Download Lab. 7 Mendelian Genetics “Law of Independent assortment”

Lab. 7
Mendelian Genetics
“Law of Independent
assortment”
Main topics:


Mendel’s Laws of Inheritance:
 Law of Independent assortment
Polygenic Trait.
Law of Independent Assortment
The Law of Independent Assortment states that different pairs of alleles are passed onto
the offspring independently of each other. Therefore, inheritance of genes at one location
in a genome does not influence the inheritance of genes at another location.
Mendelian’s Second Rule (Law of Independent Assortment)
Mendel observed that, when peas with more than one trait were crossed, the progeny did
not always match the parents. This is because different traits are inherited independently
– this is the principle of independent assortment. For example, his cross-bred pea plants
with round, yellow seeds and plants with wrinkled, green seeds. Only the dominant traits
(yellow and round) appeared in the F1 progeny, but all combinations of trait were seen in
the self-pollinated F2 progeny. The traits were present in a 9:3:3:1 ratio (round, yellow:
round, green: wrinkled, yellow: wrinkled, green).
Explanation:
The cross was made between plants having yellow and round cotyledons and plants
having green and wrinkled cotyledons. The F1hybrids all had yellow and round seeds.
When these F1 plants were self-fertilized they produced four types of plants in the
following proportion:
Yellow and round
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Yellow and wrinkled Green and round
1
Green and wrinkled
=9
=3
=3
=1
The above results indicate that yellow and green seeds appear in the ratio of (3:1)
Similarly, the round and wrinkled seeds appear in the ratio of (3:1). This indicates that
each of the two pairs of alternative characters viz. (3:1) X (3:1) = 9:3:3:1
Yellow - green cotyledon color is inherited independent of the round - wrinkled character
of the cotyledons. It means at the time of gamete formation the factor for yellow color
enters the gametes independent of R or r, i.e., gene Y can be passed on to the gametes
either with gene R or r.
Cytological explanation of the results:
In the above experiment yellow and round characters are dominant over green and
wrinkled characters which can be represented as follows:
Phenotype
Alleles of genes
(i) yellow color of cotyledons
Y
(ii) green color of cotyledons
y
(iii) round character of cotyledons
R
(iv) wrinkled character of cotyledons
r
Therefore, plants with yellow and round cotyledons will have their genotype YYRR and
those with green and wrinkled cotyledons will have a genotype yyrr.
These plants will produce gametes with gene YR and yr respectively. When these plants
are cross pollinated, the union of these gametes will produce F1 hybrids with YyRr genes.
When these produce gametes all the four genes have full freedom to assort independently
and, therefore, there are possibilities of four combinations in both male and female
gametes. (i) YR
(ii) yR
(iii) Yr
(iv) yr
This shows an excellent example of independent assortment. These gametes can unite
at random producing in all 16 different combinations of genes, but presenting four
phenotypes in the ratio of 9: 3: 3: 1.
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Figure: Dihybrid ratio
2. The Chi-Square Analysis Evaluates
the Influence of Chance on Genetics
Data “∑χ2 = (O-E) 2/ E”
Chi Square Example
Let's Test the following data to determine if it fits a 9:3:3:1 ratio.
Observed Values
Expected Values
315 Yellow,Round Seed
(9/16)(556) = 312.75 Round, Yellow Seed
108 Green,Round Seed
(3/16)(556) = 104.25 Round, Green Seed
101 Yellow, WrinkledSeed
(3/16)(556) = 104.25 Wrinkled, Yellow
32 Green, Wrinkled
(1/16)(556) = 34.75 Wrinkled, Green
556 Total Seeds
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556.00 Total Seeds
3
Probability
Degrees of
Freedom
0.9
0.5
0.1
0.05
0.01
1
0.02
0.46
2.71
3.84
6.64
2
0.21
1.39
4.61
5.99
9.21
3
0.58
2.37
6.25
7.82
11.35
4
1.06
3.36
7.78
9.49
13.28
5
1.61
4.35
9.24
11.07
15.09
Grain
Expected
Observed Expected
[Obs No. - [Obs No. - Exp
Phenotype
Ratio
Number
Number
Exp No.]
(Column 1)
(Column
(Column
(Column 5)
4)
2)
9/16
315
yellow&
round
green&
3/16
108
& 3/16
101
556 x 9/16 = 2.25
0.0161
556 x 3/16 = 3.75
0.134
556 x 3/16 = -3.25
0.101
104.25
1/16
32
wrinkled
Total No.
(Column 6)
104.25
wrinkled
green&
÷ Expected No.
312.75
round
Yellow
No.]2
381 x 1/16 = -2.75
0.217
34.75
----------
556
0
Number of classes (n) = 4
df = n-1 =(4-1) = 3
Chi-square value tab. = 7.82
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X2= 0.4681
Enter the Chi-Square Table at df = 3 and we see the probability of our chi-square
valuetabulated=7.82. By statistical convention, we use the 0.05 probability level as
our Critical Value. If the Calculated chi-square value is less than the 0.05 value, we
accept the hypothesis. If the value is greater than the value, we reject the hypothesis.
Therefore, because the calculated chi-square value is greater than we accept the
hypothesis that the data fits a 9:3:3:1 ratio.
Polygenic Trait:
Examples:
If two AaBbCc people would get married and have children. What are they get from the
following:
 Number of gametes
 Genotypes resulting
 Number of individuals
Answer:
AaBbCc X AaBbCc
N=
The resulting number of gametes = 2N
=
heterozygous The number of genotypes resulting = 3N
The resulting number of individuals = 4N =
References:
http://www.uic.edu/classes/bios/bios101/x202_assortment.pdf
http://www.houstonisd.org/cms/lib2/tx01001591/centricity/domain/5363/s_bi_mendeliangenetics09_t.pdf
http://www.ndsu.edu/pubweb/~mcclean/plsc431/mendel/mendel3.htm
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23=
33=
43=
Exercise 7
1- How many different kinds of gametes could an individual with the following
genotypes produce?

HhMm

HHMmoo

AaBbCC

VvGgSsEe
2- When you cross the following a true-breeding strain of corn with purpled,
smooth kernels to a true-breeding strain of corn with yellow, wrinkled kernels you
find that the F1 progeny all have purple, smooth kernels.
- You are unsure of the genotype of an individual F2 corn plant that has purple,
smooth kernels and would love to know its genotype.
A. What cross would allow you to determine the genotype of the purple, smooth
corn plant?
B. Describe the four different possible scenarios from your cross above. Indicate
phenotypes and genotypes.
4- Use Punnett squares to show the genotype and phenotype of the offspring that
would result from the following crosses:

GgRr x GgRr

GgRr x GgRR
5- From a cross between two plants, one completely heterogyzous tall green (Ssyy)
and the other heterozygous tall yellow (SsYy), the following results were observed:
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52 tall green
55 tall yellow
26 short green
27 short yellow
A.Show by Punnett square (or other labeled diagrams) the cross between the
two original plants and the expected phenotypic ratio of offspring.
B.Use chi square calculations to determine whether or not the variations in
the observed results could be due to chance. ( at level of frequency 5%)
6- Cross between two pea plants one of them is yellow seed with round shape
(heterozygous traits) and other is green seed with wrinkled shape (homozygous
traits). What is genotype and phenotype of first generation?
7- For the following crosses, determine the probability of obtaining the indicated
genotype in an offspring
Cross
Offspring
AAbb x AaBb
AAbb
AaBB x AaBb
aaBB
AABbcc x
AaBbCc
aabbCC
AaBbCc x
aabbcc
AaBbcc
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Probability