Monohybrid InheritanceTest Cross or BackCrossDihybrid InheritanceStatistical test - Chi squaredCo - DominanceSex DeterminationSex LinkageLinkage

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  • Created by: Ahmra
  • Created on: 19-03-13 17:02

Monohybrid Inheritance

Mendel's early experiments were based on selecting pea plants of two varities which showed clearly seperable characteristics such as tall and dwarf plants and round and wrinkled seeds. From his results he formulated his first law of inheritance, the Law Of Segregation, which states that :

" The characteristics of an organism are determined by factors ( Alleles ) which occur in pairs. Only one of a pair of factors ( Alleles ) can be present in a single Gamete. "

The inheritance of a single pair of contrasting characteristics is known as Monohybrid Inheritance. In monohybrid crosses, two heterozygous individuals will produce offspring with a phenotype ratio of three dominant to one recesive, 3:1

Mendel was lucky in his choice of characters in pea plant because these particular characters are controlled by single genes. Pea plants are either tall or dwarf, flower colour are clear-cut and easy to tell apart. This is an example of Discontinuous Variation. However, most characters are controlled by a number of genes, for instance height in humans. People are not just tall or short but show a range of heights. This is an example of Continuous Variation.

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Instructions for carrying out a genetic cross

  • Look carefully at the information you are given in the question, e.g. about phenotypes of the parents, whether they are homozygous or heterozygous
  • Choose suitable symbols for the alleles, unless they are provided in the question
  • Choose a single letter to represent each characteristic
  • Choose the first letter of one of the contrasting feature
  • If possible, choose a letter in which the higher and lower case forms differ in shape as well as size
  • Let higher case letter represent the dominant feature and the lower case letter as the recessive one
  • Represent the parents with the appropriate pairs of letters
  • Label them clearly as 'parents' and state their phenotypes
  • State the gametes produced by each parent.
  • Circle the gametes and label them clearly
  • Use a matrix called Punnet Square to show the results of the random crossing of the gametes
  • State the phenotype of each different genotype and indicate the numbers of each type
  • Make sure that you have actually answered the question.
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Test Cross or BackCross

This is a method used in genetics to determine whether a particular dominant characteristic observed in an organism is determined by one or two dominant alleles.

For example, a prize black bull bought at a market by a farmer would be expected to be pure-breeding, homozygous dominant (BB). Or if not a pedigree bull it would be heterozygous (Bb). The phenotype is identical in both cases.

The backcross consists of crossing the bull with known 'recessive' genotype, a white cow. (The double recessive phenotype has a known genotype beacuse only one allele combination can produce it.)

If resulting offspring are all black then the bull is pure-breeding or homozygous. If the resulting offspring include both black and white forms then the bull was heterozygous.

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DiHybrid Inheritance

This involves the inheritance of two seperate genes. Mendel knew from his early experiments with monohybrid crosses that round seed shape was dominant to wrinkled, and that yellow colour was dominant to green. He used plants which differed by having two pairs of contrasting characters. He crossed homozygous pea plants with the two dominant characters, round and yellow seeds with homozygous plants with the two recessive characters, wrinkled and green.

  • Homozygous plants with round and yellow seeds were crossed with homozygous plants with wrinkled and green seeds
  • He found that all the F1 generation had round, yellow seeds
  • When plants grown from these seeds were self-pollinated the seeds produced were of four different types of shapes and colour of seed coat
  • He collected and counted the seeds and found there were 4 types shown in the table

Characteristics -  Round Yellow  -  Round Green  -  Wrinkled Yellow   -   Wrinkled Green

Totals              -         315                  108                    101                          32

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DiHybrid Inheritance Continued........

Each total can be divided by the double recessive total to give an approximate whole number ratio between the phenotyps, i.e. 9:3:3:1 is the ratio between the 4 totals in the table.

The ratio proportions are known as the dihybrid ratio.

  • This led to Mendels Second Law, which states that :

    " Either one of a pair of contrasted characters may combine with either of another pair.'

  • With or preswnt knowledge of genetics this statement can be rewritten as:

    " Each member of an allelic pair my combine randomly with either of another pair."

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A Statistical Test - Chi Squared

The expected ratio of phenotypes in the offspring of a dihybrid cross is 9:3:3:1. This ratio represents the probability of getting these phenotypes. It would be surprising if the numbers came out exactly in this ratio.

So how close do the observed results have to be to the expected, and have the differences between them happened by chance, or are thet so different that something unexpected is taking place? To answer this question scientists use a statistical test called the Chi-Squared Test. The chi squared test is used to compare the observed results with those expected. It is a way of estimating the probability that differences between observed and expected results are due to chancealone and not some other factor influencing the results.

Consider the following F2 results obtained from the cross between plants with round, yellow seeds and those with wrinkled green seed.

Characteristic   -     Round Yellow     Round Green     Wrinkled Yellow     Wrinkled Green

Totals              -           315                   108                      101                        32

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Chi - Squared procedure

1 ) Calculate expected values (E) . This is total number of seeds divided by the number of possible types, for example, for roundand yellow (556 x 9) / 16 = 312.75

2 ) Calculate the differences between the observed (O) and expected (E) results

3 ) Square the differences

4 ) Use the formula : Sum Of  (O - E) squared / E and complete the table

Phenotype         Observed (O)    Expected (E)    Difference (O-E)    (O-E)^2    (O-E)^2/E

Round Yellow           315                313                       2                       4             0.01

Round Green            108                104                       4                      16            0.15

Wrinkled Yellow          101                104                      -3                       9             0.08

Wrinkled Green           32                  35                        3                       9             0.26

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Chi - Squared procedure Continued.....

5 ) Total the values in the last column     0.01 + 0.15 + 0.08 + 0.26 = 0.50 = X^2

6 ) Work out the degrees of freedom. This is a measure of the spread of the data. It is always one less than the number of classes of data. In the example there are 4 different phenotype combinations, so there are 3 degrees of freedom

7 ) To find out if this value is significant or non-significant it is necessary to use a Chi-Squared Table

8 ) Statisticians consider that if the probability is greater than 5% the deviation is said to be non-significant. In other words, the deviation is due to chance alone. If the deviation is less than 5% level, the deviation is said to be significant. That is, some factor other than chance is influencing the results

9 ) Looking along the column for 3 degrees of freedom it can be seen that Chi^2 value of 0.50 lies between 2.37 and 0.12 which is equivalent to a probability between 0.50 (50%) and 0.99 (99%). This means that the deviation from the 9:3:3:1 ratio is non-significant and is simply the results of statistical chance.

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Not all Characters are controlled by single genes which behave independently, as was the case in Mendel's experiments.

The monohybrid and dihybrid crosses considered so far involve alleles that are either dominant or recessive. Sometimes both alleles are expressed and neither is dominant. When alleles express themselves equally in the phenotype this is knwon as codominance. In most cases the heterozygote shows a phenotype intermediate between those of the two homozygotes.

Examples of codominance are :

  • Snapdragon plants have the homozygous genotypes RR and WW and produce Red or White flowers. However, if the two homozygous plants are crossed, the offspring are pink. That is, the two parents produce an intermediate offspring.
  • Similarly, shorthorn cattle have the genotypes and phenotypes RR (red), RW (roan) and WW (white) coat colour.
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Co-Dominance Continue......

The genetic diagram for these crosses is the same as that illustrating Mendel's First Law but in the F1 generation, all individuals have the intermediate phenotype.

Parental Phenotype       Red Flowers        x       White Flowers

Parental Genotype             RR                 x             WW

Gametes                            R                  x               W

F1 Genotype                                         RW

F1 Phenotype                                     All Pink

Ratio                                                   100 %

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Sex Determination

Most sexually reproducing animals show two morphologically distinct types, Male and Female, which are associated with the chromosomes found in the two types. One pair, known as the Sex Chromosomes, is similar in one sex and dissimilar in the other. The non-sex chromosome are known as Autosomes.

  • Humans have 46 chromosome arranged in 23 pairs. The first 22 pairs are autosomes; the last pair are the sex chromsome
  • In humans the males has a dissimilar pair of chromosomes, called X and Y, whilst the female has a pair of two similar X chromosomes
  • All the female's eggs contain an X chromosome. Half the male's sperm contains an X chromosme and the other half contains a Y chromosome
  • At fertilisation the egg may join with either an X sperm or a Y sperm. This gives an equal chance of the child of being a girl or a boy.
  • Parental Phenotype:         Male                 Female
  • Parental Genotype:            XY                      **
  • Gametes:                           X Y                       X
  • Offspring genotypes :                  ** , XY
  • Offspring Phenotype:      50% male and 50% female
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Sex Linkage

Some Alleles carried on the X chromosome, so they are described as sex linked. The Y chromosome is much smaller than the X and carries very few genes. Therefore in the males any recessive genes carried on the X chromosome will express themselves in the phenotype. This is because they are unpaired and so there is no dominant gene present. This special form of inheritance is known as sex linkage, an important fetaure of which is that the male connot hand on the genes to his sons as they must recieve the Y chromosome to become a male.

On the other hand, all his daughters must recieve the recessive gene from him. Females who are heterozygous for sex-linked recessive traits are known as carriers and have a 50% chance of handing on the recessive to their sons.

An example of a sex-linked trait in humans is Haemophilia. It is caused by a recessive allele on the X chromosome. The gene that codes for factor VIII, an important protein involved in blood clotting, is a sex-linked gene located on the X chromosome.

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  • Haemopilia is a potentially lethel condition. It is the result of an individual being unable to produce one of the many clotting factors. The inability of the blood to clot leads to slow and persistant bleeding.
  • It is now possible to extract the particular clotting factors from donated blood allowing haemophiliacs to lead near-normal lives. The risk of the disease passing to their offspring still remains. This condition occurs most exclusively in males.
  • When the recessive allele occurs in males it expresses itself because the Y chromosome cannot carry any corresponding dominant allele.
  • For the condition to arise in females it requires the double recessive state and as the recessive allele is relatively rare in the population this is unlikely to occur.
  • To obtain an affected femal, the father must be affected and the mother must either be affected or a carrier.
  • H is the allele for Normal bllod clotting, h is the allele for haemophilia
  • XHXH - Female normal
  • XHXh - female carrier
  • XhXh - female haemophiliac
  • XHY - male normal
  • XhY -  male haemophiliac.
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Mendel was fortunate in his choice of characters in pea plants because these particular characters are controlled by single genes. Pea plants are either tall or dwarf, flower colours are clear-cut and easy to tell apart.

However, linked genes on the same chromosomes pass into the gamete and then into the offspring together. The crosses and ratios considered so far have involved two pairs of contrasting characters found on different chromosomes. Crosses involving linkage do not follow the typical Mendelian Pattern. These ratios will not be obtained if the genes are found on the same chromosome.

Linkage takes place when two different genes are located on the same chromosome. The genes are inherited together, because they move together during meiosis and appear in the same gamete.

Recombination takes place when alleles are exchanged between homologous chromosomes as a result of crossing over. The further apart two genes are on a chromosome, the more chance there is of cossing over taking place.

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Linkage Continued......

In meiosis linked genes are passed togther to the gametes so that moast of the F2 progeny inherit the characterstics determined by the linked genes.

When crossing over occurs, an opportunity is provided for linked genes to be seperated and give rise to recombinants.

However, crossing over may take place in only a small proportion of the cells undegoing meiosis, between 5 and 10%. Therefore few gametes will contain recombinants genes.

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Marta Kohls


excellent! :D

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