ANIMAL GENETICS AND PROBABILITY

Lesson B1–1

ANIMAL GENETICS ANDPROBABILITY

Unit B. Animal Science

Problem Area 1. Animal Genetics and Biotechnology

Advanced Life Science Area: Animals

Standard AS.2.23 Molecules and Cells: Explain the importance of DNA and dif-ferentiate the following terms, genome, gene, chromatin, chromosome, and chromatids.

Standard AS.4.5 Animal Genetics and the Environment: Describe the relation-ship between genotype and phenotype.

Standard AS.4.15 Animal Genetics and the Environment: Describe the Mende-lian theory (law of segregation and the law of independent assortment) and understand itsimportance.

Standard AS.4.16 Animal Genetics and the Environment: Describe geneticdiversity and natural selection and their contributions to a population’s ability to adapt to envi-ronmental change. Describe the relationship between genes and alleles, and define terms includ-ing incomplete dominance, partial dominance, codominance, pleiotropy, and sex linkage.

Biological Science Applications in Agriculture Lesson B1–1 • Page 1

Student Learning Objectives. Instruction in this lesson should result in stu-dents achieving the following objectives:

1 Explain the importance of understanding genetics.

2 Explain how genotype and phenotype are different.

3 Explain how to estimate the heritability of certain traits.

4 Describe sex determination, linkage, crossover, and mutation.

List of Resources. The following resources may be useful in teaching this lesson:

Baker, Meecee and Mikesell, Robert E. Animal Science Biology & Technology.Danville, IL: Interstate Publishers, Inc., 1996 (Chapter 6).

Ensminger, M.E. Animal Science. 9th Edition. Danville, IL: Interstate Publishers,Inc., 1991 (Chapter 3).

Gillespie, J.R. (2002) Modern Livestock & Poultry Production, 6th Edition. Albany,NY: Delmar. (Unit 9)

Lee, Jasper S., et al. Introduction to Livestock & Companion Animals. 3rd Edition.Upper Saddle River, NJ: Pearson Prentice Hall Interstate, 2004 (Chapter 6).

List of Equipment, Tools, Supplies, and Facilities

� Writing surface

� Overhead projector

� Transparencies from attached masters

� Copies of student lab sheets

� Copies of technical supplement

Terms. The following terms are presented in this lesson (shown in bold italics):

� Alleles

� Chromosome

� Codominance

� Crossover

� Deoxyribonucleic acid

� DNA

� Dominant

� Genetic code

� Genome


� Genotype

� Heredity

� Heritability

� Heritability estimate

� Heterozygous

� Homozygous

� Incomplete dominance

� Linkage

� Mutation

� Phenotype

� Probability

� Punnett Square

� Qualitative traits

� Quantitative traits

� Recessive

� Sex chromosomes

Interest Approach. Use an interest approach that will prepare the students for thelesson. Teachers often develop approaches for their unique class and student situations. A possi-ble approach is included here.

Ask the students a series of questions: (write the outcome of each answer on the

board)

How many students can roll their tongue?

How many students have blue eyes?

How many students have brown eyes?

How many students have a widow’s peak?

How many students have attached earlobes?

How many students have unattached earlobes?

How many students have a second toe larger than their big toe?

How many students have connected eyebrows?

Discuss how the student’s traits compare with their relatives. Go on to explain that

this is genetics.


SUMMARY OF CONTENT AND

TEACHING STRATEGIES

Objective 1: Explain the importance of understanding genetics.

Anticipated Problem: Why is it important for a livestock producer to understand genetics?

I. Genetics is the study of the laws and processes of biological inheritance. The study of genet-ics is concerned with the transfer of traits. Gregor Mendel discovered that these traits areinherited through units called genes. Mendel further discovered that genes were found inpairs and half of the inherited traits come from the father and half from the mother. Thispassing of traits from parents to offspring is called heredity. Not all differences in animalsare caused by genetics. Some are caused by the environment, or conditions under which theanimal is raised.

A. A chromosome is a tiny threadlike part in a cell that contains the genetic material.1. Chromosomes are found in the nucleus of cells. The genetic material found in the

chromosomes is called the genome of the organism. When animals mate, thegenome of the offspring is a combination of the traits from the mother and thefather. All of the cells within the animal are genetically identical. Each cell containsidentical numbers of chromosomes. The number found in a cell varies betweenspecies. Chromosomes are made of DNA or deoxyribonucleic acid, which containsegments called genes. DNA is a protein-like nucleic acid that controls inheritance.Each DNA molecule consists of two stands shaped as a double helix or spiral struc-ture. These strands are nucleotides bonded together by pairs of nitrogen bases. Thenucleotides are made up of sugar molecules held together by phosphates. There arefour nitrogen bases found in DNA. They are: cytosine, guanine, adenine, and thy-mine.

2. The genetic code is the sequence of nitrogen bases in the DNA molecule. Thissequence code is for amino acids and proteins. The ability of DNA to replicate itselfallows for the molecule to pass genetic information from one cell generation to thenext.

There are many techniques that can be used to assist students in mastering this

material. Students need text material to aid in understanding the importance of

understanding genetics. Chapter 6 in Introduction to Livestock and Companion Ani-

mals is recommended. Use TM–A and TM–B to aid in the discussion on this topic.


Objective 2: Explain how genotype and phenotype are different.

Anticipated Problem: How do genotype and phenotype differ?

II. Resulting offspring of reproduction have both genotype and phenotype heredity.

A. Genotype is the actual genetic code. It controls physical and performance traits. Thegenotype of an organism cannot be changed by environmental factors.

B. Phenotype is the organism’s physical or outward appearance. This is the part of thegenotype the organism expresses or shows. In some instances, phenotype may bealtered by the organism’s environment.

C. A homozygous organism is one having similar alleles or genes on the DNA molecule fora particular trait. While a heterozygous organism is one having different alleles for a par-ticular trait.

A variety of techniques can be used to help students master this objective. Students

need text material to help understand the difference between genotype and pheno-

type. Chapter 6 in Animal Science Biology and Technology is recommended to assist

your students in mastering this information.

Objective 3: Explain how to estimate the heritability of certain traits.

Anticipated Problem: How can I estimate which traits will be inherited by offspring?

III. Estimating is based on probability. Probability is the likelihood or chance that a trait willoccur. Mating animals of particular traits does not guarantee that the traits will be expressedin the offspring. Heritability is the proportion of the total variation (genetic and environ-mental) that is due to additive gene effects. A heritability estimate expresses the likelihood ofa trait being passed on from parent to offspring. If a trait has a high heritability, the offspringare more likely to express that same trait.

A. The genes contained in an animal control traits of that animal. Some traits are con-trolled by only one pair of genes, while others require several pairs.

1. Qualitative traits are traits controlled only by a single pair of genes and cannot bealtered by the environment. Their phenotype is either one thing or the other. Thesetraits most easily show how genes are inherited. An example is coat color.

2. Quantitative traits are traits controlled by several pairs of genes. These traits areexpressed across a range. These traits can also be altered by environment. Examplesinclude rate of gain, growth rate, backfat depth, etc.

3. Not all traits contained within an organism are expressed. Dominant traits cover upor mask the alleles for recessive traits. In some organisms there are cases ofcodominance of traits in which both dominant and recessive genes are expressed.Incomplete dominance happens when a blending of the allele pair is expressed.


4. The Punnett Square is a technique for predicting genotype. It considers the domi-nant and recessive genes of the male and female parents for one trait.


need text material to help understand how to estimate the heritability of certain

traits. Chapter 6 in Introduction to Livestock and Companion Animals is recom-

mended to assist your students in mastering this information. Use TM–C, TM–D, and

TM–E to aid in discussion on this topic.

Objective 4: Describe sex determination, linkage, crossover, and mutation.

Anticipated Problem: What are sex determination, linkage, crossover, and mutation andwhy are they important?

IV. There are several other factors that are important for livestock producers to understand.Some of them are:

A. Sex determination—sex chromosomes determine the sex of zygote. The process differsslightly among species.1. Mammals—Male sex chromosomes are either X or Y. A zygote that receives a Y

chromosome from sperm will be male. A zygote that receives an X chromosomefrom sperm will be female. The male makes sex determination as all eggs fromfemale receive an X chromosome. Therefore, a female zygote will have two X chro-mosomes (XX) while a male zygote will have one X and one Y chromosome (XY).

2. Poultry—The female determines the sex of the offspring. The male carries two sexchromosomes (ZZ). The female carries only one sex chromosome (ZW). Aftermeiosis, all the sperm cells carry a Z chromosome. Only half of the egg cells carry aZ chromosome; the other half carries a W chromosome.

3. Linkage—The tendency for certain traits to appear in groups in the offspring iscalled linkage. Early studies in genetics were based on the idea that all genes areredistributed in each mating. It was found, however, that some groups of traitsseemed to stay together in the offspring.

4. Crossover—Crossover is the formation of new chromosomes resulting from thesplitting and rejoining of the original chromosome. This explains why the predictedresults of a mating do not always happen. During one stage of meiosis the chromo-somes line up together. They are very close to each other. Sometimes the chromo-somes cross over one another and split. This forms new chromosomes with differ-ent combinations of genes.

5. Mutation—Mutation is the appearance of a new trait in the offspring that did notexist in the genetic makeup of the parents.


need text material to help understand sex determination, linkage, crossover, and

mutation. Unit 9 in Modern Livestock & Poultry Production is recommended to assist


your students in mastering this information. Use TM–F, TM–G, and TM–H to aid in

discussion on this topic.

Review/Summary. Use the student learning objectives to summarize the lesson.Have students explain the content associated with each objective. Student responses can be usedin determining which objectives need to be reviewed or taught from a different angle.

Application. Use the following transparency masters, lab sheets, and technical supple-ment to apply the information.

� TM–A: Number of Chromosomes for Selected Animal Species� TM–B: DNA Structure� TM–C: Punnett Squares—Monohybrid and Dihybrid� TM–D: Examples of Heritability Estimates for Cattle� TM–E: Examples of Heritability Estimates for Swine� TM–F: Chromosome Combinations Determine Sex� TM–G: Gene Linkage� TM–H: Chromosome Crossover� LS–A: Animal Genetics and Probability� LS–B: Monohybrid Punnett Squares� LS–C: Dihybrid Punnett Squares� TS–A: Animal Genetics and Probability

Evaluation. Evaluation should focus on student achievement of the objectives for eachlesson. Various techniques can be used, such as performance on the application activities. A sam-ple written test is attached.

Answers to Sample Test:

Part One: Matching

1. e2. f3. b4. c5. g6. d7. h8. a


Part Two: Fill-in-the-Blank

1. DNA2. different3. codominance4. mutation5. linkage6. replicate

Part Three: Multiple Choice

1. d2. b3. c4. a5. b

Part Four: Short Answer

1. Refer to Objective 4 in the lesson for scoring this question.2. 50% Bb (black), 50% bb (red)


Name ________________________________________Test

ANIMAL GENETICS AND PROBABILITY

� Part One: Matching

Instructions: Match the word with the correct definition.

a. Crossover d. Heritability estimate g. Genotypeb. Heritability e. Genome h. Qualitative traitsc. Quantitative traits f. Phenotype

_______1. Genetic material found in the chromosomes

_______2. The organism’s physical or outward appearance

_______3. The likelihood of a trait being passed of from parent to offspring

_______4. Traits controlled by several pairs of genes. These traits are expressed across a range. These traits canalso be altered by environment

_______5. The actual genetic code

_______6. The proportion of the total variation (genetic and environmental) that is due to additive gene effects

_______7. Traits controlled only by a single pair of genes and cannot be altered by the environment

_______8. The formation of new chromosomes resulting from the splitting and rejoining of the original chromo-some

� Part Two: Fill-in-the-Blank

Instructions: Complete the following statements.

1. ______________ is a protein-like nucleic acid on genes that controls inheritance.

2. A heterozygous organism is one having ___________ alleles for a particular trait.

3. ____________________ of traits in which both dominant and recessive genes are expressed.

4. _____________ is the appearance of a new trait in the offspring that did not exist in the genetic makeupof the parents

5. The tendency for certain traits to appear in groups in the offspring is called __________________.


6. The ability of DNA to _________________ itself allows for the molecule to pass genetic information fromone cell generation to the next.

� Part Three: Multiple Choice

Instructions: Circle the letter of the correct answer.

_______1. The genetic make-up of an individual is known as the

a. phenotypeb. homozygousc. heterozygousd. genotype

_______2. This is the ratio of the specified events to the total number of events.

a. quantityb. probabilityc. statisticsd. all of the above

_______3. Which is an example of a homozygous trait?

a. RRb. Rrc. Rrd. Rb

_______4. In guinea pigs, short hair is dominant. What would a short haired guinea pig be?

a. LL or Llb. Llc. LL or lld. none of the above

_______5. What is the probability of having a homozygous recessive offspring from two heterozygous parents?

a. 50%b. 25%c. 90%d. 75%

� Part Four: Short Answer

Instructions: Answer the following statements.

1. Explain how sex of offspring is determined in mammals and in poultry.

2. Use the Punnett Square method to estimate the possible gene combinations for the following situation.Mating a black cow (Bb) to a red bull (bb).


TM–A

NUMBER OF CHROMOSOMES FORSELECTED ANIMAL SPECIES

SpeciesNumber of

Chromosomes

Cat 38

Cattle 60

Chicken 78

Dog 78

Donkey 62

Horse 64

Human 46

Mule 63

Sheep 54

Swine 38


TM–B

DNA STRUCTURE


T AG C

C G

T A

T A

G C

C

TAGC

CG

TA

TM–C

PUNNETT SQUARES—MONOHYBRID AND DIHYBRID


Red Sire(RR)

White Dam(rr) r

r

R R

Rr

Roan

¼

Rr

Roan

¼

Rr

Roan

¼

Rr

Roan

¼

BS

BS

Bs

bS

bs

BBSSblack,short

BBSsblack,short

BbSSblack,short

BbSsblack,short

BBSsblack,short

BBssblack,long

BbSsblack,short

Bbssblack,long

BbSSblack,short

BbSsblack,short

bbSS

shortbrown,

bbSs

shortbrown,

BbSsblack,short

Bbssblack,long

bbSs

shortbrown,

bbssbrown,

long

Bs bS bs

Guinea Pig — Female

Gu

ine

aP

ig—

Male

9 dominant (black), dominant (short)

3 dominant (black),

3 , dominant (short)

1 ,

recessive (long)

recessive (long)

recessive (brown)

recessive (brown)

MONOHYBRID

DIHYBRID

TM–D

EXAMPLES OF HERITABILITYESTIMATES FOR CATTLE

Trait Heritability (%)

Number born 5

Calving interval(fertility) 10

Percent calf crop 10

Services per conception 10

Conformation score at weaning 25

Cancer eye susceptibility 30

Gain on pasture 30

Weaning weight 30

Yield grade 30

Carcass grade 35

Age at puberty 40

Birth weight 40

Body condition score 40

Carcass—percent lean cuts 40

Conformation score at slaughter 40


Trait Heritability (%)

Cow maternal ability 40

Efficiency of gain 40

Preweaning gain 40

Yearling frame size 40

Yearling weight 40

Fat thickness 45

Feedlot gain 45

Dressing percent 46

Marbling score 50

Mature weight 50

Scrotal circumference 50

Tenderness 50

Final feedlot weight 60

Retail yield 60

Rib eye area 70


TM–E

EXAMPLES OF HERITABILITYESTIMATES FOR SWINE

Trait Heritablility (%)

Litter survival to weaning 5

Litter size 10

Number farrowed 10

Number of pigs weaned 12

Weaning weight (3 weeks) 15

Birth weight 20

Five month weight 25

Number of nipples 25

Conformation 30

Feed efficiency 30

Age at puberty 35

Percent lean cuts 45

Probe backfat (live at 200 pounds [99.8 kg]) 45

Carcass length 50

Loin muscle area 50

Percent of shoulder 50

Percentage carcass muscle 50

Percent ham 55

Percent fat cuts 60


TM–F

CHROMOSOME COMBINATIONSDETERMINE SEX


II Ir

I

I

I

I I r r

II

I r

Ir

I

Female

Germ Cell

Male

Germ Cell

Egg Sperm

FEMALE MALE

NOTE: If sperm 1 or 2 unites with the egg,

the progeny will be a female, but if

sperm 3 or 4 unites with the egg,

the progeny will be a male.

Key: I

r

=

=

Female Chromosome

Male Chromosome

1 2 3 4

TM–G

GENE LINKAGE


A

C

B

D

Gene Linkage

Genes A and B will tend to stay

together when the chromosomes

divide, as will genes C and D.

Genes A and D are not as likely

to stay together because they are

farther apart.

TM–H

CHROMOSOME CROSSOVER


New combinations of genes are

formed when chromosomes

cross over and split.

B

b

b

b

b

B

B

B

A

A

A

A

a

a

a

a

LS–A: Teacher Information


Agricultural Applications and Practices

Animal producers use genetics to improve their herd/flock mainly through reproduction.By breeding particular animals to each other, producers can improve on a particular qual-ity or trait, eliminate unwanted characteristics and improve the overall production of theanimals. Researchers have linked several diseases in animals to genetic inheritance such asspider syndrome in sheep, D.U.M.P.S. in cattle and creeper syndrome in chickens, toname a few. By understanding the principles behind genetics, animal producers are capa-ble of continually improving their animal’s production levels. Producers are also able toformulate breeding programs to get the best traits of their animals. In dairy cows forinstance, producers can mate a high milk-producing female with a male from high pro-ducing parents and who has a good progeny record. In doing so, producers will improvetheir herd’s milk production by their own cows.

Another advantage to understanding the rules of genetics is in crossbreeding. By matingtwo different breeds of species producers can take advantage of the different positive traitsof breeds. For example in beef cattle, by mating an Angus male to a Brahman female, theproducer can take advantage of the heat tolerance and parasite resistance of the Brahmanbreed but still maintain the carcass qualities and mothering ability of Angus cattle.

Science Connections—Questions for Investigation

1. How are scientists utilizing genetics to improve the quality of food products andeliminate genetic disorders?

2. What are some ways to determine the genetic makeup of an animal?

3. Why is it more difficult to determine the genotype of an animal as compared to itsphenotype?

Research Problem

How is the genotype of an offspring determined?


Purpose of Laboratory and Student Performance Objectives

The purpose of this experiment is to familiarize students with the use of probability todetermine the genetic make-up of an animal under various situations. Through thisexperiment and related discussions students will be able to:

1. understand how traits are distributed;

2. understand the importance of genetics and reproduction in animal improvement;

3. determine the probability of a genetic phenotype and genotype of an animal; and,

4. understand the difference between dominant and recessive traits.

Materials and/or Equipment

� Piece of round cardboard about 8 inches in diameter

� Plastic spinner

� Marker

� Pencil eraser

Procedure

1. Divide the piece of cardboard with the marker into 4 equal sections so it resembles apunnet square.

2. Place the plastic spinner in the center of the cardboard circle.

3. Write the symbols (in pencil) on the cardboard for the problems on the studentworksheet.

4. Give each student or group of students a copy of the worksheet.

5. The students should spin the spinner 4 times, once for each offspring, and solve theproblems.

Helpful Hints

� When teaching monohybrid and dihybrid Punnett squares, encourage students towrite the capital letters (dominant) before lowercase letters (recessive). This makes itmuch easier when trying to predict the genotypes and phenotypes of the offspring.


Name ________________________________________LS–A: Student Worksheet


Example

This example will help you understand homozygous traits. Spider syndrome is a geneticproblem in sheep. The sheep develop a roman nose and the legs bend inward. The sheepafflicted with this syndrome generally do not live past 2 or 3 years of age. Spider syndromeis a homozygous recessive trait. If a ram with this syndrome is mated to known carriers ofthe syndrome but not showing the physical characteristic, what is the genotypic andphenotypic ratio of offspring?

n—represents gene for spider syndromeN—represents normal dominant gene

The anticipated result of this particular example is 2Nn and 2nn, hence the genotypicratio of 1:1. A similar disk should then be done for the phenotype.

Problems

1. White Bengal tigers lack some pigment. The Brookfield Zoo has a male, white Ben-gal (ww) and he was mated to 4 homozygous dominant females (WW) who are allorange. What will the offspring look like? What will the genotypic ratio be?


2. In guinea pigs, short hair (L) is dominant over long hair (1). Two heterozygousshorthaired guinea pigs are mated. Measure their genotypic and phenotypic ratio.

3. A roan colored cow has a combination of red and white hairs and its phenotype is acodominant color. A red bull (RR) is mated to 5 roan (Rr) females and 3 whitefemales (rr). Determine the genotypic and phenotypic ratios for each of these mat-ings.

4. The creeper syndrome in chickens results in the severe shortening of legs. A chickenproducer mates a rooster (Cc) to 5 hens that are carrying the creeper gene also (Cc).What is the genotypic and phenotypic ratio of the offspring.


Name ________________________________________LS–B: Student Worksheet

MONOHYBRID PUNNETT SQUARES

Crossing Animal Coats

In a certain species of animal, black fur (B) is dominant over brown fur (b). Show yourwork.

1. Using a Punnett square, predict the genotypes and phenotypes of the offspringwhose parents are both Bb or have heterozygous black fur.

Genotypes

_____% Homozygous black fur (BB)

_____% Heterozygous black fur (Bb)

_____% Homozygous brown fur (bb)

Phenotypes

_____% black fur

_____% brown fur

2. Now do the same when one parent is homozygous black and the other is homozy-gous brown.

Genotypes




Phenotypes

_____% black fur

_____% brown fur


3. Repeat this process again when one parent is Heterozygous black and the other isHomozygous brown.

Genotypes




Phenotypes

_____% black fur

_____% brown fur

Blood Type and Inheritance

In blood typing, the gene for type A and the gene for type B are codominant. The gene fortype O is recessive. Using Punnett squares, determine the possible blood types of the off-spring when:

1. Father is type O; Mother is type O.

_____% O

_____% A

_____% B

_____% AB

2. Father is type A, homozygous; Mother is type B, homozygous.

_____% O

_____% A

_____% B

_____% AB

3. Father is type A, heterozygous; Mother is type B, heterozygous.

_____% O

_____% A

_____% B

_____% AB


4. Father is type O; Mother is type AB.

_____% O

_____% A

_____% B

_____% AB

5. Father and Mother are both type AB.

_____% O

_____% A

_____% B

_____% AB


Name ________________________________________LS–C: Student Worksheet

DIHYBRID PUNNETT SQUARES

Directions

Please show all your work on a separate sheet of paper. Include both genotypes and phe-notypes in your answer.

Symbols

LL/Ll long eyelashes

ll short eyelashes

TT/Tt broad lips

tt thin lips

PP/Pp polydactyly (extra fingers)

pp normal fingers

SS/Ss syndactyly (webbed fingers)

ss normal fingers

BB/Bb dark eyes

bb light eyes

DD/Dd dark hair

dd light hair


Problems

1. heterozygous for eyelashes and lips × short eyelashes, thin lips

2. heterozygous for polydactyly and syndactyly × normal for both traits

3. homozygous dominant for eye color, light hair × light eyes and hair

4. homozygous polydactyly, normal for webbed fingers × normal for extra fingers,homozygous syndactyly

5. TtPp × ttPp

6. BbDd × Bbdd

7. LLdd × llDd


TS–A

Technical Supplement


1. How are scientists utilizing genetics to improve the quality of food prod-ucts and eliminate genetic disorders?

Scientists are using genetics to improve food quality through controlled reproduc-tion. One example would be cross-breeding different breeds of a species to takeadvantage of stress tolerance and still maintain good carcass quality. Scientists arealso working on eliminating genetic disorders. Technology is now available so manygenetic disorders can be detected early in gestation. At this point, the genetic disor-der can either be corrected or the pregnancy aborted early in gestation to reducestress on the dam.

2. What are some ways to determine the genetic makeup of an animal?

Since genes are passed from parent to offspring, the characteristics of the offspringcan be predicated if enough is known about parents. Genes are small specific spotson the chromosomes. Each spot controls a specific function in the animal. The newoffspring always gets half of its genes from each parent. Therefore if you know whatgenes the parents have, you can predict what gene the offspring will have.

Genetics can be determined by observing the phenotype (visible traits) and per-forming test crosses. In order to do this effectively, good records are essential.Although this method is not 100% accurate, it is the most economical. Many traitsare controlled by more than one set of genes and therefore are more difficult todetermine. However, there are simple traits controlled by one set of genes. Oneexample of a simple trait is polledness (absence of horns) in cattle. Polled (no horns)


is a dominant trait. The best test for homozygous polledness is to mate a polled bullto a horned cow (at least 10, preferably 14 or more). If one or more calves havehorns, the bull carries the genes for horns and is heterozygous polled. If no calveshave horns, the bull is more likely homozygous polled.

3. Why is it more difficult to determine the genotype of an animal as com-pared to its phenotype?

The genotype is the gene makeup of the animal. The phenotype is the visible char-acteristics. Each trait is controlled by at least two genes (one from each parent). Forexample, a cow may have one gene for polledness and one gene for horns (thiswould be the genotype). Since polledness is dominant, the cow would be polled(this would be the phenotype). Methods of determining genotype are throughblood testing or test crossing. The phenotype is more easily determined because itcan be visibly seen.


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ANIMAL GENETICS AND PROBABILITY