HUMANS AS TO PROVIDE FOR THE RESULT OF CROSSING: Mendel's laws of probabilities and predictions. THE LAW OF INDEPENDENT
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Sunday, July 25, 2010
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If we know the genetic basis of a character, Mendel's laws can be used to predict the outcome of the crosses.
The Punnett square
If one or two genes are involved we can use a method called square punnet, thanks to it possimo write gametes and crossed to generate all possible zygote genotypes. Following exploiting the laws of dominance we can determine the phenotypes.
The Punnet square is a simple way to display the results of segregation and random union of alleles during fertilization, and then gives us an easy way to track the types of gametes products and possible combinations that may occur during fertilization. As shown in above each hybrid produces two types of gametes and b B in a 1:1 ratio so half pollen and eggs carry the B allele and the other half of the allele b insemination with 1 / 4 BB, 1 / 2 Bb, 1 / 4 bb. The punnet square, however, shows us two simple rules of probability
that are essential in genetic analysis. There are rules that can be very useful in the prediction of occurrence of a certain event. Using these rules offers an alternative method, and certainly faster than the Punnett square, especially when we are dealing with multiple alleles.
The law of product : the probability that two independent events occurring simultaneously is equal to the product of individual probabilities.
To illustrate, suppose there is posed the following question: From the intersection of two plants heterozygous for two different genes Aa X Aa, what is the probability that a zygote is AA? The answer is simply the probability that each of the gametes that have joined or contains A, then that is one chance in 2 1 / 2. So whereas the cross Aa X Aa is the probability that a zygote is an AA / 2 X 1 / 2 = 1 / 4 because the two gametes are produced independently, the same goes for AA.
If two events A and B are independent, the likelihood of future together
P (AxB)
P (AxB)
is given by P (A) x (B). 
So for two independent events will have a chance event chance event x 2. generally in the field of probability, two events are defined as independent when each of them does not provide information on the other. For example what is the probability that a cross between two heterozygous hybrids (YySs YySs X) are produced yellow seeds and smooth, if smooth and yellow are the dominant characters? The question we are faced with two contemporary events, that are obtained by crossing yellow seeds and smooth. To answer this question we must take account of:
1) probability that the seeds are yellow and 3 / 4 (YY and Yy). 2)
3)
likelihood of both events is 3 / 4 yellow x 3 / 4 smooth = 9 / 16. Same goes for consecutive throws of a coin, of course, are independent events, the fact that after a throw bait head does not reduce the likelihood of exit or cross head again in the next launch or the launch of another currency, consequently the probability that both coins come out with head is the product of their independent probabilities.
As we said before, this concept we apply it to genes. Formation of eggs and pollen are independent events, fertilization occurs at random, the probability that a particular combination of maternal and paternal alleles occur in the same zygote is the product of the probability of being independent of these alleles in the sperm and egg. 
intersection is the probability that X Aa Aa Aa heterozygote to form a?
The answer is 1 / 2, there are two ways to get a heterozygous A can come from the egg or sperm, and vice versa. Each of these events has possibilitàò of 1 / 4 to occur, the possibility of obtaining a total heterozygote is ¼ + ¼ = ½ 
intersection is the probability that X Aa Aa Aa heterozygote to form a? This is the additive rule
: the probability of occurrence of any of the two events that exclude each other is the sum of individual probabilities : P (A or B) = P ( A) + P (B) - [P (A) x P (B)]. If the two events do not overlap this rule we can reduce to a simple sum, such as: what is the probability that the progeny of a
Aa is a hybrid similar to the parents? In this case we must add 1 / 4, the probability that the maternal allele to join with the paternal allele and vice versa, ie the probability that the paternal allele to be united with the maternal allele that is 1 / 4, the sum is 1 / 2.
Take another example: What is the probability that by crossing two heterozygous hybrids YySs xYySs yellow seeds are produced or smooth, if smooth and yellow are the dominant characters? In this
If possible combinations are: 1) Yellow and smooth = 3 / 4 x 3 / 4 = 9 / 16 YS- 2) yellow and wrinkled = 3 / 4 x 1 / 4 = 3 / 16 Y-ss
3) Green and smooth = 1 / 4 x 3 / 4 = 3 / 16 yyS- 4) Green and wrinkled = 1 / 4 x 1 / 4 = 1 / 16 yyss
So the chances of getting yellow or smooth is: 9 / 16 + 3 / 16 + 3 / 16 = 15/16.
To give an idea of \u200b\u200bthe importance of these simple rules enough to make a somewhat more complex example in which use of the Punnett square becomes complex and we must refer to the above. In a cross between two plants heterozygous for four different genes assort independently, which fraction of the offspring will be homozygous recessive alleles for the four? For the first gene to the population of homozygous recessive offspring will be 1 / 4 as well as for the second, third and fourth. So for the independent assortment law
, the progeny will quadruple omozigoi 1 / 4 x 1 / 4 x 1 / 4 x 1 / 4 = 1 / 256. Surely it is easier to apply the method of probability rather than build a 256-square Punnett squares.
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