Biology Practical 12

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Aim: To determine the phenotypic ratios for various crosses (monohybrid and dihybrid) of maize

cobs and explain the heredity of the inherited characters.

Part A- Monohybrid Cross

Hypothesis: The specimen has a mixture of yellow and white kernels. Hence, we deduce that it is

probably maize cob of the F2 generation Zea mayswhose parents are heterozygous for the colour of

the maize kernels. As such, we expect the phenotypic ratio of yellow kernels to white kernels to be

3:1.

Results

Table 1.1 showing the observed and expected number and ratios of yellow and white kernels andthe total number of kernels of the specimen

Phenotype

Yellow White

Observed number ofkernels

382 115

Observedphenotypic ratio

3.32 1

Expected number ofkernels

373 124

Expectedphenotypic ratio

3 1

Total number ofkernels

497

As stated earlier in the hypothesis, for the given specimen, we expect the phenotypic ratio of yellow

kernels to white kernels to be 3:1. Hence,3

4of the kernels are expected to be yellow and

1

4of the

kernels are expected to be white.

We can calculate the expected number as follows:

Expected number =Total number Expected probability

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Therefore,

the expected number of yellow kernels3

497 3734

and

the expected number of white kernels 1 497 1244

.

The Chi-Square goodness-of-fit test

To determine if the observed results are close enough to the expected results, we must submit them

to a chi-square analysis. The Chi-square test is a statistical hypothesis test that can be used to

determine whether the observed results are close enough to the expected results by a hypothesis.

This test makes use of a very logical assumption: the smaller the difference between the observed

results and the expected results, the more confidence we can have that our original hypothesis is

correct. Chi-square tells us how many times out of 100 a deviation from the expected results is due

to chance alone. It is the probability (expressed in percent) that chance alone has caused the

deviation from the expected results. If chance has caused the difference between observed and

expected results, then the results support the hypothesis.1

Calculations of Chi-Square values

Null hypothesis: There is no difference between the observed data and the expected data.

H0: observed = expected

For the case where the observed data is not equal to the expected data, we consider an alternative

hypothesis,

H1: observed expected

The Chi-Square value can be determined using the following the equation:

22 (observed exp ected)x

expected

For monohybrid crosses where the degree of freedom is 1 and the probability is 0.05, the2

0.05x

value obtained from the Chi-Square table is 3.841.

1http://www.ableweb.org/volumes/vol-19/mini.8.collins.pdf

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If 2 3.841x , we can say that our hypothesis is supported by our data and we can accept the null

hypothesis. Differences between the observed and expected results are due only to chance.

If 2 3.841x , we can say that our original hypothesis is not supportedby our data and we reject the

null hypothesis. The differences between the observed and expected results are not due to chance.2

Using the observed and expected data as shown in table 1.1, we can now calculate the2

x value.

2x

( )

( )

=

382 - 373( )2

373+

115- 124( )2

124

0.870

< 3.841

Therefore,2 2

0.05x x , our hypothesis is supported by our data. Hence, the null hypothesis is

accepted. Differences between the observed and expected results are due only to chance.

Discussion

According to Mendels first law of segregation, alternative versions of genes (alleles) account forvariations in inherited characters. For each characters, an organism inherits two copies of a gene,

one from each parents (these are also called alleles of that genes). If the two alleles at a locus differ,

then one, the dominant allele, determines the organisms appearance; the other, the recessive allele,

has no noticeable effect on the organisms appearance. The two alleles for a heritable character

then segregate (separate from each other) during gamete formation and end up in different

gametes.3

Based on Mendels first law of segregation, each Maizeplant(Zea mays)carries a pair of alleles for

kernel colour. The P (parental) generation (true breeding parents) Zea mays with yellow-kernelmaize cobs are homozygous for yellow-kernel allele (Y) and Zea mayswith white-kernel maize cobs

are homozygous for white-kernel allele (y). The alleles segregate when gametes are formed. Hence,

2stuy.enschool.org/pdf/bio_lab/second_semester//LAB%2019.doc

3Campbell, Neil A., and Jane B. Reece. "Chapter Mendel and the Gene Idea." Campbell Biology.

9th ed. Boston [u.a.: Pearson, 2011. 310-11. Print.

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the offspring receive one allele from each parent. The allele for yellow kernels (Y) is dominant over

the allele for white kernels (y). Recessive allele is only expressed when in homozygous form (yy)

and in heterozygous form (Yy), only the dominant allele is expressed. Therefore, the F 1 generation

Zea mayswhich are heterozygous will have maize cobs with all yellow kernels. If the F1 generation

(hybrids) Zea mays are self crossed, the offspring, F2 generation Zea mayswill have maize cobs

with a mixture of yellow and white kernels. The ratio of yellow kernels to white kernels is 3:1.4

The monohybrid cross can be explained as shown:

Let Y be the dominant allele for yellow kernel and let y be the recessive allele for white kernel.

Conclusion

Since2 2

0.05x x , we accept the null hypothesis. Differences between the observed and expected

results are due only to chance. Our hypothesis holds true as supported by the data and the Chi-

Square test. We can now confidently conclude that the phenotypic ratio of yellow kernels to white

kernels of the specimen is 3:1 and the parents that have given rise to the above results are F 1

generation hybrids which are heterozygous for the colour of the maize kernels (Yy).

4http://www.ableweb.org/volumes/vol-19/mini.8.collins.pdf

P Generation (true-breeding parents)

Phenotypes: Yellow kernels White kernels

Genotypes: YY yy

Gametes : Y Y y y

F1 Generation (hybrids)

Genotype: Yy Yy Yy Yy

Phenotype: Yellow kernels

Gametes : Y y Y y

F2 Generation

Genotype: YY Yy Yy yy

Phenotype: Yellow kernels Yellow kernels Yellow kernels White kernels

Phenotypic ratio: 3:1

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Part B- Dihybrid cross

Hypothesis: The specimen shows the F2 results of a cross between a pure breeding smooth, black

Zea mays with a pure breeding wrinkled, yellow Zea mays. If the F1 Zea mays which are

heterozygous for both colour and texture of kernels are self-crossed, the F2 dihybrids will have

mixture of smooth black, wrinkled black, smooth yellow and wrinkled yellow kernels in the ratio of

9:3:3:1.

Table 2.1 showing the observed and expected number and ratios of smooth black, wrinkled black,smooth yellow and wrinkled yellow kernels and the total number of kernels of the specimen

Phenotype

Smoothblack

Wrinkledblack

Smoothyellow

Wrinkledyellow

Observednumber ofkernels

368 117 140 50

Observedphenotypic ratio

7.36 2.34 2.8 1

Expectednumber ofkernels

380 127 127 42

Expected

phenotypic ratio

9 3 3 1

Total numberof kernels

675

As stated earlier in the hypothesis, for the given specimen, we expect the phenotypic ratio of smooth

black kernels to wrinkled black kernels to smooth yellow kernels to wrinkle yellow kernels to be

9:3:3:1. Hence,9

16of the kernels are expected to be smooth black,

3

16of the kernels are expected

to be wrinkled black, 316

of the kernels are expected to be smooth yellow and 116

of the kernels are

expected to be wrinkled yellow.

We can calculate the expected number as follows:

Expected number =Total number Expected probability

Therefore,

the expected number of smooth black kernels 9 675 38016

,

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the expected number of wrinkled black kernels3

675 12716

,

the expected number of smooth yellow kernels3

675 12716

and

the expected number of wrinkled yellow kernels1

675 4216

.

Calculations of Chi-Square values

Null hypothesis: There is no difference between the observed data and the expected data.

H0: observed = expected

For the case where the observed data is not equal to the expected data, we consider an alternative

hypothesis,

H1: observed expected

The Chi-Square value can be determined using the following the equation:

22 (observed exp ected)x

expected

For dihybrid crosses where the degrees of freedom are 3 and the probability is 0.05, the2

0.05x

value obtained from the Chi-Square table is 7.815.

If 2 7.815x , we can say that our hypothesis is supportedby our data and we can accept the null

hypothesis. Differences between the observed and expected results are due only to chance.

If 2 7.815x , we can say that our original hypothesis is not supportedby our data and we reject the

null hypothesis. The differences between the observed and expected results are not due to chance.

Using the observed and expected data as shown in table 2.1, we can now calculate the2

x value.

2x

( )

( )

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( )

( )

2 2 2 2

368 380 117 127 140 127 50 42

380 127 127 42

4.02

7.815

Therefore,2 2

0.05x x , our hypothesis is supported by our data. Hence, the null hypothesis is

accepted. Differences between the observed and expected results are due only to chance.

Discussion

According to Mendels law of independent assortment, each pair of alleles segregates independently

of each other pair of alleles during gametes formation.5

The allele for black kernels (B) is dominant to the allele for yellow kernel (b), and the allele for

smooth kernels (R) is dominant to the allele for wrinkled kernels(r). The alleles for colour and texture

are located on different chromosomes (non-homologous chromosomes). True breeding parents (the

P Generation) of Zea maysare homozygous dominant (BBRR) and homozygous recessive (bbrr)for both colour and texture of kernels. Crossing a pure breeding smooth black Zea mayswith a pure

breeding wrinkled yellow Zea mayswill results in F1 Generation which are heterozygous (BrRr) for

both colour and texture of kernels. Since the alleles assort themselves independently when gametes

are formed, self-crossing the F1 dihybrids will yield offspring in the F2 Generation which have the

following phenotypic ratio:6

9 round black kernels: 3 wrinkled black kernels: 3 smooth yellow kernels: 1 wrinkled yellow kernels

The dihybrid cross can be explained as shown:

Let B be the dominant allele for black kernels, b be the recessive allele for yellow kernels, R be the

dominant allele for smooth kernels and r be the recessive allele for wrinkled kernels.

5

Campbell, Neil A., and Jane B. Reece. "Chapter Mendel and the Gene Idea." Campbell Biology. 9th ed.Boston [u.a.: Pearson, 2011. 313-15. Print.6 http://www.ableweb.org/volumes/vol-19/mini.8.collins.pdf

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BR Br bR br

BR BBRR BBRr BbRR BbRr

Br BBRr BBrr BbRr Bbrr

bR BbRR BbRr bbRR bbRr

br BbRr Bbrr bbRr bbrr

= Smooth black kernels 9

= Wrinkled black kernels 3

= Smooth yellow kernels 3

= Wrinkled yellow kernels 1

Phenotypic ratio is 9:3:3:1.

It must be noted that gene which codes for colour and gene which codes for texture are both

expressed at the same time. As mentioned earlier, recessive allele is only expressed when in

homozygous form (bb or rr) and in heterozygous form (Br or Rr), only the dominant allele is

P Generation (true-breeding parents)

Phenotypes: Smooth black kernels Wrinkled yellow kernels

Genotypes: BBRR bbrr

Gametes : BR BR br br

F1 Generation

Genotype: BbRr BbRr BbRr BbRr

Phenotype: Smooth black kernels

F2 Generation (dihybrids)

The alleles assort themselves independently when gametes are formed.

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expressed. Therefore, the F1 generation Zea mayswhich are heterozygous (BbRr) for both colour

and texture of kernels will have maize cobs with smooth black kernels. The genotypes BBRR, BBRr,

BbRr and BbRR result in smooth black kernels, the genotypes BBrr and Bbrr result in wrinkled

black kernels, the genotypes bbRR and bbRr result in smooth yellow kernels and the genotype bbrr

results in wrinkled yellow kernels.

Conclusion

Since2 2

0.05x x , we accept the null hypothesis. Differences between the observed and expected

results are due only to chance. Our hypothesis holds true as supported by the data and the Chi-

Square test. We can now confidently conclude that the phenotypic ratio of F 2 Generation is 9:3:3:1and the parents that have given rise to the above results are F1 generation dihybrids which are

heterozygous for both colour and texture of the maize kernels (BbRr).

We also investigated another specimen whose kernels are of similar character as F 2 Generation

investigated earlier, however, the phenotypic ratio is different.

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Hypothesis: If the F1 dihybrids is crossed with pure breeding wrinkled yellow Zea mays

(homozygous recessive), the offspring will have mixture of smooth black, wrinkled black, smooth

yellow and wrinkled yellow kernels in the ratio of 1:1:1:1.

Results

Table 2.2 showing the observed and expected number and ratios of smooth black, wrinkled black,smooth yellow and wrinkled yellow kernels and the total number of kernels of the specimen

Phenotype

Smoothblack

Wrinkledblack

Smoothyellow

Wrinkledyellow

Observednumber ofkernels

59 56 58 63

Observedphenotypic ratio

1.05 1 1.04 1.13

Expectednumber ofkernels

59 59 59 59

Expectedphenotypic ratio

1 1 1 1

Total numberof kernels

236

As stated earlier in the hypothesis, for the given specimen, we expect the phenotypic ratio of smooth

black kernels to wrinkled black kernels to smooth yellow kernels to wrinkle yellow kernels to

be1:1:1:1. Hence,1

4of the kernels are expected to be smooth black,

1

4of the kernels are expected

to be wrinkled black,1

4of the kernels are expected to be smooth yellow and

1

4of the kernels are

expected to be wrinkled yellow.

We can calculate the expected number as follows:

Expected number =Total number Expected probability

Therefore,

the expected number of smooth black kernels1

236 594

,

the expected number of wrinkled black kernels1

236 59

4

,

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the expected number of smooth yellow kernels1

236 594

and

the expected number of wrinkled yellow kernels1

236 594

.

Calculations of Chi-Square values

Null hypothesis: There is no difference between the observed data and the expected data.

H0: observed = expected

For the case where the observed data is not equal to the expected data, we consider an alternative

hypothesis,

H1: observed expected

The Chi-Square value can be determined using the following the equation:

22 (observed exp ected)x

expected

For dihybrid crosses where the degrees of freedom are 3 and the probability is 0.05, the2

0.05x

value obtained from the Chi-Square table is 7.815.

Similarly, if 2 7.815x , we can say that our hypothesis is supportedby our data and we can accept

the null hypothesis. Differences between the observed and expected results are due only to chance.

If 2 7.815x , we can say that our original hypothesis is not supportedby our data and we reject the

null hypothesis. The differences between the observed and expected results are not due to chance.

Using the observed and expected data as shown in table 2.2, we can now calculate the2

x value.

2x

( )

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( )

( )

( )

2 2 2 2

59 59 56 59 58 59 63 59

59 59 59 59

0.441

7.815

Therefore,2 2

0.05x x , our hypothesis is supported by our data. Hence, the null hypothesis is

accepted. Differences between the observed and expected results are due only to chance.

Discussion

The cross between the F1 dihybrid and P Generation homozygous recessive Zea mays (true

breeding parent) can be explained as shown:

Let B be the dominant allele for black kernels, b be the recessive allele for yellow kernels, R be the

dominant allele for smooth kernels and r be the recessive allele for wrinkled kernels.

F1 dihybrid (BbRr)

BR Br bR br

br BbRr Bbrr bbRr bbrr

br BbRr Bbrr bbRr bbrr

br BbRr Bbrr bbRr bbrr

br BbRr Bbrr bbRr bbrr

P

Generationtru

ebreedingparent(bbrr)

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= Smooth black kernels 4

= Wrinkled black kernels 4

= Smooth yellow kernels 4

= Wrinkled yellow kernels 4

Phenotypic ratio is 4:4:4:4 = 1:1:1:1.

The genotype BbRr results in smooth black kernels, the genotype Bbrr results in wrinkled black

kernels, the genotype bbRr results in smooth yellow kernels and the genotype bbrr results in

wrinkled yellow kernels. As mentioned earlier, the gene which codes for colour of kernels and the

gene which codes for texture of kernels are both expressed at the same time. This gives rise to the

two characters of the kernels (colour and texture).

Conclusion

Since2 2

0.05x x , we accept the null hypothesis. Differences between the observed and expected

results are due only to chance. Our hypothesis holds true as supported by the data and the Chi-

Square test. We can now confidently conclude that the phenotypic ratio of the specimen is 1:1:1:1

and the parents that have given rise to the above results are F 1 dihybrid and P Generation

homozygous recessive Zea mays(true breeding parent).

The genes for colour and texture kernels are not linked as linked genes go not give a phenotypic

ratio of 9:3:3:1. The genes for colour and texture are located on different chromosomes (non-

homologous chromosomes). According to Mendels law of independent assortment, each pair of

alleles segregates independently of each other pair of alleles during gametes formation.

Recombination of unlinked genes is achieved via independent assortment of chromosomes.7

Hence,

each pair of alleles for a trait have a separate effect on the phonotype of the plant.

7elysciencecenter.com/.../Gene_Mapping_Linked__Unlinked_Genes.1231548.ppt

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Evaluation

The sample size taken for this experiment is small. This might affect the ratio obtained as the small

sample might not be representative of the F2 generation as a whole. The experiment should be

repeated with different maize cobs as to increase the sample size and to obtain a more reliable ratio

which is representative of the F2 generation as a whole.

The number of kernels might have been miscounted. Some of the maize cobs have hundreds over

kernels and it is difficult the count the number of kernels accurately. We could have count the

number of kernels several times and take an average number to minimise the errors and to obtain

more reliable and accurate results.

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Bibliography

elysciencecenter.com/.../Gene_Mapping_Linked__Unlinked_Genes.1231548.ppt

http://www.ableweb.org/volumes/vol-19/mini.8.collins.pdf

stuy.enschool.org/pdf/bio_lab/second_semester//LAB%2019.doc

Campbell, Neil A., and Jane B. Reece. "Chapter14 Mendel and the Gene Idea." Campbell Biology.9th ed. Boston [u.a.: Pearson, 2011. 310-15. Print.