Probability can be sued to predict the types fo progeny that will result from a monohybrid or dihybrid cross. The Punnett square is a graphical representation of these possible outcome. Phenotypes are the result of the genotype of an organism, more than one genotype may result in the same phenotype. Distinct segregation patterns result from monohybrid, dihybrid, and test-crosses.
Probability (expected frequency): probability of an outcome…# of times event is expected to happen/# of opportunities (trials)
The sum of all the probabilties of all possible events = 1
General rules:
Step 1: For outcome “A,†what events must happen? Multiply the probability of each event.
Step 2: Are there different ways to get outcome “A� Calculate the probabilities of each different way and add them up.
Multiplication Rule (Product Rule). The probability of two independent events, A and B, being realized simutaneously is given by the product of their separate probability
Prob{A & B} = Prob{A}â‹…Prob{B}
The probability of indepdent events occuring together is the product of the possibilties of the individual events: p(A and B) = p(A)p(B)
If you roll two dice, what is the chance of getting two 5′s? A 5 on the 1st die and a 5 on the 2nd die? 1/6*1/6
Addition Rule (Sum Rule). The probability of the realization of one or the other of two mutually exclusive events, A and B, is the sum of their separate probability. The probability of either of two mutually exclusive events occurring is the sum of their individual probabilites.
Prob{A or B} = Prob{A}+Prob{B}
If you roll two dice, what is the probabiltiy of two 5′s or two 6′s? Probability of a 5 on 1st die and a 5 on 2nd die or a 6 on 1st die and a 6 on 2nd die.
1/36+1/36=1/18
The Punnett Square is a way of depicting the product rule. Using Mendel’s Law of segregation, we know that both alleles are equaly likely to occur. So for a cross:
1/2R 1/2 r
1/2R 1/4RR 1/4 Rr
1/2r 1/4 rr 1/4 Rr
1/4 RR 1/4 Rr 1/2 Rr
t T
t tt tT
T Tt TT
| 60 red gum bals nd 40 green gum balls. If you buy one gum ball, what is the probability of getting a red one? |
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| # of red gum balls/total # of gum balls = 60/100 = .6 (.06 x 100% = 60%) |
| How many (ratio) progeny will have the same phenotypes of one of their parents from such a cross: AaBBCcDdee x aaBbccDdEE? |
| They must be either A_B_C_D_ee, or aaB_ccD_E_ … (½)(1)(½)(¾)(0) + (½)(1) (½)(¾)(1) = 3/16 |
| What is the probability of one green and one red gum ball if we have two quarters? Green first then red or red first then green. When not specifying order, we must figure out each way of getting the outcome. |
| p(green, then red) = .4*.6 and p(red, then green) = .6*.4 .24+.24=.48 |
| What fraction of the progeny will have large, smooth, purple fruit? SsPpLl x ssPPLl Texture: S-smooth and s-rough; P-purple and p-pink; L-large and l-small. |
| large: p(LL or Ll) = 1/4 + 2/4 = 3/4 smooth: p(Ss) = 1/2 purple: p(Pp or PP) = all = 1 large, smooth, purple p(L_S_P_) = 1/2 * 1 * 3/4 = 3/8 |
| What fraction of the progeny will not have large, smooth, purple fruit? SsPpLlxssPPLl |
| large, smooth, purple = p(L_P_S_) = 3/8. Not large, smooth, purple = 1-3/8 = 5/8 |
These are extension to the 3:1 Ratio, the ratio of monohybrid crosses (1 gene):
Definition: The phenotype of the heterzygote is different from the phenotype of either of the homozygote, and the heterzygote’s phenotype lies somewhere in the range between the phenotypes of the homozygotes.
Example: Four O’Clock flower
Red x White ⇒ Pink
Pink x Pink ⇒ 1:2:1 Red:Pink:White
Instead of seeing 3:1 ratio, we observe a 1:2:1 phenotypic ratio.
(Also note that the phenotypic ratio is the same as the genotypic ratio)
Definition: Codominance is the condition under which the heterozygote manifests phenotypically the features of two different alleles, rather than an intermediate phenotype between the homozygous dominant and homozygous recessive phenotypes.
Although there is 1:2:1 phenotypic ratio, as in incomplete dominance, the heterozygote does not represent an intermediate phenotype, but rather a phenotype in which the traits from the two different alleles coexist.
Example 1: ABO blood types. IA, IB, and IO are the alleles for a gene that produces blood antigen:
IA and IB alleles encode for antigen A and B, respectively.
IO alleles does not encode for any antigen.
An AB individual has both A and B antigen. So allele IA and IB are codominant.
Example 2: The L and M antigens are similar to the ABO system. The ability to produce M and N antigens is determined by alleles of single gene.
Homozygotes for M allele LMLM produce only M antigens.
Homozygotes for N allele LNLN produce only N antigens.
Heterozygotes LMLN produce both M and N antigens.
Question: If two AB blood type individuals mate, what type of blood will the offspring have?
Answer: There will be a 1 :2: 1 ratio, in which 1/4 of the offspring will be type A, Y2 will be type AB, and 1/4 will be type B.
AB x AB ⇒ 1A : 2AB : 1B
They are often multiple forms of the same gene present in a population. As a diploid organism, only two of the alleles are expressed in the organism. Many genes have more than two alleles: for example: blood groups (discussed above) have 3 (A, B, O).
Example: Multiple alleles determine hair color in rabbits.
Genotype Phenotype
cc
Albino -white hair allover the body
Ch Ch Himalayan- Black hairs on the extremities and white hair on the body
Cch Cch Chinchilla- White hair with black tips over the entire body CC colored hair(wild type)
Cch and Ch show codominance: the genotype Cch Ch has white hair with black tips on the body but black hair at the extremities.
Cch and c show incomplete dominance: Crossing a chinchilla with an albino produces a rabbit with white hair with gray tips (light chinchilla). It’s genotype is Cch c.
Some alleles turn out to be lethal if carried by individuals. Most of these lethal alleles are only lethal when homozygous. Such alleles may distort the expected 3:1 ratio a 2:1 ratio may be observed because the individuals that are homozygous for the lethal allele die.
Example: coat color in mice.
Yellow coat color (Y) is dominant over brown coat color (y). When two heterozygous (Yy) mice mate, 2/3 of the offspring are yellow (Yy) and 1/3 are brown (yy). The 2:1 ratio results because the homozygous yellow (YY) mice die.
Yy x Yy ⇒ 1 YY (dead!) : 2 Yy (yellow) : 1 yy (brown)
Question: What happens if the lethal allele is sex-linked?
Answer: The offspring will skew toward having more females than males.
| Cross | Resulting Genotypes |
| AaBb x AaBb | 9: A_ B_ 3: A_ bb 3: aa B_ 1: aa bb |
The 3:1 ratio deals with one gene. The 9:3:3:1 ratio deals with two genes. However, there are many extensions to this ratio. The classical 9:3:3:1 ratio is produced when we cross two double heterozygote individuals exhibiting traits with complete dominance. The table below explains patterns which result via extensions of the 9:3:3:1 ratio.
| Ratio | Term | Meaning | ||||||||||||||||||||||
| 9:3:3:1 | No Effect | The expression of gene A has no effect on the expression of gene B. Interaction between gene A and B will lead to the variation of this ratio. There are 9 possible variations to this ratio. The most common ones result from epistasis, duplication, complementation, and suppression, and are listed below. | ||||||||||||||||||||||
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| 9:7 | Complementation |
Complementation implicates different genes needed for a particular trait. If two recessive mutations are alleles of the same gene, then the phenotype of an organism containing both mutations is mutant; if they are alleles of different genes, then the phenotype of an organism containing both mutations is wild type (non mutant). The flower color in sweet peas. If you cross two mutant white flowers and end up with purple wildtype F1 progeny, complementation occurs.
At each pair of loci in the F1 heterozygous progeny, there is at least one dominant allele present. Therefore, all of these flowers will be purple. Would you expect to see complementation if the mutations were in the same genes? If the mutations are on the same gene, crossing two mutations will not restore the wildtype phenotype. In the other words, of the two mutations don’t complement. (Complementation is a commonly used test in genetics to examine whether the mutations are on the same gene). |
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| 12:3:1 | Dominant epistasis | |||||||||||||||||||||||
| 9:4:3 | Recessive Epistasis |
Recessive epistasis, a condition where gene A masks the effect of gene B when A is homozygous aa. For example, the production of pigment (i.e. Albinism) in a species of snake. We are looking at two genes (P and G) which affect coloration in these snakes:
Cross two heterozygotes (green snakes): PpGg x PpGg
The phenotypic ratio observed here is 9:3:4. (9 green: 3 orange: 4 albino). In this case, the P gene determines whether pigment will be produced (P) or not produced (p). The G gene determines if the color produced will be green or orange. If the genotype is pp, then no pigment is produced, and the allele of the color (G) allele is irrelevant. |
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| 9:7 | Suppression |
Suppression, where One genotype suppresses the expression of a second genotype. This usually results in only two phenotypes observed. Sometimes one gene suppresses the effect of an abnormal (mutant) allele of another gene, resulting in the normal phenotype. The wild type phenotype is therefore restored. |
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| 13:3 | ||||||||||||||||||||||||
| 15:1 | Duplicate Genes |
Duplicate genes, where only one dominant allele of either gene is needed in order to produce a given phenotype (two genes share the same phenotype). This results in a 15: 1 ratio, where only the homozygous recessive are different from the rest of the offspring. Question: The plant shepard’s purse has 2 different fruits (heart shaped or narrow). A cross between 2 heart-shaped heterozygotes produces offspring with heart-shaped fruits and narrow fruits in the ratio of 15: 1. How might you explain this? Using an example, discuss why duplication is an important precursor to evolution. Answer: The observation that there are only two phenotypes might make you think that there is only 1 gene involved, which isn’t the case since you should see a 3:1 ratio in a monohybrid cross. However, if that gene becomes duplicated and you cross two individuals that are heterozygous for both copies of the gene, only l out of 16 will not have a single dominant allele. The dominant phenotype appears if a dominant allele is present for either copy of the gene. This is true because the genes have identical actions. Al a1 A2 a2 x Al al A2 a2 If there is a dominant allele for either Al or A2 a heart shaped fruit is produced. If a mutation occurs in one of the copies and causes production of an enzyme slightly different from the original, its carrier has not lost the gene’s function. Moreover, if the duplicated gene mutates again, causing it to produce a novel gene product that is valuable the individual may actually be at an advantage relative to other individuals. Duplication is an important precursor to evolution because with two identical genes one is free to evolve into something better. |
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| 9:6:1 | ||||||||||||||||||||||||
The following questions refer to offspring ratios after the mating of individuals heterozygous for the suppressor (gene A) and heterozygous for the gene it suppresses (gene B), e.g. AaBb x AaBb
Question 1: What ratio of offspring phenotypes would you expect if suppression occurs when “gene A” possesses a dominant allele and the target gene produces a mutant phenotype when “gene B” is homozygous recessive?
A = suppressor gene: suppresses mutant pigment when genotype is AA or Aa
B = pigment gene: produces mutant pigment when genotype is bb
9 A_B_ normal
3 A_bb normal (mutant pigment is suppressed)
3 aaB_ normal
1 aabb mutant (not suppressed)
Ratio is 15 : 1 (normal : mutant).
Question 2: What ratio of offspring phenotypes would you expect if suppression occurs when “gene A” is homozygous recessive and the target gene produces a mutant phenotype when “gene B” possesses a dominant allele?
A= suppressor gene: suppresses the mutant pigment when genotype is aa
B= pigment gene: produces a mutant pigment when genotype is BB or Bb
9 A_B_ Mutant (not suppressed)
3 A_bb normal
3 aaB_ normal (mutant pigment is suppressed)
1 aabb normal
Ratio is 7 : 9 (normal : mutant).
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