Mendelian Principle of Inheritance

MENDELIAN GENETICS

Gregor Johan Mendel did studies of genetics using the Pisum sativan (garden peas). He was trying to find the laws that govern the passage of characters from one generation to another. He established that Pisum sativum had the following advantages over other species:

1. They were several varieties available which had quite district characteristics.
2. The plants were easy to cultivate
3. The reproductive structures were enclosed by the petals, this made the plant self pollinating and hence producing varieties of the some characteristics (pure breading).
4. Artificial cross – breeding between varieties was possible and resulting hybrids were confertile.

Mono hybrid inheritance and the principle of segregation: Monohybrid inheritance is a pattern of inheritance which involves two contrasting variations of only one characteristic. Example: Tall Vs short (height). Red Vs White (colour). Rough Vs Smooth (texture). Common genetic terms:

1. 1. Gene: The basic unit of inheritance for a given characteristic.
   2. Allele : One of number of alternative forms of the same gene responsible for determining contrasting characteristics e.g. A or a (pared genes).
   3. Locus: Position of an allele within a DNA molecule. Alleles of one gene are on one locus.
   4. Homozygous: The diploid condition in which the alleles at a given locus are identical e.g. AA or aa.
   5. Heterozygous: The diploid condition in which the alleles at a given locus are different e.g. Aa.
   6. Phenotype: The observable characteristics of an individual usually resulting from the interaction between the genotype and the environment in which development occurs e.g. Red, blue.
   7. Genotype: The genetic constitution of an organism with respect to the allele under consideration e.g. AA2, A2, or did.
   8. Dominant:  The allele which influence the appearance of the phenotype even in the presence of an alternative allele e.g. A
   9. Recessive: The allele which influence the appearance of the phonotype only be or in the presence of another identical allele e.g. a
   10. F1 generation: The generation produced by crossing homozygous parents.
   11. F2 generation: The generation produced by crossing two F1 organisms.

Basic Monohybrid ratioThis is the phenotypic ratio contained in the F2 generation of the original pure parents. The ratio is always 3:1 Mendel’s experiment and the Monohybrid ratio

1. In one of his experiments, Mendel crossed a red flowered plant from a pure line with a white flowered plant also from a pure line. Al the resulting F1 plants had red flowers.
2. When the F1 members were selfed, the resulting F2 were a mixture of red and white phenotypes in the approximate ratio of 3:4.
3. This is the basic monohybrid ratio obtained from a cross between two heterozygous individuals.

Illustration:-   **Non coding DNA.**Though human DNA contains large number of genes, the problem is about 95% of the DNA appears to have no obvious function because it is non – coding. In other words does not code for proteins or RNA.

1. The factor for redness was dominant over that for whiteness which was red.
2. The factor for whiteness was present in the F1 though not expressed effect was obscured by the factor for redness.
3. The characteristic red and the characteristic white remained unchanged. I.e.: There was no an intermediate colour.
4. Each characteristic is controlled by a pair of factors that segregate during gamete formation.

This observation, suggested to Mendel the formulation of his first law “the law of segregation.”   Assumptions:

1. Let ‘R’ be factor redness and ‘r’ factor for whiteness.
2. Let ‘R’ dominate ‘r’ so that when the two are together, only R is expressed.
3. Let each character be controlled by a pair of factors that segregate gametes formation

Consider the following cross: Parental phenotypes         Pure breeding Red flower x Pure bleeding   White flower     Phenotypic ratio           3Red : 1White Mendel’s 1^st^ law of inheritance (Law of segregation) The law states that: “The characteristics of an organism are determined by internal factors which occur in pairs. Only one of a pair of such factors can be represented in a single gamete. Meiotic explanation of Mendel’s first law.

1. Although Mendel knew nothing about Meiosis, but his first law is explained by Meiosis as follows:-
2. During Meiosis, the paired homologous chromosomes, separate from each other as a result of which the gametes receive only one type of chromosome instead of the normal two.
3. Alleles also occur in pairs at the homologous chromosomes, thus separation of homologous chromosomes occurs currently with the separation of alleles.
4. Thus, there is similarity between separation of homologues chromosomes in Meiosis and segregation of Mendelian factors.

We know that Mendel’s factors are specific portion of a chromosome called genes. We also know that the process which produces gametes with only one of each pairs of factors is Meiosis. On the basis of his results, Mendel had effectively predicted the existence of genes and Meiosis. Methods used to solve Mendelian problems:-

1. Algebraic method.
2. Punnet square/chequer board method.
3. Mendelian crosses/genetics diagrams

Algebraic method Consider a cross between two tall plants both heterozygous for height.   Symbols used in genetics:

1. In genetics any symbol can be used to represent any characteristics provided it is defined.
2. However, it is common that a dominant characteristic is represented by the first latter of its name. Eg. R for red, T for tall, G for green etc. The characteristics will take the lower case letter of the dominant one e.g r white where R red is dominant to white. The symbol P1 stands for parents and F1 and F2 are filial generations 1 and 2respectively

Example: One of the causes of dwarfism in man is the inheritance of dominant gene D. The allele for a normal height is d, Given that the genotype for Kijeba a man suffering from dwarfism is Dd, work out the genotype and phenotype rations of the offspring if he marries.

1. A normal woman
2. A dwarf woman

Solution:- Given: D – allele for dwarfness d – allele for tallness

1. If he marries, the genotypes and phenotypes of there child will depend on the genotype of the woman.
2. If she is homozygous tall, then half the offspring will be phenotype tall and the half short(dwarf) above reveals.

If she is homozygous dwarf, then the products will be. The genotype ratio will be 1 DD: 1 Dd If she is heterozygous dwarf, then the products will be. Genotype ratio is 1DD : 2Dd: 1dd Phenotype ratio is 3 Dwarf : 1 tall

BACK CROSS AND TEST CROSS

1. Back cross :This is a cross between an organism and either of its parents.
2. Test cross :This is a cross between an experimental organism with a dominant phenotype and that of a recessive phenotype, of its parent so as to determine the genotype of that experimental organism.

Explanations: One common genetic problem is that an organism which shows a dominant character has two possible genotypes. Example A plant producing seeds with round coats could either be homozygous dominant (RR) heterozygous (Rr). The appearance of the seeds (phenotype) is identical in both cases. However it is often necessary to determine the genotype accurately. This involves the use of a technique known as Test mass in which an organisms is unknown genotype is crossed with the one whose genotype is accurately known. A genotype which can positively be identified from its phenotype alone is one which shows recessive features. In the case of the seed coast, any pea seed with a linked coast must have the genotype “rr”. By crossing the dominant character, the unknown genotype can be identified. Let   R = allele for round seeds R = allele for wrinkled seeds If the plant producing round seeds have the genotype RR.     Conclusion: The only possible offspring are plants which produce round seeds, thus the unknown   genotype is RR. If the plant producing round seeds have the genotype Rr. This 1:1 ratio is the monohybrid test cross ratio obtained from a cross investigation between heterozygous dominant and a homozygous recessive. NON – MONOHYBRID INHERISTANCE This is a pattern of inheritance which involves more than one character. These may be two three etc.   Dihybrid inheritance and Mendel’s Law of Independent assortment. Dihybrid inheritance is the pattern of inheritance which involves inheritance of two characters simultaneously. In one of his experiments Mandel investigated the inheritance of the seed shape (size Vs Wrinkled) and seed colour (Yellow Vs green) at the same time. He knew from the monohybrid crosses that the round seeds were dominant to wrinkled ones and yellow seeds were dominant to green. He chose to cross plants with both dominant seed (round and yellow) with one that were recessive for both (Wrinkled and green). The F1 generation yield plants all of which produced round, yellow seeds – hard surprising as these are two dominant features. F1 seeds were planted and then allowed to self pollinate. The resulting members were a mixture of phenotype in the following proportions: 315       Round yellow (Two dominant features). 701       Wrinkled yellow (recessive and Dominant). 108     Round green (Dominant and recessive). 32        Wrinkled green (Two recessive features).

• Those numbers represent an appropriate ratio of 9:3:3:1. This is the basic dihybrid ratio.
• In a dihybrid cross, characteristic behaves independently of the other i.e Each characteristics behaves as if it is in the monohybrid cross.

Now, treating each characteristic separately we have:-   (a)       Considering seed texture (Ignore colour) Round                       Winkled 315 + 108                         101 + 32 423                                 133 133                                 133 3                 :                 1 (b)       Considering colour (Ignore seed texture) Yellow                           Green 315 + 101                     108 + 32 416                               140 140                           140 =   3                  :           1

• Thus, in the F2 generation of a dihybrid cross each characteristics, has a phenotype ratio of 3:1
• The binomial of the two ratios renders.

(3:1) (3:1) = 9:3:3:1 Thus, the dihybrid ratio is a binomial expression of two bases monohybrid rations.   Genetic representation of the dihybrid cross:- Let      R =      allele for round seed r =      allele for wrinkled seed G =      allele for yellow seed g =      allele for green seed.   Punnet square to show the fusion of gametes:-

Gametes	RG	Rg	rG	rg
RG	RRGG	RRGg	RrGG	RrGg
Rg	RRGg	RRgg	RrGg	Rrgg
rG	RrGG	RrGg	rrGG	rrGg
rg	RrGg	Rrgg	rrGg	rrgg

rrGg   =   Wrinkled, Yellow seed – 2 rrgg   =   Wrinkled, green seed     –   1 rrGG =   Wrinkled, Yellow seed –   1 Hence the ratio 9:3:3:1   How to calculate the genotype and phenotype ratio of a dihybrid cross. There are two alternative ways:-

1. By counting the number of boxes on the Punnet square containing the genotype and phenotype of interest.
2. Using a method based on the probability principle that:-

“The chances that a number of independent events will occur together, is square to the product of the chances that each event occur separately.” From above example, there is a1 in 4 chance of any gamete containing any of the F2 allele combination shown above. From a consideration of monohybrid inheritance where ¾ of the F2 phenotypes show the dominant allele and ¼ the recessive allele, the probability of the four alleles appearing in any F2 phenotype as follows: Round (dominant) ¾. Yellow (dominant) ¾. Wrinkled (Recessive) ¼. Green (Recessive) ¼. Hence the probability of the following combinations of alleles appearing in the F2 phenotypes is as follows:- Round and Yellow   =   ¾ x ¾ = 9/16. Round and green       =   ¾ x ¼ =   3/16. Round and yellow     =   ¼ x ¾ = 3/ 16. Wrinkled and green   = ¼ x ¼ = 1/16.

Mendel’s 2nd law of Inheritance (Low of Independent assortment)

In the dihybrid inheritance, Mendel realized that during gametes formation in each sex either one or another pair of factors may enter the same gametes cell (random combination) with either one or another cell. The law states that:-“Any one of a pair of characteristics may combine with either one of another pair”   Meiotic explanation of Mendel’s second low Mendel’s second law is explained by Meiosis as follows:-

• During gametes formation, the distribution of each allele from a homologous chromosome pair, is entirely independent of the distribution of alleles of another pair. It is the random alignment of the homologous chromosomes on the equator spindle in “Metaphase 1” and their subsequent separation in “Anaphase I” that leads to a variety of alleles in the gametes.

Examples

1. In the guinea pig (Cavia), there are two alleles for hair colour, black and white, and two alleles for hair length short and long. In a breeding experiment the F1 phenotypes produced from a cross between pure – breeding short black haired and pure – breeding, long white – haired parents had short black hair. Explain;

2. Which alleles are dominant, and

3. The expected proportions of F2 phenotypes.

Answer

1. If short black hair appeared in the F1 phenotypes, then short hair must be dominant to long hair and black hair must be dominant to white.
2. Let B represent black hair

b represent white hair S represent short hair s represent long hair. F1 phenotypes                         Short black hair x short black hair F1 genotypes (2n)                     SbBb             SsBb gametes                              SB             Sb       sB       sb.

Gametes	SB	Sb	sB	sb
SB	SSBB	SSBb	SsBB	SsBb
Sb	SSBb	SSbb	SsBb	Ssbb
sB	SsBB	SsBb	ssBB	ssBb
sb	SsBb	Ssbb	ssBb	ssbb

9 short black hair : 3 short white hair : 3 long black hair : 1 long white hair

2. Flower colour in sweet pea plants is determined by two allelomorphic pairs of gene (R,r and S,s). If at least one dominant gene from each allelomorphic is present in the flowers are purple. All other

genotypes are white. If two purple plants, each having the genotype RrSs, are crossed, what will be the phenotypic ratio of the offspring? Parental phenotype:             Purple x Purple Parental genotype:               RrSs     x RrSs gametes                                RS             Rs                         rS                       rs

Gamete	RS	Rs	rS	rs
RS	RRSS	RRSs	RrSS	RrSs
Rs	RRSs	RRss	RrSs	Rrss
Rs	RrSS	RrSs	rrSs	rrSs
Rs	RrSs	Rrss	rrSs	rrss

Offspring phenotype:          9 purple : 7 white

3. Consider a pea plant with round yellow seeds of the genotype Rr Yy. This means there are two pairs of homologous chromosomes. One pair carrying the allele for the colour and another pair carrying the allele for the seed form (texture). Thus chromosomes carrying the alleles for seed colour are homologous with another as those for seed form.

– At Meiosis, the homologous chromosomes come together (assort), but they carry themselves on the spindle independently of each other. They may arrange themselves in one of the following way or

RrYy: Ry, RY, rY, ry

Examples

1. In guinea pig, rough coat is dominant over smooth coat and black coat is dominant over white coat. When a rough black guinea pig was crossed with a rough white guinea pig the offspring obtained were.

328 rough black 311 rough black 111 smooth black 110 smooth white What were the genotypes of the parents? Soln Let      –     R- rough coat r- smooth coat B- black coat b- white coat – In the dihybrid cross, each character behaves independently of the other. Thus considering coat texture we have:- Rough                             Smooth 328 + 311                       110 + 111 639                                        221 221                                         221 =   3         :           1 This is a basic monohybrid ratio obtained from a cross involving two heterozygous individuals. Thus, the genotype of the rough coat with respect to this gene was Rr. Considering coat colour. Black                       White 328 + 111               311 + 110 439                               421 421                               421 =   1       :         1

• This (1:1) is a monohybrid test cross ratio obtained from a cross of a heterozygous dominant and a homozygous recessive.
• Therefore the genotype of a black coat was Bb and that of a white coat
• Therefore, the genotypes of the parents were:-

Rough black : RrBb Rough white : Rrbb 2. A tall plant with red flowers, form a true breeding line was crossed with a short plant with white flowers. One of the resulting plants was crossed in short red flowered plant unknown parentage. This cross gave the following results:- 109 – short white 38 – tall red 29 – tall white 100 –   short red

1. Interprete the results.
2. What was the phenotype of the plants produced by cross I?.

Solution According to Mendel’s 2^nd^ law, in a dihybrid cross, each characteristic behaves independently of the other. – Thus, treating each characteristic separately we have: Short                           Tall 109 + 100                         38 + 29 209                             67 67                               67 = 3               :                 1 – This is a basic monohybrid ratio obtained from a cross between two heterozygous plants – From this ratio, short is dominant over tall. Colour White                                        Red 109 + 29                                 38 + 100 138                                        188 138                                         138 1                   :                         1 – This is a monohybrid test cross ratio obtained when a homozygous recessive is crossed with a heterozygous dominant. Consider the two crosses for colour only From the above cross, the red colour was recessive to white. Defn: of symbols Let: W = White w = red S = short s = tall Punnett square to show the fusion of gametes.

Gametes	SW	Sw	sW	Sw
Sw	SSWw	SSww	SsWw	Ssww
sw	SsWw	SSww	ssWw	ssww

The phenotypes are:- 3 short white 3 short red 1 Tall white 1 Tall red From cross 1 above, the phenotypes of the product was short white.

6. Two form IV students Sophia and Issa were eager to put into practice their genetic knowledge. They carried out the following crosses:-

CROSS I A pure breed plant for terminal purple flowers was crossed with a home plant for axial white flowers. CROSS II A plant with axial purple flowers of unknown percentage was crossed with one of the products of the first cross. This cross produced the following results. 338 axial white flowers. 109 terminal purple flowers. 84 terminal white flowers. 304 axial purple flowers. – Due to their elementary knowledge in genetics, Sophia and Issa failed to interprete their results. – Using your advanced biology knowledge, show how Issa and Sophia could;

1. Interprete their results
2. Identify the genotypes and phenotypes of the plants produced in the first cross.

Solution;- According to Mendel’s second law, each characteristic in a dihybrid cross behaves independently of the other. Thus, treating each characteristic separately we have. – Considering position of the flowers, we have:- Axial                                 Terminal 338 + 304                               109 + 34 642                                               193 =             3                     :                   1 This is a basic monohybrid ratio obtained given a cross between two heterozygous individuals. From this ratio, axial flowers are dominant over terminal flowers. Considering colour of the flowers:- Purple                           White 304 + 109                     338 + 84                          =             1               :                   1 This is a monohybrid test cross ratio obtained from a cross between a heterozygous dominant and homozygous recessive. Considering the two crosses for flower colours only. Since this ratio is obtained when a heterozygous dominant is crossed with homozygous recessive, then purple was recessive and white was dominant. Definition of symbols:- Let:- A – axial a – terminal W – white w – purple Punett square to show the fusion of gametes:-

Gametes	AW	Aw	aW	aw
Aw	AAWw	AAww	AaWw	Aaww
Aw	AaWw	Aaww	aaWw	aaww

The phenotypes are in the following proportions- 3 Axial white 3 Axial purple 1 Terminal white 1 Terminal purple The results Issa and Sophia have been interpreted since the ratio obtained corresponds with the figures given. From Gross 1 above, the genotype and phenotype of the products of the flowers cross are AaWw and axial white respectively.   MERITS AND DEMERITIS OF MENDEL MERITS:-

1. Mendel was successful in his work where others had failed.
2. He was very systematic and scientific in his researches and data analysis and for this reason he managed to come out with the laws of inheritance.
3. He realized the role of gametes in the transfer of genetic information from parents to the offspring.
4. The secret behind Mendel’s success is within the following facts:-
5. Preliminary investigations were carried out to obtain familiarity with experimental organisms.
6. He paid attention to one characteristic at a time.
7. He used organisms with limited continuous variations
8. Meticulous care was taken during data collection and analysis so as to avoid introduction of contaminating variables.
9. He collected sufficient data to have statistical significance

DEMERITS: The shortfalls of Mendel include the following:-

1. His gametes describe only the diploid sexually reproducing organism. The haploid organisms such as Bryophylum are not explained.
2. His gametes is only based on the dominating- excessiveness principle’s but not all the time that one characteristic is dominant over the other.
3. Not all the time genes assert freely. Linkage interferes was free assortment.
4. Mendel did not consider gene interaction such as epistasis collaboration, lethal genes etc. all of which interfere with his basic ratio.

More examples:

1. The position of starch in pollen grains in maize is controlled by the presence of one allele of certain gene. The other allele of that gene results in starch being deposited. Explain in terms of reasons why half the pollen grains produced by a heterozygous plant contain starch.

Solution: The two alleles segregate during metaphase I and anaphase I.

1. Calculate the number of different combination of chromosomes in the pollen grains of the cross (cross balance) which has a diploid number of six (2n = 6).

Solution The number of different combination of chromosomes in the pollen grants cell is calculated, using the formula 2n, whore n is the haploid number of chromosomes. Since 2n = 6, n = 3 Therefore, combination = 23 = 8