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Mendelian Genetics of corn kit Lab Report
Genetics (University of the West of Scotland)
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UNIT 18 ASSIGNMENT 2
TASK 1
LAB REPORT
MENDELIAN GENETICS OF CORN KIT
AIM:
The aim of this experiment to study the inheritance of the colour and shape of corn seeds
using monohybrid cross and dihybrid cross.
INTRODUCTION:
Mendelian features refer to phenotypic features whose inheritance pattern is consistent with
Mendel's theories of feature inheritance. Corn is a diploid organism that has been used for
studying and illustrating Mendelian traits. In corn, the dominant Y gene is determining the
presence of colored aleuron. Individuals with one copy of the gene will show a purple colour.
Recessive phenotypes will show a yellow colour. Also in maize, the dominant S gene is
producing a smooth phenotype. Smooth shapes seem hard and starchy. The recessive
phenotype is producing wrinkled shapes that appear shrunken.
The chi-square test is used to investigate the role of chance in generating differences between
observed values and expected values. The test relies on an external hypothesis, since it
involves calculation of theoretical expected values. The test demonstrates the probability that
the deviation between the expected and the observed values was only produced by chance.
When the probability determined from the chi-square test is high the difference is presumed
to have been generated by chance alone. Likewise, if the probability is small, it is presumed
that the variance has been caused by a significant factor other than chance.
In this lab, we will learn the results of some inheritance crosses similar to those that Mendel
did in his classic experiments on heredity in the colour and shape of corn seeds. We will use
monohybrid cross and dihybrid cross. In addition we will learn how to use a statistical test to
determine the validity of data obtained by monohybrid cross and dihybrid cross.
HYPOTHESIS:
Each phenotype is inherited as an allele through a single gene locus.
EQUIPMENT:



4 purple corns
3 yellow corns
2 corn parental cross cards
METHOD:
1. Get three yellow corns.
2. Count all seeds yellow and smooth as well as yellow and shrunken.
3. Write down the numbers of yellow and smooth as well as yellow and shrunken and
calculate their sum.
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4. Next, get four purple corns.
5. Count all seeds for purple and smooth, purple and shrunken, yellow and smooth, and
yellow and shrunken.
6. Write down the numbers for purple and smooth, purple and shrunken, yellow and smooth,
and yellow and shrunken and calculate their sum.
RESULTS:
MONOHYBRID CROSS
First generation
Phenotype: smooth x shrunken
Genotype: SS x ss
s
S
Ss
S
Ss
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s
Ss Ss
This punnet square indicates that their offspring would all be heterozygous (Ss) for smooth.
This means there is a 100 % chance that their offspring will be smooth.
Second generation
Phenotype: smooth x smooth
Genotype: Ss x Ss
S
s
S
SS Ss
s
Ss ss
This punnet square indicates that their offspring would show a ratio of 3:1. There is 75 % of
smooth and 25 % of shrunken.
Second generation phenotype count for smooth x shrunken
Phenotype
Corns
Total
All phenotypes counted
Smooth
Corn D = 381
Corn G = 422
Corn B = 465
465
1649
Shrunken
Corn D = 125
Corn G = 132
Corn B = 124
381
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DIHYBRID CROSS
First generation
Phenotype: purple, smooth x yellow, smooth
Genotype: YYSS x yySS
Gametes: All YS x all yS
YS
YS
YS
YS
yS
YySs YySs YySs YySs
yS
YySs YySs YySs YySs
yS
YySs YySs YySs YySs
yS
YySs YySs YySs YySs
All of the F1 generation would all be heterozygous for both characteristics. Since the
genotype for all offspring are the same, all the YySs genotype encodes a purple colour and a
smooth shape, then 100 % of the phenotypes will be purple and smooth.
Second generation
Phenotype: purple, smooth x purple, smooth
Genotype: YySs x YySs
Gametes: YS, Ys, yS, ys x YS, Ys, yS, ys
YS
YS
YYSS
Ys
YYSs
yS
YySS
ys
YySs
Ratio 9:3:3:1
Ys
YYSs
YYss
YySs
Yyss
yS
YySS
YySs
yySS
yySs
ys
YySs
Yyss
yySs
yyss
12 purple and smooth : 3 Purple and shrunken : 3 Yellow and smooth : 1 Yellow and
shrunken
Second generation phenotype count for purple, smooth x purple smooth
Phenotype
Purple, smooth
Corn
Corn 1 = 252
Purple,
shrunken
Corn 1 = 79
Yellow, smooth
Corn 1 = 79
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Yellow,
shrunken
Corn 1 = 28
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Total
All phenotypes
counted
Corn 6 = 314
Corn 10 = 295
Corn 13 = 266
1127
1917
Corn 6 = 77
Corn 10 = 99
Corn 13 = 71
326
Corn 6 = 95
Corn 10 = 104
Corn 13 = 75
353
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Corn 6 = 24
Corn 10 = 39
Corn 13 = 20
111
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CALCULATIONS FOR CHI-SQUARE TEST
MONOHYBRID CROSS
Phenotype: smooth
Expected count: 1236,75 ≈ 1237
Phenotype: shrunken
Expected count: 412,25 ≈ 412
Total: 1649
Expected ratio: 3:1
1649/4 = 412,25 ≈ 412
412,25 x 3 = 1236,75 ≈ 1237
H0 = there is no statistically significant difference between the observed frequency and the
expected frequency.
HA = there is a significant difference between the observed frequency and the expected
frequency.
If the value for χ2 exceeds the critical value (P = 0.05), then I can reject the null hypothesis.
Phenotype
Smooth
O
1268
E
1237
Shrunken
381
412
O-E
1268–1237 =
31
381–412 =
-21
(O-E)2
31 x 31 =
961
-21 x (-21) =
441
(O-E)2/E
961/1237 =
0.8
441/412 =
1.1
Σ(O-E)2/E = 0.8 + 1.1 = 1.9
X2 = 1.9 [degrees of freedom 2-1 = 1 df]
1 d 3.84 (P = 0.05)
3.84 > 1.9
The value for x2 of 1.9 does not exceed the crtitical value of 3.84 (P = 0.05), so I can accept
the null hypothesis.
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As I accepted the null hypothesis that there was no statistically significant difference between
the observed frequency of maize grains and the expected frequency (the 3:1 ratio) then this
supports the theory that phenotypes will be inherited in the predictable ratios that I calculated
from my Punnet square.
DIHYBRID CROSS
Expected ratio: 12:0:4:0
1917/16 = 119,8125
119,8125 x 12 = 1437 = 1437,75 ≈ 1438 (purple, smooth)
119,8125 x 0 = 0 (purple, shrunken)
119,8125 x 4 = 479,25 ≈ 479 (yellow, smooth)
118,8125 x 0 = 0 (yellow, shrunken)
H0 = there is no statistically significant difference between the observed frequency and the
expected frequency.
HA = there is a significant difference between the observed frequency and the expected
frequency.
If the value for χ2 exceeds the critical value (P = 0.05), then I can reject the null hypothesis.
Phenotype
O
Expected
ratio
9
Purple,
1127
smooth
Purple,
326
3
shrunken
Yellow,
353
3
smooth
Yellow,
111
1
shrunken
Σ(O-E)2/E = 2.2 + 3 + 0.1 +0.7 = 6
E
O-E
(O-E)2
(O-E)2/E
1078
1127-1078
= 49
326-359 =
-33
353-359 =
-6
111-120 =
-9
49 x 49 =
2401
-33 x (-33)
= 1089
-6 x (-6) =
36
-9 x (-9) =
81
2401/1078 =
2.2
1089/359 =
3
36/359 = 0.1
359
359
120
81/120 = 0.7
X2 = 100.4 [degrees of freedom 4-1 = 3df]
3 df = 7.82
7.82 > 6
The value for x2 of 6 does not exceed the crtitical value of 7.82 (P = 0.05), so I can accept the
null hypothesis.
As I accepted the null hypothesis that there was no statistically significant difference between
the observed frequency of maize grains and the expected frequency (the (9:3:3:1 ratio) then
this supports the theory that phenotypes will be inherited in the predictable ratios that I
calculated from my Punnet square.
DISCUSSION:
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For the first experiment I used B corn parental cross card. I used a monohybrid cross because
B-1 and B-2 differed only in one feature. Both corns had the same colour-yellow but the
differed in shape. B-1 corn was shrunken whereas B-2 corn was smooth. For the first
generation I was able to find out that their offspirng would all be heterozygous (Ss) for
smooth. This means that there is 100 % chance that their offspirng will be smooth. For the
second generation I was able to discover that offspring would show a ratio of 3:1. So, there
would be 75 % that offspring will be smooth and 25 % that offspring will be shrunken.
A monohybrid cross was an evidence that the phenotypes I was investigating were actually
being inherited. A monohybrid cross showed how genes and their alleles were passed from
one generation to the next generation. A punnet square that I used could predict potential
combinations of offspring from two parents with a known genotype. I also used a Punnett
square to predict the possible genetic outcomes for the second generation (F2) based on
probability.
I can accept my hypothesis that each phenotype is inherited as an allele through a single gene
locus. If there were several genes involved , there would be no ratio of 3:1 for the second
generation (F2) and there would be many more changes.
As I predicted the possible genetic outcomes for the second generation (F2) I got an F2 ear
and counted the number of grains of each phenotype. I used corn D, G and B to find out
second generation count for smooth x shrunken. Total phenotype count for smooth was 1268
and for shrunken was 381, which gave me 1649 of the total number of phenotypes counted.
After that, I was able to perform Chi-square test. Chi-square test is used to test if the observed
frequency fits the frequency expected or predicted. Using Σ(O-E)2/E formula I was able to
calculate degrees of freedom and I was able to find out if I can accept or reject the null
hypothesis. In this case, the value for x2 of 1.9 did not exceed the critical value of 3.84 (P =
0.05), so I accepted the null hypothesis. Therefore, there was no statistically significant
difference between the observed frequency of maize grains and the expected frequency (the
3:1 ratio. It supported the theory that phenotypes were iherited in the predictable ratios that I
calculated from my Punnett square.
For the second experiment I used two parental cross cards to cross A-1 with B-1. I used a
dihybrid cross as two corns differed in several ways. A-1 corn was purple and smooth
whereas B-1 corn was yellow and shrunken. For the first generation I was able to find out that
their offspring would all be heterozygous for both characteristics. Since the genotype for all
offspring was the same, and the YySs genotype encodes for a purple colour and a smooth
shape, then 100 % of the phenotypes will be purple and smooth. For the second generation I
was able to discover that offspring would show a ratio of 9:3:3:1. So, there would be 9 purple
and smooth : 3 purple and shrunken : 3 yellow and smooth : 1 yellow and shrunken.
I treated it as dihybrid cross because two corns differed in several ways. Dihybrid crosses are
being used when considering the inheritance of two characteristics at the same time. In this
case, there could be four possible combination of the two characteristics. The crons could be:
purple and smooth, purple and shrunken, yellow and smooth or yellow and shrunken.
As I predicted the possible genetic outcomes for the second generation (F2) I got an F2 ear
and counted the number of grains of each phenotype. I used corn 1, 6, 10 and 13 to find out
second generation count for purple, smooth x purple smooth. Total phenotype count for was
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1127 for purple and smooth, 326 for purple and shrunken, 353 for yellow and smooth, and
111 for yellow and shunken. I was able to calculate the total number of phenotypes counted
which was 1917.
After that, I was able to perform Chi-square test again. Chi-square test is used to test if the
observed frequency fits the frequency expected or predicted. Using Σ(O-E)2/E formula I was
able to calculate degrees of freedom and I was able to find out if I can accept or reject the null
hypothesis. In this case, the value for x2 of 6 did not exceed the critical value of 7.82 (P =
0.05), so I accepted the null hypothesis. Therefore, there was no statistically significant
difference between the observed frequency of maize grains and the expected frequency (the
9:3:3:1 ratio). It supported the theory that phenotypes were iherited in the predictable ratios
that I calculated from my Punnett square.
EVALUATION:
I would say that the experiment was successful and I did not perform many errors during the
whole investigation. I accepted the null hypothesis for monohybrid cross and dihybrid cross
which means that there was no statistically significant difference between the observed
frequency of maize grains and the expected frequency. As my x2 did not exceed the critical
value (P = 0.05) I could reject the alternative hypothesis which states: there is a significant
difference between the observed frequency and the expected frequency.
However, there are some improvements that could improve the quality of the results for this
experiment. Accuracy should be considered in this evaluation of the data performed. Each
corn was counted only once which leads to unawareness of the numbers obtained. This is a
random error which can be improved by counting each corn several times and taking the
mean of the various measurements. Accuracy data could also increase the reliability of this
experiment. Precision should also be considered in this evaluation to improve the experiment.
The experiment should be repeated at least one more time to increase precision and
reliability. Precision would show us how close measurements of the same item are to each
other. If the repeat me asurements are close to each other, the data is precise and repeatable.
CONCLUSION:
As I performed the experiment of Mendelian genetics of corn kit I can confirm my hypothesis
that each phenotype is inherited as an allele through a single gene locus. I was able to find out
how colours and shapes are passed on from one generation to another. Even though there are
some improvements that could be made the experiment met my expectations. I was also able
to perform a monohybrid cross obtaining a ratio of 3:1 and dihybrid cross obtaining the
9:3:3:1 ratio. Finally, I used a chi-square test that confirmed the null hypothesis for
monohybrid cross and dihybrid cross. Therefore, there was no statistically significant
difference between the observed frequency of maize grains and the expected frequency. It
supported the theory that phenotypes were iherited in the predictable ratios that I calculated
from my Punnett square.
In Mendelian genetics of corn kit I was investigating dihybrid cross using maize different in
colour - purple and yellow. While counting I observed that some individual grains were
purple with white mottling. The mottling effect is contrary to the basic principles of Mendel's
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genetics, because individual grains can be two-coloured instead of just one colour. Colourful,
colourless and a variety of grains that do not match the traditional Mendelian ratios based on
the assortment during meiosis and random combination of gametes can be caused by the
movement of transposons on chromosomes. "Gene jumping" or transposons is the
explanation for this phenomenon.
Transposons are genes that are moving from one place to another on a chromosome. The
position of transposons can inhibit the production of pigment in some cells in the pigmented
aleuronic layer of corn kernels. For instance, if the transposon moves to a position adjacent to
the pigment producing gene, the cells will not be able to produce the purple pigment. This
creates white mottling instead of a uniform purple grain. The duration of the transposon in
this "turned off" position is affecting the degree of mottling. If the pigmentation gene is
turned off long enough by transposon, the grain will be completely unpigmented.
The appearance of this phenomenon during the experiment means that the observed ratio is
not accurate as would be expected. In this case, it affected our ratio becuase this contrary to
the basic principles of Mendel’s genetics. Purple with white mottling grains were treated and
counted as purple colour. However, this affects the ratio and the number of grains counted as
the individuals were two-coloured. Grains of corn normally should have only one colour – in
this case, purple or yellow. Colourful grains does not fit the traditional Mendelian ratios.
https://www2.palomar.edu/users/warmstrong/transpos.htm
CYSTIC FIBROSIS
Cystic fibrosis is caused by gene mutations creating the cystic fibrosis transmembrane
conductance regulator (CFTR) protein. It is this protein that controls the movement of salt
and fluids in and out of the cells in various parts of the body. Mutations in the CFTR gene in
individuals with CF may disrupt the normal functioning of CFTR protein found in lung cells
and other parts of the body. Mutations in the CFTR gene is causing the CFTR protein to
malfunction or not be made at all, resulting in thick mucus build-up, that in turn causes to
persistent lung infections, pancreatic destruction and complications in other organs.
Cystic fibrisis is caused by a recessive allele.
Phenotype: carrier x affected
Genotype: Cc x cc
c
c
C
Cc
Cc
c
cc
cc
This punnet square indicates that their offspring would show a ratio of 1 carrier (Cc) : 1
affected (cc).
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This means there is a 50 % chance that their offspring would have cystic fibrosis, since half
of those bron are homozygous recessive (cc). This also mean there is 50 % chance that their
offspring would be a carrier, since half of those born are heterozygous (Cc).
https://www.cff.org/What-is-CF/Genetics/CF-Genetics-The-Basics/
Lobo Ingrid, 2008. Genetics and Statistical Analysis | Learn Science at Scitable. Nature.com.
[Online]. [Accessed 18 January 2020]. URL:
https://www.nature.com/scitable/topicpage/genetics-and-statistical-analysis-34592/
Merriam J., 2001. Encyclopedia of Genetics. Mendelian Inheritance - an overview |
ScienceDirect Topics. Sciencedirect.com. [Online]. [Accessed 18 January 2020]. URL:
https://www.sciencedirect.com/topics/biochemistry-genetics-and-molecularbiology/mendelian-inheritance
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