AGRY 320 Assignment #1: Solutions
Mendelian Genetics
1)Define the following terms: true-breeding, monohybrid experiment, Punnett square, parental generation, F1 generation, F2 generation, locus, allele, genotype, phenotype, dominant, recessive, homozygous, heterozygous,.
true-breeding: an individual that is homozygous and produces identical homozygous offspring when self-fertilized or hybridized (mated) with an individual with the same homozygous genotype.
monohybrid experiment: an experiment where two true-breeding parents differ for only one trait
Punnett square: a graphic model where the phenotypic and genotypic proportions of zygotes (fertilized eggs) can be predicted.
parental generation: the varieties that serve as parents for a cross.
F1 generation: the first filial generation; the first generation produced after mating between parents that are homozygous for different alleles.
F2 generation: the second filial generation; the generation produced by intermating of F1 individuals.
locus: the position of a gene on a chromosome; used synonymously with the term gene in many instances.
allele: a particular form of a gene, presumably reflecting a certain DNA sequence. Different DNA sequences at the same locus are called different alleles.
genotype: the genetic makeup of an individual organism.
phenotype: the observable outward appearance of an organism, which is controlled by the genotype and its interaction with the environment.
dominant: an allele or trait that expresses its phenotype when heterozygous with a recessive allele. OR: the ability of one allele to phenotypically mask the effect of another allele in a heterozygote.
recessive: an allele or trait that does not express is phenotype when heterozygous with a dominant allele.
homozygous: a condition where the two alleles on homologous chromosomes are identical.
heterozygous: a condition where the alleles at the same locus on homologous chromosomes are different.
2)Briefly outline the principle of segregation and principle of independent assortment.
Principle of Segregation: because homologous chromosomes segregate from each other during meiosis, alleles at the same locus on homologous chromosomes also segregate from each other, so half the gametes receive one allele and half receive the other.
Principle of Independent Assortment: because genes located on nonhomologous chromosomes assort independently during meiosis, the inheritance of alleles at one locus does not influence the inheritance of alleles at another locus.
3) For the two parents A1A1B1B1; A2A2B2B2 give the frequency of the progeny for a
a) backcross
b) F2
c) F3
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a)
A1A1B1B1 x
A2A2B2B2
gametes A1B1 A2B2 gametes
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b)
A F2 is the intermating of the F1. From part a we know that F1 is 100% A1A2B1B2 and the genotype of the gametes and the probability associated to them. Then, the Punnett square looks as follows:
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Male gametes
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¼ A1B1 |
¼ A1B2 |
¼ A2B1 |
¼ A2B2 |
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Female gametes |
¼ A1B1 |
1/16 A1 A1B1 B1 |
1/16 A1 A1B1 B2 |
1/16 A1 A2B1 B1 |
1/16 A1 A2B1 B2 |
|
¼ A1B2 |
1/16 A1 A1B1 B2 |
1/16 A1 A1B2 B2 |
1/16 A1 A2B1 B2 |
1/16 A1 A2B2 B2 |
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|
¼ A2B1 |
1/16 A1 A2B1 B1 |
1/16 A1 A2B1 B2 |
1/16 A2 A2B1 B1 |
1/16 A2 A2B1 B2 |
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|
¼ A2B2 |
1/16 A1 A2B1 B2 |
1/16 A1 A2B2 B2 |
1/16 A2 A2B1 B2 |
1/16 A2 A2B2 B2 |
|
Adding up repeated genotypes from the table (same colors/patterns), the F2 is:
1/16 A1A1B1B1
1/16 A1A1B2B2
1/16 A2A2B1B1
1/16 A2A2B2B2
2/16 A1A1B1B2
2/16 A1A2B1B1
2/16 A1A2B2B2
2/16 A2A2B1B2
4/16 A1A2B1B2
c) From part b, we can get the genotypes of the gametes that each of the individuals in the F2 produces:
1/16 A1A1B1B1 : 100% gametes A1B1
1/16 A1A1B2B2 : 100% gametes A1B2
1/16 A2A2B1B1: 100% gametes A2B1
1/16 A2A2B2B2 : 100% gametes A2B2
2/16 A1A1B1B2 : ½ gametes A1B1 , ½ gametes A1B2
2/16 A1A2B1B1: ½ gametes A1B1 , ½ gametes A2B1
2/16 A1A2B2B2 : ½ gametes A1B2 , ½ gametes A2B2
2/16 A2A2B1B2 : ½ gametes A2B1 , ½ gametes A2B2
4/16 A1A2B1B2 : 1/4 gametes A1B1 , 1/4 gametes A1B2 , 1/4 gametes A2B1 , 1/4 gametes A2B2
Adding up the probabilities for the possible ways of getting a particular gamete:
A1B1 = ¼
A2B1 = ¼
A1B2 = ¼
A2B2 = ¼
Example: for gamete A1B1, there are the following possibilities:
Probability (gamete A1B1) = probability (A1A1B1B1 and produces gamete A1B1) + probability (A1A1B1B2 and produces gamete A1B1) + probability (A1A2B1B1 and produces gamete A1B1) + probability (A1A2B1B2 and produces gamete A1B1)
Probability (gamete A1B1) = 1/16*1 + 2/16*1/2 + 2/16*1/2 + 4/16*1/4 = 4/16 = 1/4
Finally, the frequency of each of the possible gametes is ¼. The Punnett square will look as follow:
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Male gametes
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¼ A1B1 |
¼ A1B2 |
¼ A2B1 |
¼ A2B2 |
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Female gametes |
¼ A1B1 |
1/16 A1 A1B1 B1 |
1/16 A1 A1B1 B2 |
1/16 A1 A2B1 B1 |
1/16 A1 A2B1 B2 |
|
¼ A1B2 |
1/16 A1 A1B1 B2 |
1/16 A1 A1B2 B2 |
1/16 A1 A2B1 B2 |
1/16 A1 A2B2 B2 |
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¼ A2B1 |
1/16 A1 A2B1 B1 |
1/16 A1 A2B1 B2 |
1/16 A2 A2B1 B1 |
1/16 A2 A2B1 B2 |
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¼ A2B2 |
1/16 A1 A2B1 B2 |
1/16 A1 A2B2 B2 |
1/16 A2 A2B1 B2 |
1/16 A2 A2B2 B2 |
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This table is the same as the one we constructed in part b to get the F2. The genotypic frequencies of the F3 are then:
1/16 A1A1B1B1
1/16 A1A1B2B2
1/16 A2A2B1B1
1/16 A2A2B2B2
2/16 A1A1B1B2
2/16 A1A2B1B1
2/16 A1A2B2B2
2/16 A2A2B1B2
4/16 A1A2B1B2
which are exactly the same frequencies we got in part b for the F2.
F3 has the same genotypic frequencies as the F2. If we repeat the process to get the F4 from the F3 we are going to get the same results.
4) For the two parents A1A1B2B2C1C1; A3A3B1B1C2C2 the frequency of the progeny for a
a) backcross
b) F2
c) F8 (Challenge)
a) The genotype of the F1 is A1A3B2B1C1C2
To get the backcross offspring, we have to cross the F1 by one the parents (let’s assume A1A1B2B2C1C1). The gametes from the parent are all A1B2C1 whereas the gametes from the F1 individuals are:
1/8 A1B2C1
1/8 A1B2C2
1/8 A1B1C1
1/8 A1B1C2
1/8 A3B2C1
1/8 A3B2C2
1/8 A3B1C1
1/8 A3B1C2
Then, the frequency of the progeny for the back cross is:
1/8 A1A1B2B2C1C1
1/8 A1A1B2B2C1C2
1/8 A1A1B1B2C1C1
1/8 A1A1B1B2C1C2
1/8 A1A3B2B2C1C1
1/8 A1A3B2B2C1C2
1/8 A1A3B1B2C1C1
1/8 A1A3B1B2C1C2
b) Because each individual in the F1 can give 8 different gametes (part a), a Punnett square would consist of 8 rows and 8 columns. In order to avoid making the computations for 64 cells, let’s divide the genotype of the F1 in individual genes A, B and C (each with two alleles).
Then, for gene A, the genotypic frequency in the F2 is ¼ A1A1 ½ A1A3 ¼ A3A3 (these frequencies come from intermating individuals A1A3 which is the genotype of the F1 in this exercise if we only look at gene A. Doing the same for the three genes individually, the F2 are:
¼ A1A1 ½ A1A3 ¼ A3A3
¼ B1B1 ½ B1B2 ¼ B2B2
¼ C1C1 ½ C1C2 ¼ C2C2
Because the each gene segregates independently of one another, we can multiply the frequencies computed above to get the segregation ratios in the F2:
A1A1B1B1C1C1 : 1/64 (1/4*1/4*1/4)
Triple homozygotes: 1/64 (8 combinations)
Double homozygotes:2/64 (12 combinations)
Simple homozygote: 4/64 (6 combinations)
Heterozygotes: 8/64
c) The frequencies in the F8 are the same as the frequencies in the F2.
5)You have two true-breeding plants and you are focusing on the trait of thorn type, one plant is homozygous for the dominant allele of large thorns and the other is homozygous for the recessive allele of small thorns. What will be the segregation ratio of the F1, the F2, and the F3 generation?
Assume T represents large thorns, and t represents small thorns (or any letter). Since both parents are homozygous for the given allele, all F1 offspring will be heterozygous for the trait. That is, all progeny will have a Tt genotype. To generate the F2, F1 plants are intermated. The Punnett square looks as follows:
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T |
t |
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T |
TT |
Tt |
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t |
Tt |
tt |
Genotypic ratio: 1 TT, 2 Tt, 1 tt
Since T is dominant over t, ¾ of the offspring have large thorns, and ¼ have small thorns.
Segregation ratio in the F3 is the same as the segregation ratio in the F2: 1 TT, 2 Tt, 1 tt
6) Given two genes, one coding for scale texture and one coding for type of exhaling, and the alleles (rough or smooth scales, fire or smoke breathing): develop symbols for the genes and the alleles. If you have a rough scaled, fire-breathing plant, what are the possible genotypes? Assume that rough scales are dominant over smooth scales and fire-breathing is dominant over smoke-breathing.
Any letters are fine to use, with capital letters denoting the dominant version of a given allele, and the lower-case denoting the recessive allele. For example, scale texture can be R for rough scales (the dominant allele) and r for smooth scales. Fire-breathing can be F, while f would denote smoke-breathing.
A rough scaled, fire-breathing plant can be any of the following four genotypes:
RR FF, Rr FF, RR Ff, or Rr Ff.
7) Construct the F2 generation if your parental generation was made up of a rough scaled, fire-breathing plant and a smooth-scaled, smoke-breathing plant. Assume that both parental plants were double heterozygotes. Write out the genotypic and phenotypic ratios of the progeny.
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RF |
Rf |
rF |
rf |
RF |
RR FF |
RR Ff |
Rr FF |
Rr Ff |
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Rf |
RR Ff |
RR ff |
Rr Ff |
Rr ff |
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rF |
Rr FF |
Rr Ff |
rr FF |
rr Ff |
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rf |
Rr Ff |
Rr ff |
rr Ff |
rr ff |
Genotypic Ratio:
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RR FF |
1 |
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RR Ff |
2 |
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Rr FF |
2 |
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Rr Ff |
4 |
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RR ff |
1 |
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Rr ff |
2 |
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rr FF |
1 |
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rr Ff |
2 |
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rr ff |
1 |
Phenotypic Ratio:
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rough scales, fire-breathing |
9 |
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rough scales, smoke-breathing |
3 |
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smooth scales, fire-breathing |
3 |
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smooth scales, smoke-breathing |
1 |
8)Assume that for a given breed of dog, for the trait tooth type, huge sharp teeth are dominant over little sharp teeth and that for the trait tooth shape and size are together determined by a single gene. You are a dog breeder and all of your customers want dogs with huge, sharp teeth. To maximize profits, therefore, you want to ensure all future puppies have huge, sharp teeth. You have two litters of puppies from two different sets of parents. One litter all has little, sharp teeth and the other has all huge, sharp teeth. Which puppies would you use to breed and is the litter with all little, sharp teeth of any use to you? Show the proper analysis and all possible results that led you to your conclusions.
Puppies with huge, sharp teeth will not necessarily produce offspring that also have huge sharp teeth. This is because two heterozygous dogs with huge, sharp teeth will produce puppies with little, sharp teeth ¼ of the time. You therefore need to find the puppies that are homozygous for huge, sharp teeth and use them in future breeding. This is done by doing a test-cross, where the dogs showing the dominant phenotype are mated with a homozygous recessive dog.
Assume H represents huge, sharp teeth and h represents little, sharp teeth. There are two possible genotypes of dogs with huge, sharp teeth: HH and Hh. Each of these is mated to a homozygous hh dog, with the following possible results:
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H |
H |
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h |
Hh |
Hh |
Mating an HH dog with an hh dog:
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Result: all offspring have huge, sharp teeth if the parent with the huge, sharp teeth was homozygous (HH) for this trait.
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H |
h |
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h |
Hh |
hh |
Mating an Hh dog with an hh dog:
Result: half the offspring will have huge, sharp teeth and the other half will have little, sharp teeth if the parent with the huge, sharp teeth was heterozygous (Hh) for this trait.
Therefore the litter with the little, sharp teeth is useful to you, since it allows you to perform the test-cross to determine the genotype of the dogs with the huge, sharp teeth.
9)A number of researchers repeated Mendel’s crosses in order to confirm his genetic principles. For example, Carl Correns and Erich von Tschermak crossed yellow and green cotyledon colored varieties of peas and self-fertilized the subsequent F1 progeny to get F2 progeny arrays. Correns observed 1,394 yellow and 453 green, while Tschermak observed 3,580 yellow and 1,190 green. Are these values consistent with the principle of segregation? Do the test.
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Researcher |
Pea Color |
Observed Numbers |
Expected Numbers |
[(O – E)/E]2 |
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Correns |
Yellow |
1 394 |
1 847 x (0.75) = 1 385.25 |
0.055 |
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Green |
453 |
1 847 x (0.25) = 461.75 |
0.166 |
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Total: |
1 847 |
1 847 |
χ2 = 0.221 |
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Tschermak |
Yellow |
3 580 |
4 770 x (0.75) = 3 577.5 |
0.002 |
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Green |
1 190 |
4 770 x (0.25) = 1 192.5 |
0.032 |
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Total: |
4 770 |
4 770 |
χ2 = 0.074 |
In this case, there is 1 degree of freedom (2 categories (yellow and green) minus 1 = 1 degree of freedom). As a result, the χ2 values are not significant because both are less than 3.84 (see the partial Chi-square table, below), making the observations consistent with the principle of segregation.
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Probability |
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Degrees of Freedom |
0.9 |
0.5 |
0.1 |
0.05 |
0.01 |
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0.02 |
0.46 |
2.71 |
3.84 |
6.64 |
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2 |
0.21 |
1.39 |
4.60 |
5.99 |
9.21 |
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3 |
0.58 |
2.37 |
6.25 |
7.82 |
11.34 |
10) There is a tetraploid that uses bivalent random pairing in meiosis. What are the possible F2 progeny of the parents (challenge)
a) A1A1A1A1; A2A2A2A2
b) A1A1A2A2;A3A3A4A4
Note: same of the solutions presented below includes the segregations frequencies for all generations, which are not necessary to obtain full grade.
a)
There is only one possible genotype for the F1: A1A1A2A2. To get the F2, individuals from the F1 are intermated. The genotypic frequencies of the gametes produced by F1 individuals are:
1/6 A1A1
1/6 A2A2
4/6 A1A2
So that the Punnett square would look like:
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1/6 A1A1 |
4/6 A1A2 |
1/6 A2A2 |
1/6 A1A1 |
1/36 A1A1A1A1 |
4/36 A1A1A1A2 |
1/36 A1A1A2A2 |
4/6 A1A2 |
4/36 A1A1A1A2 |
16/36 A1A1A2A2 |
4/36 A1A2A2A2 |
1/6 A2A2 |
1/36 A1A1A2A2 |
4/36 A1A2A2A2 |
1/36 A2A2A2A2 |
If we add up repeated cells in the Punnet square we will get:
1/36 A1A1A1A1
8/36 A1A1A1A2
18/36 A1A1A2A2
8/36 A1A2A2A2
1/36 A2A2A2A2
that is the segregation ratio of the F2 progeny.
b)
From part a we know that the possible gamete from an individual A1A1A2A2 are:
1/6 A1A1
1/6 A2A2
4/6 A1A2
Using the same reasoning, the possible gametes from an individual A3A3A4A4 are:
1/6 A3A3
1/6 A4A4
4/6 A3A4
The Punnett square to get the F1 would look like:
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|
1/6 A3A3 |
4/6 A3A4 |
1/6 A4A4 |
1/6 A1A1 |
1/36 A1A1A3A3 |
4/36 A1A1A3A4 |
1/36 A1A1A4A4 |
4/6 A1A2 |
4/36 A1A2A3A3 |
16/36 A1A2A3A4 |
4/36 A1A2A4A4 |
1/6 A2A2 |
1/36 A2A2A3A3 |
4/36 A2A2A3A4 |
1/36 A2A2A4A4 |
The segregation ratio for the F1 is then:
1/36 A1A1A3A3
4/36 A1A1A3A4
1/36 A1A1A4A4
4/36 A1A2A3A3
16/36 A1A2A3A4
4/36 A1A2A4A4
1/36 A2A2A3A3
4/36 A2A2A3A4
1/36 A2A2A4A4
To get the F2, the F1 has to be intermated. Instead of starting to compute all possible gametes that can be produced in the F1, notice that all combinations of two alleles are possible because we are considering a random bivalent pairing:
gametes A1A1 from individuals A1A1A3A3, A1A1A3A4 and A1A1A4A4,
gametes A1A2 from individuals A1A2A3A3, A1A2A3A4 and A1A2A4A4,
and so on.
Then, if all possible gametes are present for intermating, all combinations of four alleles (there are 35 combinations) can be obtained in F2. Namely:
A1A1A1A1 A2A4A4A4 A1A2A2A4
A2A2A2A2 A3A4A4A4 A2A2A3A4
A3A3A3A3 A1A1A2A2 A1A2A3A3
A4A4A4A4 A1A1A3A3 A1A3A3A4
A1A1A1A2 A1A1A4A4 A2A3A3A4
A1A1A1A3 A2A2A3A3 A1A2A4A4
A1A1A1A4 A2A2A4A4 A1A3A4A4
A1A2A2A2 A3A3A4A4 A2A3A4A4
A2A2A2A3 A1A1A2A2 A1A2A3A4
A2A2A2A4 A1A1A2A3
A1A3A3A3 A1A1A2A4
A2A3A3A3 A1A1A3A4
A1A4A4A4 A1A2A2A3
…
11) For the parents A1A2; A3A4 what is the segregation of the F2 generation?
Parents are diploid. The frequencies and the gamete genotypes from parent A1A2 are:
½ A1
½ A2
For parent A3A4:
½ A3
½ A4
The F1 is the result of crossing those two parents:
¼ A1A3