Probability

• Probability is a branch of mathematics that deals with calculating the likelihood of a given event’s occurrence, which is expressed as a number between 1 and 0.
• An event with a probability of 1 can be considered a certainty: for example, the probability of a coin toss resulting in either “heads” or “tails” is 1, because there are no other options, assuming the coin lands flat.
• An event with a probability of 0.5 can be considered to have equal odds of occurring or not occurring: for example, the probability of a coin toss resulting in “heads” is 0.5, because the toss is equally as likely to result in “tails.”
• An event with a probability of 0 can be considered impossibility: for example, the probability that the coin will land (flat) without either side facing up is 0, because either “heads” or “tails” must be facing up. A probability may be expressed as a decimal (0.75), a percentage (75%), or a fraction (3/4).

Probability in mendelian inheritance

• The probability concept has great utility in genetics. For instance, the chance of inheriting one of two alleles from a parent is 50%.
• Genetic ratios are most properly expressed as probabilities – for example, ¾ round seed: ¼ wrinkled seed.
• These values predict the outcome of each fertilization event, such that the probability of each zygote having the round seed is ¾, whereas the probability for having the wrinkled seed is ¼.
• Probabilities range from 0, when the event is certain not to occur, to 1.0 when the event is certain to occur.
• Being able to predict ratios of genotypes and phenotypes of a cross from known parental genotypes is but a first step in understanding Mendelian genetics.

Methodology to calculate Probability

• One of the easiest ways to calculate the mathematical probability of inheriting a specific trait was invented by an early 20^th century English geneticist named Reginald Punnett.
• His technique employs what we now call a Punnett square. 
• This is a simple graphical way of discovering all of the potential combinations of genotypes that can occur in children, given the genotypes of their parents. 
• It also shows us the odds of each of the offspring genotypes occurring.
• Setting up and using a Punnett square is quite simple once you understand how it works. 

Steps

• First draw a grid of perpendicular lines

• Now, put the genotype of one parent across the top and that of the other parent down the left side.  For example, if parent pea plant genotypes were YY (Dominant) and GG (recessive) respectively, the setup would be

• Note that only one letter goes in each box for the parents.   It does not matter which parent is on the side or the top of the Punnett square.  

• Next, all you have to do is fill in the boxes by copying the row and column-head letters across or down into the empty squares.  This gives us the predicted frequency of all of the potential genotypes among the offspring each time reproduction occurs.

• In this example, 100% of the offspring will likely be heterozygous (RY).  Since the R (Red) allele is dominant over the Y (green) allele for pea plants, 100% of the RG offspring will have a red phenotype as Mendel observed in his breeding experiments.
• In another example (shown below), if the parent plants both have heterozygous (RY) genotypes, there will be 25% RR, 50% RY, and 25% YY offspring on average.  These percentages are determined based on the fact that each of the 4 offspring boxes in a Punnett square is 25% (1 out of 4).  As to phenotypes, 75% will be R and only 25% will be Y.  These will be the odds every time a new offspring is conceived by parents with RY genotypes. 

Predicting results of a monohybrid cross

• A cross between individuals that involves one pair of contrasting traits is called a monohybrid cross.
• For example, a cross between a pea plant that is homozygous for producing purple flowers (genotype PP) and one that is homozygous for producing white flowers (genotype pp).
• The application of the probability concept provides a simple method to calculate probable results of a genetic cross.
• Let’s practice once more

Homozygous x Homozygous

• The progeny that result from a cross between genotype PP and genotype pp will all have Pp, and therefore, show purple flowers. Thus, there is a 100% probability that the offspring will have the Genotype Pp (Heterozygous Dominant) and the phenotype purple flower colour.

Homozygous x Heterozygous

• Consider the results of a cross between Genotype BB and Genotype Bb.
• The gametes from the genotype BB will all be B, since the genotype is homozygous.
• However, from the genotype Bb, half the gametes will be B and half will be b.
• By considering the combination of these alleles from the male and female parents, the possible genotypes that can result from this cross will be BB and Bb.
• The predicted genotype BB is 2/4 or 50% and the genotype Bb is 2/4 or 50%.
• That means, there is a 50% probability that the offspring will have the genotype BB (Homozygous Dominant) and the phenotype Black.
• There is a 50% probability that the offspring will have the genotype Bb (heterozygous dominant) and the phenotype Black. The probability of the phenotype of Black coat in every case is 4/4 or 100%.

Heterozygous x Heterozygous

• Consider the results of a cross between genotype Bb and genotype Bb.
• Half of the gametes from the genotype Bb will be B and the other half will be b.
• By considering the combination of these alleles from the male and female parents, the possible genotypes that can result from this cross will be BB (¼); Bb (½) and bb (¼).
• The probability of the phenotype of black coat in this case is 3/4 (BB + Bb) or 75%, and the probability of the phenotype of brown coat is ¼ (bb).
• The ratio of the genotypes that appear in offspring is called the genotypic ratio (1:2:1 in this case), and the ratio of the phenotypes that appear in offspring is called the phenotypic ratio (3:1 in this case).

Test Cross

• Test cross is used to determine the genotype of an individual. You perform a test cross in which an individual of unknown genotype is crossed with a Homozygous Recessive (bb) individual.
• A testcross can determine the genotype of any individual whose phenotype is dominant.
• For example, if we do not know whether the genotype of a plant is BB or Bb, we may conduct a test cross with homozygous recessive (bb) for that trait. If the unknown genotype is BB (homozygous dominant) all the offspring from the test cross will be black (Bb). If the genotype is heterozygous (Bb), about 1/2 the offspring will be black.

Predicting results of a dihybrid cross

• Predicting the results of a dihybrid cross is more complicated than predicting the results of a monohybrid cross because there are more possible combinations.
• Consider the following example in the pea plants.

R = Dominant Round; r = Recessive Wrinkled

Y = Dominant Yellow; y = Recessive Green

• Suppose you want to predict the results of a cross between a pea plant that is homozygous for genes responsible for round seed shape and yellow seed colour (RRYY) with another pea plant homozygous for genes responsible for wrinkled seed shape and green seed colour (rryy).
• The independently assorted alleles in the gametes from one parent will all be RY, and the independently assorted alleles in the gametes from another parent will all be ry.
• Therefore, genotype of all the offspring resulting from this cross will be heterozygous for both genes, that is, RrYy, and the phenotype of all the F1 offspring will be round and yellow-coloured seeds.
• Consider next a cross between two pea plants heterozygous for round and yellow seeds, with genotype Rr Yy. The gametes for both parents will be RY, Ry, rY, ry.
• If the segregation of the alleles of the gene responsible for round seed shape (R and r) is independent of the segregation of the alleles of the gene responsible for the seed colour (Y and y), and there is no interaction between the two genes (R and Y), the offspring of this dihybrid cross will show the following phenotypic segregation pattern:

¾ round (with genotype RR or Rr) x ¾ yellow (with genotype YY or Yy) = 9/16 round, yellow (R-Y-)

¾ round (with genotype RR or Rr) x ¼ green (with genotype yy) = 3/16 round, green (R- yy)

¼ wrinkled (with genotype rr) x ¾ yellow (with genotype YY or Yy) = 3/16 wrinkled, yellow (rr Y-)

¼ wrinkled (with genotype rr) x ¼ green (with genotype yy) = 1/16 wrinkled, green (rryy)

• In the above case, we simply used the combination of genetic knowledge with the probability rules to predict the results.
• This is based on the logic that for a single gene with two alleles showing dominance-recessive relationship, the F2 phenotypic segregation will be 3:1 (in the above case, ¾ round: ¼ wrinkled, and ¾ yellow: ¼ green).
• Therefore, for predicting the probability of occurrence of a pea plant in the F2 generation, from this dihybrid cross, with both round seed shape and yellow seed colour, we need to multiply the individual probability for getting round seed shape with that of the yellow seed colour.