. RW – genotype for pink colour (flower).   The F2 members are 1292 red flowered, 2570 Pink flowered, 1290 White flowered

1. The name of the mode of inheritance is Incomplete dominance inheritance.
2. The observation differ from Mendelian principles in that, the inheritance the flower colour does not follow the dominance – recessiveness principle.

All the F1 offsprings are pink flowered instead of them to show the dominant colour from either of the experiment is fully dominance or fully recessive. In the Mendelian experiment, when F1 are selfed the resulting one in the dominance-recessive phenotypes ratio is of 3:1. But in this experiment, selfing the F1 individuals give 1:2:1. The genetical test to be carried out is back test cross in which the pink flowers plant will be test crossed with either of the homozygous plants say red flowered plant RR. The result of the test cross above will give products of red (RR) and pink (RW) in the ratio of 1:1.

• The above results show a siltation of RW (Pink) and RR (red). This can prove that there has been no true blending, has occurred in F1 generation then we could expect offspring which were again all pink flowers.
• The appearance of pink colour trait in the F1 generation is not a true deviation from Mendel’s laws; otherwise the gene for red could not be reviewed unchanged.
• It was just observed when in the presence of WW (White) while its identity was being retained.

Thus, the inheritance of the flower colour in the experiments precisely obeys Mendel’s principle of inheritance only the phenotype ratios are different. Partial dominance (Co dominance & Incomplete dominance) Sometimes both alleles express themselves in the phenotype, but one more so than another. This is an intermediate stage between complete dominance and co dominance.   LINKAGEFor just 23 pairs of chromosomes to determine the many thousands of different human characteristics, it follows that each chromosome must possess many different genes. Any two genes which occur in the same chromosome are said to be linked. All the genes on a single chromosome form a linkage group. Under normal circumstance, all the linked genes remain together during cell division and so pass into the gamete, and hence the offspring, together. They not therefore segregate in accordance with Mendel’s law of Independent Assortment. The figure below shows the different gametes produced if a pair of genes A and B are linked rather than on separate chromosomes. If gene A and B occur on separate chromosomes i.e.. are not linked NOTE: Linked genes do not confirm to Mendel’s principle of independent assortment, therefore they fail to produce the expected 9:3:3:1 ratio in a breeding situation involving the inheritance of two pairs of contrasted characters (dihybrid inheritance).   **Crossing over and crossover values (COV)**During cross over (in chiasmata formation), the alleles of parent linked group separate and new associations of alleles are formed in the gamete cells, a process known as genetic recombination. Offspring formed from these genes showing ‘new’ combinations of characteristics are known as recombinants. Hence crossing over is a source of variations. The recombination frequency (COV) is calculated using the formula if individual showing recombination x 100 no. of offspring Gene mappingCalculation of COV enables geneticists to produce maps showing the relative positions of genes on chromosome. Chromosome maps are constructed by directly converting the COV between genes into hypothetical distances along the chromosomes.   Sex determinationSex is a state of being male or female. In human there are 23 pairs of chromosomes of these 22 pairs are identical in both sexes. The 23^rd^ pair, however is different in the male from the female. The 22 identical pairs are called autosomes, the 23^rd^ pairs are referred to as sex chromosomes or heterosomes. In females the two sex chromosome are identical (X chromosomes) are said to be Homogametic, while in males the two chromosomes are non- identical (Y – chromosome is smaller in size than X – chromosome) and are said to be Heterogametic. Unlike other features of an organization, sex is determined by chromosomes rather than genes. Humans of genotype XXY are phenotypically male, while genotypes with just one X chromosome (XO) are phenotypically female. This suggests that the presence of the Y chromosome which makes the human male, in its absence the sex is female. How does the Y chromosome determine maleness? The Y chromosome possesses several copies of a testicular differentiating gene which codes for the production of a substance that causes the undifferentiating gonads to become testes. In the absence of the gene and hence this substance, the gonads develop into ovaries.

• In humans, sex is determined by the type of sex chromosome contained in the spermatozoa that fuses with an X – chromosome of the egg cell.
• If the X – chromosome bearing sperm fuses with an X- chromosome of the egg cell, the resulting zygote will develop into a female (XX).
• If the Y- chromosome bearing sperm fuses with an X – chromosomes of the egg cell, the resulting zygote will develop into a male (XY).

Consider a cross below From this cross it is evident that;

1. The chances of a zygote develop into a male or female are 50%
2. The sex of the individual before to be born is determined by the father.

Sex determination differs in other organisms. In birds, most reptiles and fish and all butterflies, the male is the homogametic sex (XX) and the female is from the cross, all the resulting individuals are phenotypically normal with all females beings carriers. A homozygous purple flowered short-stemmed plant was crossed with a homozygous stemmed-flowered long stemmed plant and the F1 phenotypes had purple flowers and short stems. When the F1 generation was test crossed with a double homozygous recessive plant the following progeny were produced. 52 Purple flower, short stem. 47 Purple flower, long stem. 49 red flower, short stem. 45 red flower, long stem. Explain these results fully. The F1 phenotypes show that purple flower and short stem are dominant and red flower and long stem are recessive. The approximate ratio of 1:1:1:1 in a dihybrid cross suggests that the two genes controlling the characteristics of flower colour and stem length are not linked and the four alleles are situated on different pairs of chromosomes (see below). Let      P- Purple flower p – Red flower S – Short stem s – Long stem Since the parental sticks were both homozygous for both characteristics the genotypes must be PpSs. Test cross phenotypes:       Purple flower short stem                x          red flower, long stem Test cross genotype:                           PpSs                                                                 ppss

gametes	PS	Ps	ps
ps	PpSs	Ppss	ppss

Meiosis: Gametes (n): Random fertilization (shown inpunett square): Offspring genotypes (2n) (Listed in each square): Offspring phenotypes: 1 purple flower, short stem : 1 purple flower, long stem. 1 red flower, short stem : 1 red flower long stem.   MULTIPLE ALLELISMMultiple alleles are those alleles of a single locus when there are more than alternatives in a population. In humans, the inheritance of the ABO blood groups is determined by a gene which has different alleles. Any two of these can occur at a single locus at one time. Allele A causes production of antigen A on red blood cells. Allele B causes production of antigen B on red blood cells. Allele O causes no production of antigens on red blood cells. Alleles A and B are codominant and allele O is receive to both. The transmission of these alleles occurs in normal Mendelian fashion. A cross between an individual of group AB and one of group O therefore gives rise to individuals non of whom possess either parents blood group.       Phenotype:                               blood group AB   x   blood group O

		gametes
I^A^	I^B^
gametes	I^O^	I^A^I^O^	I^B^I^O^

Paternity suits. Although blood group cannot prove who a father of a child is, it is possible to use inheritance to show that an individual could not possibly be the father. Imagine a mother who is blood group B having a child of blood group O. She claims the father is man whose blood group is found to be AB. As the child is group O, its only possible genotype is I^O^I^O^. It must therefore herited one I^O^allele from each parent. The mother, if I^B^I^O^ could donate such an allele. The man with blood group AB can only have the genotype I^A^I^B^. He is unable to donate an I^O^ allele and cannot therefore be the father. Dominance seriesCoat colour in rabbits is determined by a gene C which has four possible alleles. Alleles C^F^ determines full coat colour and is dominant to Allele C^CH^ which determines chinchill a coat and is in turn dominant to Allele C^H^ which determines Himalayan coat and is in turn dominant to Allele C^A^ which determines albino coat colour. There is therefore a dominance series and each type has a range of possible genotypes. Inheritance is once again in normal Mendelian fashion. Other characteristics controlled by multiple alleles are coat colour in mice, eye colour.   Worked examplesA woman of blood group A claims that a father of blood group B is the father of her child whose blood group is O. How far are the woman’s claims valid? Answer. If both are heterozygous for their blood groups, then the woman’s claims are valid, but if either of them is homozygous for the blood group then women’s claims are invalid. Anna is a woman married to John. This couple once had a child, Kitto was one day discovered that his parents were in bad terms. John claims that Kitto is an illegitimate child but Anna is opposing the case. Blood tests rewards that John is of type A and Kitto is of blood group O. Anna’s mother blood type is B and Anna’s father blood group type is AB. Using this information alone

1. Suggest the possible genotype for Anna, show how you determine genotype.
2. Show dearly whether Kitto is or he is not an illegitimate child of the said family.

**Solution:**Since Kitto is of blood group O, then his genotype is of no doubt. This implies that the genotype for Anna should have an allele I^O^ considering Anna’s parents we have: Mother: Blood group B whose possible genotypes are I^B^I^B^ or I^B^ I^O^ Father: Blood group AB whose possible genotype is I^A^ I^B^

• But for Anna to have an allele I^O^ in her genotypes, the mother should bear the allele I^O^. Thus, the genotypes should be I^B^I^B^

A cross between Anna’s parent reveals the following;   Since Anna is heterozygous for her blood group and John is of blood group A, then the legitimacy of Kitto will depend on John’s genotype. If he is heterozygous for his blood group, then Kitto is a legitimate child as the cross below reveals. If John is homozygous for his blood group, then for sure, Kitto is an illegitimate child of the family.   **Sex Limited and sex influenced characters:**Sex limited characters are those characters that are concerned to only one sex eg. baldness, beards and Adam’s apple in males and enlarged breasts and hips in females. The development of such character is controlled by sex hormones, they are thus said to be sex influenced characters. PEDIGREE ANALYSISPedigree is a sequential arrangement of individual in a given family to show the passage of certain character from one generation to another. In analyzing a pedigree the first individual to show the characters of interest is called a propositous Features of a pedigree.

1. Circles represent females, squares represent males.
2. Shaded figures show a phenotypic expression of the character, open figures represent a normal phenotype.
3. Parents are connected by a horizontal line as children are connected to parents by vertical line.

Examples

1. Study the pedigree shown below, circles represent females, squares represent males, shaded figures represent colour blindness, open figures represent normal phenotype

1. What is the probable genotype for 1?
2. What are the possible genotypes for 5 and 9?
3. If 8 marries a normal man, what are the chance that she will have a colour blind son?

Solution Consider the following genetic attribute; X^N^X^N^ – Normal female. X^N^X^n^ – Carrier female. X^n^X^n^ – Colour blind female. X^N^Y – Normal male. X^n^Y- Colour blind male.

1. The possible genotype for 1 is X^N^X^N^ ie: homozygous normal.
2. The possible genotypes for, 5 is X^N^X^n^ ie: heterozygous normal
3. Since 8 was born from a certain mother, she has 50% chance of receiving an allele for colour blindness.

There are also 50% chances that the allele will pass to the son. Thus, the chances of 8 to have a colour blind son are: ½ x ½ x ½ = 1/8

1. State Mendel’s laws of inheritance.
2. In dogs coat colour is determined by a series of multiple alleles. The allele A^s^ produces a uniformly dark coat, the allele a^y^ produces a tan coat and the allele a^t^ produces a spotted coat. The dominance heredity achy is A^s^> a^y^> a^t^ which means A^s^ is dominant to both a^y^ and a^t^ where a^y^ is dominant to a^t^ only.

A family tree for dogs showing their coat colors is given below;

1. State the genotype of each of the individuals 1 – 5
2. By means of genetic diagram deduce the possible genotypes and phenotypes of the puppies which could be produced by mating between individual 4 and 6

Solution (b) (i) Given dominance hierarchy A^s^> a^y^> a^t^ Phenotype:           Possible genotypes Dark:                       A^s^A,^s^A^s^a^y^,A^s^a^t^ Tan:                               a^y^a^y^,a^y^a^t^ Spotted:                           a^t^a^t^ Thus, the genotypes are:

Individual	Genotype
1	A^s^a^y^
2	A^s^a^t^
3	a^y^a^y^
4	a^t^a^t^
5	a^y^a^t^

Crossing between individual 4 and 6   GENE INTERACTIONIn dihybrid crosses, two or more genes interact to determine a single phenotype. Such an interaction may modify the basic ratios. Examples of gene interaction are:

1. Lethal genes.
2. Epistasis.
3. Collaboration (Gene complex).
4. Multiple gene interaction.
5. Complementary genes.

Lethal genes. A lethal gene is that dominant or recessive gene which when occur in the homozygous state causes death to its bearer. eg : Sickle cell anaemia in humans. Lethal genes may affect several characteristics including mortality. ExampleConsider the inheritance of fur colour in mice. Wild mice have grey coloured fur, a condition known as agouti. Some mice have yellow fur. Cross breeding yellow mice produces offspring in the ratio 2 yellow fur : 1 agouti fur. These results can only be explained on the basis that yellow is dominant to agouti and that all the yellow coat mice are heterozygous. The typical Mendelin ratio is explained by the fetal death of homozygous yellow coat mice. Explanation of this rests on the fact that examination of the uteri of pregnant yellow mice from the above crosses revealed dead yellow features. Similar examination of the uteri of crosses between yellow fur and agouti fur mice revealed no dead yellow features. The explanation is that this cross would not provide homozygous yellow (YY) mice. Let Y represent yellow fur (dominant). y represent agouti fur (recessive).   NB: the ratio 2:1 talks of lethal. The gene for yellow is dominant for fur colour of the cat, the genotype Yy produce yellow cost but it is for viability. Hence gene YY represents lethal combination. Epistasis. Epistasis arise when the allele of one gene suppress or marks the action of another. Definition; Epistasis is the type of gene interaction in which one gene (epistatic gene) effect the phenotype expression of the other (hypostatic gene). An example occurs in mice where three genes determine the coat colour. However the absence of a dominant allele at one of the loci results in no pigment being produced and the coat being albino. This occurs regardless of the genes present at the other loci, even if these produce normal coat colour. The gene at third locus clearly suppresses the action of the others. Example In white leghorn fowl, plumage colour is controlled by two sets of genes, including the following:            W (white) dominant over w (colour). B (black) dominant over b (brown). The heterozygous F1 genotype WwBb is white. Account for this type of gene interaction and show the phenotypic ratio of the F2 generation. Answer. Since both dominant alleles W, white and B, black are present in the heterozygous F1 genotype and the phenotype is white, it may be concluded that the alleles show an epistatic interaction where the white allele represents the epistatic gene. The F2 generation is shown below:- Using the symbols given in the question. F1 phenotypes:                      White cock       x                 White hen F1 genotypes:                                  WwBb                                       WwBb

Gamete	WB	Wb	wB	wb
WB	WWBB	WWBb	WwBB	WwBb
Wb	WWBb	WWbb	WwBb	Wwbb
Wb	WWBb	WWbb	WwBb	Wwbb
wb	WwBb	Wwbb	wwBb	wwbb

F2 phenotypes:                             12 white colour : 3 black colour : 1 brown colour (Symbols) Definition: Epistatic/Inhibiting gene are recessive genes which when occur in a genotype it surfers the showing up of another gene. Collaboration (complex gene/simple Gene Intervention).

• Collaboration occurs when two genes controlling the same character, interacts to produce a single character phenotype which neither could produce alone.
• An example of each collaboration genes is seen in the inheritance of the shape of the comb in domestic cowl (chicks).
• One gene, P – produces a rose comb where as its recessive allele, p produces a single comb.
• Another gene R produces a rose comb where as its recessive allele, r also produces a single comb.
• When P and R occurs together (codominance), they collaborate to produce a walnut comb.Consider the case below:-

Phenotype	Possible genotype
Pea	PPrr or Pprr
Rose	ppRR or ppRr
Single	pprr
walnut	PPRR, PPRr, PpRR or PpRr

Consider the cross below:- Selfing PpRr gives 9 Walnut comb : 3 pea comb : 3 rose comb : 1 single comb **Complementary genes.**These are genes which are mutually dependent. Neither of them produces a given phenotype in the absence of the other. ExampleIn sweet peas, purple colour of the flowers is controlled by two genes C and P. In the absence of either the flowers are white one gene (C) regulates the production of raw materials for formation of a purple pigment where as another gene (P) regulates the conversion of raw materials into a purple pigment. Consider a dihybrid cross between a purple flowered plant and a white flowered plant   From the above cross, the F2 phenotype ratio is 9:7 instead of the normal 9:3:3:1. The last three phenotype classes have been combined. Polygenic inheritance (Multiple gene interaction). Multiple gene interaction (polygenic inheritance) is a type of gene interaction in which a single character is controlled by a series of genes each exerting its effect on the present phenotype in an additive fashion. Many genes acting together are referred to as polygenes. Polygenes give rise to continuous variation.