4.3.1 Definitions
4.3.2 Monohybrid crosses
4.3.3 Multiple alleles.
4.3.4 Codominance & Multiple alleles (ABO blood groups).
4.3.5 Sex chromosomes.
4.3.6 Sex chromosomes and genes.
4.3.7 Sex Linkage.
4.3.8 Colourblindness and haemophilia.
4.3.9 Female and sex linkage.
4.3.10 Females & X-linked recessive alleles.
4.3.11 Predicting genotypic and phenotypic ratios.
4.3.12 Pedigree
It is possible using genetic crosses to determine the genotype and phenotype of the offspring. The method used is called the Punnett square which is a simple grid which allows the genotypes and phenotypes to be determined methodically.
When you begin genetic crosses it is worth writing out in full the calculation and only later start to abbreviate your calculations. This may seem very time consuming but it will prepare you properly for the questions asked in the examination.
In the following example a very long hand form is used that includes images of chromosomes and alleles to help us track what it taking place. I strongly advise that students always think about what is taking place in the stages of meiosis and fertilisation.
Monohybrid genetic crosses: genetics involving one gene.
Example: Pea plants and the texture of their seed coats.
The characteristic of seed coat texture is controlled by by one gene with two alleles.
The seed coat can be either smooth or rough.
Smooth coat is dominant to rough coat.
One parent is homozygous dominant and the other is homozygous recessive.
Step 1: Write an allele key using the upper case letter for the dominant allele and the lower case latter for the recessive. (here S and s). It is actually better to use easily distinguished letter such as L and l, Q and q and so on.
Step 2: Write out the parental phenotype cross.
Step 3: Write down the genotypes of the parents(diploid).
In this case they are both homozygous
Step 4: Write down the genotype of the gametes (haploid).
In this case since both alleles are the same in appearance they can only produce on genotype of gamete for this gene.
Step 5 : In the Punnett square write down the possible fertilisations.
Remember these are just probabilities (chance fertilisations).
Step 6: Write out the genotypes and ratio of the offspring.
Step 7. Write out the phenotypic ratio.
The drawings help to visualize where the alleles are going. However the aim is just to use the allele code letters.
F= filial (first generation of the homozygous parent cross.) Note that they are all heterozygous.
F1 Cross = F1 (heterozygote) x F1 (heterozygote)
Phenotype = Smooth coat seed x Smooth coated seed:
Step 1: use the same allele key as above.
Step 2: Smooth Coat X Smooth coat
Step 3: Ss X Ss
Meiosis will separate the homologous pairs (Ss).
Step 4: Gamete genotypes are S and s for both parents.Step 5: Enter possible fertilisations into the Punnett grid. Note this time each parent will produce two types of gamete. The chance of producing on or other type of gamete is p= 0.5 ( 50%) or 1 in 2.
Step 6. Write out the gamete genotype and ratio.
1 SS : 2Ss: 1ss
Step 7: Write out the phenotypic ratio:
3 Smooth coats: 1 Rough Coat
Remember that homozygous dominant cannot be distinguished from the heterozygote so they all appear the same.
Points to remember:
All parental alleles must segregate in meiosis to form gametes.
There is an equal probability of each allele carried occurring in the gamete.
Fertilisation is a random process with each gamete(allele) from one parent having an equal chance of fertilizing any of other gamete ( alleles) of the other parent.
The genotypes of any particular offspring genotype or phenotype are only probabilities.
One fertilisation of two gametes does not affect he probability of the other possible fertilisations.
In the previous example F1 x F1 the ratio produced is 3:1
The probability of the offspring developing to produce smooth coated seeds is 3 in 4 or 75%
The larger the population of offspring the closer the phenotypic ratio will be to 3:1
The smaller the population the more likely a larger deviation from the 3:1 ratio.
Some genes have more than two alleles.
An individual can only possess two alleles.
The population may contain many alleles for a given gene.
Multiple alleles increases the number of different phenotypes.
Multiple alleles can be dominant, recessive or co-dominant to each other.
Example: Rabbit coat colour(C) has four alleles which have the dominance hierarchy: C > cch > ch> c
This produces 5 phenotypes, Dark(C_) , Chinchilla( cchcch), light grey (cchch ,cchc), Point restricted (ch ch, chc) and albino (cc)
The ABO blood group system is an example of both a multiple allele and codominance condition.
There are three alleles the base letter = I stands for immunoglobulin
IA and IB are codominant to each other. Both these alleles are dominant to i
The Allele hierarchy is IA = IB > i
Gender in humans is controlled by the 23rd pair of chromosomes.
XX is female and XY is male.
The female possess two X chromosomes one inherited from the father the other from the mother. They are both the longer chromosomes
The male possess one X chromosome inherited from the mother and the much shorter Y chromosome inherited form the father.
The image to the left represents the difference in the XX and XY combination. The y chromosome length is greatly exaggerated in this image.
This image represent a theoretical cross between a human male and female
The female can provide only one type of chromosome (X)
The male however provides sperm cells either with and X or with a Y.
Theoretically this means that in any fertilisation there is a P=0.5 ( 50% , 1 in 2) chance of having either a boy or a girl.
This is the basis of many genetic crosses and the one adopted here.
However studies of families shows that some families tend to have more boys than girl and some families have more girls than boys. The inheritance of this tendency is explored in the excellent book about the Y-chromosome:
Sykes, B. (2003). Adams Curse. New York: Norton PaperbacK
The genetic basis of gender is associated with the SRY gene (Sex determining region of the Y chromosome) that was identified as the previously hypothesised trf gene. This gene is normally found at the very tip of the y-chromosome but has also been found on the X-chromosome due to translocation errors. In such a case it is possible to be male and yet have XX chromosomes.
Male:
Some genes are present on the X-chromosome but missing on the shorter Y-chromosome.
The image of the male 23rdpair of homologous chromosomes represent the size difference in the two chromosomes.
In the non-homologous region of the X-chromosome a male will only have one allele for any gene in this region.
Genes in the homologous region have two alleles per gene and function just as other genes already described.
Female:
The complete length of the X-chromosome has a homologous pair on the other X-chromosome.
Genes on the x-chromosome of female therefore have two alleles just like another gene on the other chromosomes.
Genes on the non-homologous region of the X - chromosome are said to be sex linked.
Phenotypes associated with recessive alleles are more common in males than in females.
Assuming that this plant species is dioecious
The recessive allele (a) is found on the non-homologous region of the X-chromosome.
Males only get one allele for this gene.
Males have a 50% chance of being recessive.
Female have a lower risk (33.3 %) since they always receive 2 alleles.
'Recessive' males can pass on this condition( X-chromosome) to the 'daughter'.
Cannot pass these conditions to the 'sons' as they pass the y-chromosome with no alleles.
Red Green colourblindness is a sex linked condition.
The gene loci is on the non-homologous region of the X-chromosomes.
Red Green colour blindness is more common in males than in females.
Males always inherit the colourblind allele form their mothers.
Males cannot pass on colourblindness to their sons since the Y-allele does not have any of the colourblindness alleles.
Inheritance of colourblindness:
Calculation: Calculate the phenotypic ratio of a cross between a female carrier for red green colour blindness and a normal vision male.
Haemophilia
Haemophilia is a recessive, sex-linked genetic disorder.
Persons suffering from haemophilia are unable to produce clotting factor.
The haemophiliac allele (Xh)is recessive to the normal allele (XH).
The gene is located on the non-homologous region of the x-chromosome.
Haemophilia is more common in men than women.
Males inherit the allele from their mother and develop the disease.
Since (until recently) the prognosis was poor such males did not survive to pass on the allele to their daughters (its on the X-chromosome). Therefore female haemophilia would be rare.
Haemophilia can occur in the children of where the mother is a carrier and a normal male.
The mother is heterozygous for the allele (XH Xh) .
The father carries the normal allele on the x-chromosome and none on the Y chromosomes (XH Y).
We can see that from such a cross the probability of being a haemophiliac male is P=0.25 ( 25% or 1 in 4).
Today with treatment haemophiliac males can survive until sexual maturity but they cannot have daughters who are normal for this condition, why?
Historically the haemophiliac allele has played a significant role in history and not least amongst the royal families of europe.
Females can be homozygous or heterozygous for the sex-linked alleles
Carrier are individuals that are heterozygous for the allele.
The have both the dominant and the recessive (disease) allele.
Carriers do not have the disease.
1.The ability to taste the chemical PTC is determined by a single gene in humans with the ability to taste given by the dominant allele T and inability to taste by the recessive allele t. Suppose two heterozygous tasters (Tt) have a large family.
a. Predict the proportion of their children who will be tasters and non-tasters.
b. Use a Punnett square to illustrate how you make these predictions.is the likelihood that their first child will be a taster?
c. What is the likelihood that their fourth child will be a taster?
d. what is the likelihood that the first three children of this couple will be non tasters.
2.In certain trees, smooth bark is dominant over wrinkled.
a. Cross two trees that are heterozygous for smooth bark.
b. If there are 100 offspring produced, how many will have wrinkled bark.
3. A rooster with grey feathers is mated with a hen of the same phenotype. Among their offspring 15 chicks are grey, 6 are black and 8 are white.
a. What is the simplest explanation for the inheritance of these colours in chickens?
b. What offspring would you expect from the mating of a grey rooster and a black hen?
4.In Mountain Boomers, the genes for length of tail exhibit co-dominance.
a. Use a Punnett Square to predict the result of a cross between a homozygous Long-tailed and a homozygous Short-tailed Mountain Boomer.
b. Suggest what the the offspring might look like?
5. In roses, red petal is dominant over white petal. Use the allele key R for the red allele and r for the white allele.
a. Cross two heterozygous red roses,
b. Describe the phenotype of the offspring.
6. In dogs, wire hair is due to a dominant gene (W) and smooth hair is due to its recessive allele (w).
a. If a homozygous wire-haired dog is mated with a smooth-haired dog, what type of offspring could be produced?
b. What type of offspring could be produced in the F2?
c. Two wire-haired dogs are mated. Among the offspring of their first litter is a smooth-haired pup. If these two wire-haired dogs mate again, what are the chances that they will produce another smooth-haired pup? What are the chances that the pup will be wire-haired?
d. A wire-haired male is mated with a smooth-haired female. The mother of the wire-haired male was smooth-haired. What are the phenotypes and genotypes of the pups they could produce?
7.In snapdragons, red flower colour is incompletely dominant over white flower colour; the heterozygous plants have pink flowers.
a.If a red-flowered plant is crossed with a white-flowered plant, what are the genotypes and phenotypes of the plants of the F1 generation?
b. What genotypes and phenotypes can be produced in the F2 generation?
c. What kinds of offspring can be produced if a red-flowered plant is crossed with a pink-flowered plant?
d.What kinds of offspring can be produced if a pink-flowered plant is crossed with a white-flowered plant?
8. In cattle, roan coat colour (mixed red and white hairs) occurs in the heterozygous (Rr) offspring of red (RR) and white (rr) homozygotes. When two roan cattle are crossed, the phenotypes of the progeny are found to be in the ratio of 1 red : 2 roan : 1 white. Which of the following crosses could produce he highest percentage of roan cattle?
(a) red x white; (b) roan x roan; (c) white x roan; (d) red x roan; (e) all of the above crosses would give the same percentage of roan.
9.Roan colour in cattle is the result of the absence of dominance between red and white colour genes. How would one produce a herd of pure-breeding roan-coloured cattle?
10.In some cats, black colour is due to a sex-linked (X-linked) recessive gene (b); the dominant allele (B) produces orange colour The heterozygote (Bb) is calico.
a. What kinds of offspring would be expected from the cross of an orange male and a black female?
11. Haemophilia is a sex-linked trait where XH gives normal blood clotting and is dominant to the haemophilia allele Xh.
a. Give the genotypes of 1) a woman with normal blood clotting whose father had haemophilia and 2) a normal man whose father had haemophilia.
b. What is the probability that a mating between these two individuals will produce a child, regardless of sex, that has haemophilia?
c. If this couple has a daughter, what is the probability that the daughter will be a carrier of the haemophilia trait? What is the probability a daughter would have haemophilia?
d. If this couple has a son, what is the probability he will have haemophilia?
12. If a woman who is red-green colour blind mates with a man with normal vision, what phenotypes would one expect their children to have?
13. Cystic fibrosis is found on the autosomal chromosomes. It is a recessive disorder in which the allele key is dominant CFand Cf for the recessive allele.
a. A child is diagnosed with cystic fibrosis show with a diagram the likely genotype of their parents?
b. What is the probability of the next child from this couple being not having the disease?
Even more monohybrid questions
TOKBIT 4.3.11
Quote:
'Statisticians are convinced that Mendel's results are too close to exact ratios to be genuine'.
'....whether it is right to discard results that do not fit a theory as Louis Pasteur is known to have done,....'
To what extent are statements like these made with the benefit of hind site or with the benefit of historical perspective?
Is it correct to judge the work of another person by the value system and methodologies of a historically later date?
External Links
Comment:
It is most important when considering the above statements that we do not lapse into an all too easily adopted cynical evaluation of the work of people like Mendel, Pasteur, Darwin and more recently Watson and Crick. Historical revision of events of course is valuable but only if it has us reflects on our own conduct. I would suggest it is more important to realize that scientific work today which seems entirely valid (by today's standards) may well fail the the quality assurance standards of future generations. The work of the scientist mentioned and so many more besides stands as pivotal moments in scientific history. Students of the IB diploma might like to reflect on how their own work will be evaluated when they participate in the group 4 projects!
Often geneticists will carry out planned experiments in which breeding pairs are selected and the offspring phenotypes counted. However this is not acceptable or possible when working with humans. Instead geneticists have to collect information form about individuals and relatives within a family and construct diagrams of inheritance(family trees) called pedigrees.
The chart show the typical symbol found in a pedigree chart.
Circles are female(1),(3),(5), (6).
Squares are male (2), (4), (7).
Black means that the individual is affected by the condition,(3).
White indicates that the individual is unaffected by the condition.
Mating: Female 1 and male 2 (Horizontal line)
Children: Female (3) and male (4) are the children of (1) and (2).
Individuals (6) and (7) are the paternal grandchildren of (1) and (2).
Phenylketonuria (PKU) is a metabolic disorder and a recessive genetic condition.
The pedigree shows the inheritance through a particular family.
Which individuals can we be sure about their genotype?
Since it was not possible to identify the condition of 12 and 13 suggest their genotype and phenotype and how the diagram may need modifying?
