Exam 2 Study Guide 

What are the two main roles of DNA?

This study guide contains material covered in class from 2/23/16-3/29/16. This study  guide contains important vocab terms, fill in the blanks, and questions that focus on the  main points learned in class. All definitions and answers to these questions can be found  within the notes weeks 6 through 10. Although an answer key is provided, I strongly  suggest filling out the study guide by yourself first, and then use the answer key to check  your answers. Happy Studying! ☺

Week 6 Review (2/23/16-2/25/16) 

1.) Define the following terms:

∙ Haploid

∙ Diploid

∙ Gametogenesis

∙ Fertilization

∙ Zygote to Gastrulation

∙ Differentiation

∙ Immortalized Cells

∙ Embryotic Stem Cells

∙ Apoptosis

∙ Cell Cycle

∙ Mitosis

∙ Meiosis

∙ Crossing Over

∙ Recombination

∙ Linkage

What is the cell cycle

∙ Linkage Disequilibrium  

∙ Hapmap

∙ Taq-SNP

∙ Phenotype

∙ Genotype

∙ Allele

∙ Locus

∙ Heredity

∙ Trait

∙ Simple Trait

∙ Complex Trait

∙ Pleiotropy

∙ Homozygosity

∙ Heterozygosity

2.) What are the two main roles of DNA?

3.) In order for DNA to accomplish its roles, what are the three things it must be able  to do?

4.) What is the cell cycle and explain what occurs during each phase? 5.) _________ is the process in eukaryotes by which ________ cells replicate their DNA  to form two identical daughter cells.

6.) _________ is the process in eukaryotes by which __________ cells divide for sexual  reproduction, leading to four genetically diverse (different) cells. 7.) Review the phases of mitosis.

8.) Review the phases of meiosis.

9.) During what stage in meiosis does crossing over occur?

What information is needed in order to construct a Punnett square?

10.) During what specific stage of prophase can chromosomal material cross-over  among a homologous pair of chromosomes?

11.) How does linkage, linkage disequilibrium, and linkage equilibrium influence  genetic diversity? Don't forget about the age old question of usc understanding america study
If you want to learn more check out hist 5057 study guide

12.) __________ is one’s physical characteristics and trait measure, such as coat  color, where _________is one’s genetic makeup.

13.) What is the difference between simple traits, complex traits, and Pleiotropy? 14.) What is the foundational information needed to understand inheritance? (This  question is a reoccurring theme throughout the notes, so know them!) 15.) What are Mendel’s principles? (This question is a reoccurring theme throughout  the notes, so know them!) We also discuss several other topics like unl nutrition science

16.) How can animal pedigrees, Al Sire Linage and Generations, and Punnett  Squares help us understand the segregation of alleles?

17.) What information is needed in order to construct a Punnett square? (This  question is a reoccurring theme throughout the notes, so know them!) 18.) Compare and contrast homozygosity and heterozygosity.

Week 7 Review (3/1/16-3/3/16) 

1.) Define the following terms:

∙ Autosomal Chromosome

∙ Sex Chromosome

∙ Locus

∙ Genotype

∙ Homozygous

∙ Heterozygous

∙ Allele

∙ Bi-allelic

∙ Multi-Allelic

∙ Deleterious Allele

∙ No Dominance

∙ Additive

∙ Complete Dominance

∙ Partial Dominance

∙ Over-Dominance

∙ Co-Dominance

∙ Epistasis

∙ Sex-Linked

∙ Sex-Limited

∙ Sex Influenced

∙ Filial

∙ Inter-Se Mate We also discuss several other topics like niu biology major

∙ Repeated Back Crossing

∙ Grading Up

∙ Phenotypic Ratio

∙ Genotypic Ratio

2.) What is the foundational information needed to understand inheritance? (This  question is a reoccurring theme throughout the notes, so know them!) 3.) What the difference between sex and autosomal chromosomes? 4.) What the different nomenclature of genotypes and alleles?

5.) What are the different types of autosomal modes of inheritance? 6.) What are the different sex-related modes of inheritance?

7.) What are some exceptions to Mendelian inheritance?

8.) Know how to draw and use a Punnett square.

9.) Find the genotypic and phenotypic ratio when you cross a male with the alleles  BB and a female with the alleles bb. Where BB equals a black coat and bb  equals a red coat.

10.) Find the genotypic and phenotypic ratio when you cross a heterozygous male  (Bb) and a heterozygous female (Bb) from the previous generation.

11.) Find the genotypic and phenotypic ratio when you cross a female that has RW  alleles and a male that has RW alleles. Where RR codes for a red coat, RW codes  for a roan coat, and WW codes for a white coat.

12.) Find the genotypic and phenotypic ratio when you cross a female with the  following alleles BBPP and a male with the following alleles bbpp. Where the  allele b represents coat color and the p allele represents the presence of horns.

13.) Find the genotypic and phenotypic ratio when you cross a male with the alleles  BbPp and a female with the alleles BbPp from the previous generation. Where  the allele b represents coat color and the p allele represents the presence of  horns. (Remember PP and Pp mean polled and pp means horned). We also discuss several other topics like econ1101 notes

14.) Find the genotypic and phenotypic ratio when you cross a female black lab  with the alleles BBEE and a male yellow lab with the alleles bbee.

15.) Find the genotypic and phenotypic ratio when you cross a male with the alleles  BbEe and a female with the alleles BbEe from the previous generation.

16.) For more Punnett square practice review the Punnett Square practice we did in  class on 3/8/16. (The worksheet along with the answer key can be found on  Canvas in the modules section under supplementary information.) Also  reviewing the homework problems is a good idea.If you want to learn more check out why do the kalabari carve wooden sculptures of spirits?

Week 8 Review (3/8/16-3/10/16) 

1.) Define the following terms:

∙ Test Cross

∙ DNA Test

∙ Introgression

∙ Precision Breeding

∙ Simple Trait

∙ Agouti Gene

2.) Know the different characteristics of various simple traits in domestic animals. Week 10 Review (3/22/16-3/24/16) 

1.) Define the following terms:

∙ Pedigree

∙ Inbreeding

∙ Line Breeding

2.) Know the structures and components of a pedigree.

3.) Know how to read a Mendelian flow chart.

4.) Know how to create a Mendelian flow chart from a pedigree. 5.) What is the foundational information needed to understand inheritance? (This  question is a reoccurring theme throughout the notes, so know them!) 6.) Probability is ranked on a scale of _____ to ____. Thus, when an event has a score  of ____ is has 0% chance of occurring, where when an event has a score of ____  then it has 100% chance of occurring.

7.) What is the rule of multiplication and when do I use it?

8.) How do you use the rules of multiplication when solving a di-hybrid problem? 9.) What is the rule of addition and when do I use it?

10.) How do I use the rules of multiplication and addition to solve complex  problems?

Exam 2 Answer Key 

Week 6 Review (2/23/16-2/25/16) 

1.) Define the following terms:

∙ Haploid (n)

o When cells have half the number of usual chromosomes.

▪ Gametes (sperm and egg cells) are the only haploid cells in  

the body that will contain a half of a set (1 of each  

chromosome) of chromosomes.

∙ Diploid (2n)

o When cells have a complete set (2 of each chromosome) of  

chromosomes.

▪ Somatic Cells which is any cell in the body excluding gamete  

cells (sperm and eggs cells) that contains a full set of one’s  

chromosome.

∙ Gametogenesis

o The formation of gametes—spermatogenesis (sperm) and  

oogenesis (egg).

∙ Fertilization

o When a haploid sperm and a haploid egg meet to form a diploid  zygote.

∙ Gastrulation

o After the zygote has been fertilized the blastomere forms into a  

placenta, which will then form into differential tissue and a fetus.

∙ Differenation

o Cells that can differentiate into specialized cells and can divide  through mitosis to produce stem cells.

∙ Immortalized Cells

o Cells that keep dividing through the process of mitosis.

▪ Ex. HeLa Cells.

∙ Embryonic Stem Cells

o Cells that can be turned into any cell within the body.

∙ Apoptosis

o Programmed cell death.

∙ Cell Cycle

o The cell cycle depicts the life cycle of a somatic cell, which is any  cell in the body except for gamete cells (sperm and egg).

∙ Mitosis

o The process in eukaryotes by which somatic cells replicate their  

DNA to form two identical daughter cells.

∙ Meiosis

o The process in eukaryotes by which gamete cells divide for sexual  reproduction, leading to four genetically diverse (different) cells.

∙ Crossing Over

o A naturally occurring way to create cell diversity during prophase I  of meiosis, by exchanging DNA between homologous  

chromosomes.

∙ Recombination

o The rearrangement of genetic material through crossing over,  causing the progeny to look different from its parents due to the  genetic diversity of the gamete cells that occur through meiosis.  ∙ Linkage

o Genes that are located closely to each other are most likely to be  inherited together during the crossing over of meiosis.

∙ Linkage Disequilibrium

o When genes are always going to be inherited together.

▪ Linkage Equilibrium

⮚ When genes are not linked and thus are not going to  

be inherited together.

∙ Hapmap

o A genome that contains haplotype blocks.

▪ Allows us to improve selective breeding.

∙ Taq-SNP

o A SNP that represents a haplotype block.

∙ Phenotype

o One’s physical characteristics and trait measure.

▪ Ex. Red Coat

▪ Ex. How fast a quarter horse can run in a quarter mile.

∙ Genotype

o One’s genetic makeup.

∙ Allele

o A variation of a gene.  

∙ Locus

o The location of a gene on a chromosome.

∙ Heredity (Inheritance)

o What genes get passed down (segregated) from the parents to  progeny.

∙ Trait

o A genetically determined characteristics.

▪ Trait doesn’t always equal phenotype.

∙ Simple Trait

o When a gene is controlled by 1 or a few loci.

o Can be qualitative or categorical.

▪ Ex. The coat color of a horse is only controlled by a few loci. ∙ Complex Trait

o When a gene is controlled by many loci.

o Can be quantitative or polygenic.

▪ Ex. Milk production in cows in controlled by many loci.

∙ Pleiotropy  

o When a single gene controls many things.

▪ Ex. The growth hormone somatotropin (rbst)

∙ Homozygosity

o When the alleles are the same.

▪ Ex. BB or bb

⮚ If an individual possess the genotype BB they are said  

to be homozygous dominant for that trait.

⮚ If an individual possess the genotype bb they are said  

to be homozygous recessive for that trait.

∙ Heterozygosity

o When the alleles are different.

▪ Ex. Bb

⮚ If an individual has the genotype Bb they a said to be  

heterozygous for that trait.

o The number of heterozygous outcomes can be calculated by using  the following equation: 2n, where n equals the number of  

heterozygous loci.

▪ Ex. BbxCc

⮚ There are two different alleles, so the equation is 22  

which gives you 4 different out comes. (BC, Bc, bC,  

bc).

2.) What are the two main roles of DNA?

∙ DNA has two main roles: code for protein and for the transmission of genes  from parent to progeny (heredity).

3.) In order for DNA to accomplish its roles, what are the three things it must be able  to do?

∙ Stable

∙ Accurately Replicate

∙ Have the capacity for diversity.

4.) What is the cell cycle and explain what occurs during each phase?

∙ The cell cycle depicts the life cycle of a somatic cell, which is any cell in the  body except for gamete cells (sperm and egg).

o Interphase

▪ The phases (G1, S, G2,) in which the cell grows, synthesizes  

(replicates) it’s DNA, and prepares for mitosis.

o G0 Phase

▪ A phase in which a somatic cell is not actively dividing.  

o G1 Phase

▪ The phase in which the somatic cell grows.

o Synthesis Phase (S Phase)

▪ The phase in which DNA synthesizes (replicates).

o G2 Phase

▪ The phase in which the somatic cell continues to grow.

o Mitosis (M Phase)

▪ The phase in which a somatic cell divides into two identical  

daughter cells.

5.) Mitosis is the process in eukaryotes by which somatic cells replicate their DNA to  form two identical daughter cells.

6.) Meiosis is the process in eukaryotes by which gamete cells divide for sexual  reproduction, leading to four genetically diverse (different) cells.

7.) Review the phases of mitosis.

∙ Prophase

o The chromatin condenses into chromosomes.

o Each centrosome makes it way to the opposite sides of the cells  and starts forming mitotic spindles.

▪ Centrosome

⮚ Helps in the formation of mitotic spindles.

∙ Prometaphase

o The nuclear envelope disappears.  

o Mitotic spindles attach to each side of the chromosome’s  

centromeres by attaching to the kinetochore.

▪ Centromere

⮚ Where two sister chromatids are closely attached.  

✔A chromatid is an individual chromosome.

▪ Kinetochore

⮚ A protein structure on the centromere where mitotic  

spindles attach.

∙ Metaphase

o The chromosomes are aligned along the metaphase plate with the  help of mitotic spindles and centrosomes.

∙ Anaphase

o With the help of the centrosomes and the mitotic spindles the two  sister chromatids separate from each other and are pulled to the  opposite ends of the cell.

∙ Telophase

o Each daughter cell now has the same number of chromosomes as  its mother cell.

▪ A nuclear envelope starts to redevelop around the  

chromosomes.

▪ The chromosomes start to de-condense into chromatin.

▪ A cleavage furrow starts to develop between the two  

daughter cells.

⮚ Cleavage Furrow

✔The dividing of the two daughter cell’s  

cytoplasm.

∙ Cytokinesis

o The division of the two daughter’s cell cytoplasm.

o Once complete the cells are now two separate identical daughter  cells.

8.) Review the phases of meiosis.

∙ Meiosis I: Homologous Chromosomes Separate 

o Prophase I

▪ Just as in mitosis the centrosomes move to opposite sides of  

the cell, and chromatin condenses into chromosomes.

▪ Crossing over occurs.

⮚ Crossing Over

✔In which two different sister chromatids  

exchange DNA, leading to the genetic  

diversity of gamete cells.

o Metaphase I

▪ A chromosome randomly pairs up with another chromosome  

along the metaphase plate with the help of mitotic spindles  

and centrosomes, creating what is called homologous  

chromosome pairs.

o Anaphase I

▪ With the help of the centrosomes and the mitotic spindles  

each pair of homologous chromosome separate from each  

other and are pulled to the opposite ends of the cell.

o Telophase I

▪ Each cell is now made up of single chromosomes, that each  

contain two sister chromatids.

▪ A cleavage farrow forms.

o Cytokinesis

▪ The cytoplasm between the two cells is separated, forming  

two haploid daughter cells.

∙ Meiosis II: Sister Chromatids Separate 

o Prophase II

▪ The two haploid daughter cells from meiosis I will then each  

separate again to form four genetically diverse cells.

▪ Centrosomes move to opposite sides of the cell, and  

chromatin condenses into chromosomes.

▪ The two sister chromatids move towards the metaphase plate. o Metaphase II

▪ The two sister chromatids randomly line up along the  

metaphase plate.

o Anaphase II

▪ With the help of the centrosomes and the mitotic spindles  

each sister chromatid separate from each other and are  

pulled to the opposite ends of the cell.

o Telophase II

▪ Each cell is now made up of single chromatids.

o Cytokinesis

▪ The cytoplasm between the cells is separated.

▪ There are now four genetically diverse haploid cells.

9.) During what stage in meiosis does crossing over occur?

∙ Prophase I

10.) During what specific stage of prophase can chromosomal material cross-over  among a homologous pair of chromosomes?

∙ Pachytene

11.) How does linkage, linkage disequilibrium, and linkage equilibrium influence  genetic diversity?

∙ Linkage

o Genes that are located closely to each other are most likely to be  inherited together during the crossing over of meiosis.

∙ Linkage Disequilibrium

o When genes are always going to be inherited together.

▪ Linkage Equilibrium

⮚ When genes are not linked and thus are not going to  

be inherited together.

12.) Phenotype is one’s physical characteristics and trait measure, such as coat  color, where genotype is one’s genetic makeup.

13.) What is the difference between simple traits, complex traits, and Pleiotropy?

∙ Simple Trait

o When a gene is controlled by 1 or a few loci.

o Can be qualitative or categorical.

▪ Ex. The coat color of a horse is only controlled by a few loci.

∙ Complex Trait

o When a gene is controlled by many loci.

o Can be quantitative or polygenic.

▪ Ex. Milk production in cows in controlled by many loci.

∙ Pleiotropy  

o When a single gene controls many things.

▪ Ex. The growth hormone somatotropin (rbst)

14.) What is the foundational information needed to understand inheritance? ∙ Need to know the genotypes of the parents.

∙ Need to know how genes interact with each other.

∙ Mendel’s principles and how they work.

∙ How genotype and allele frequency work.

15.) What are Mendel’s principles?

∙ Segregation

o The separation of paired alleles during meiosis.

▪ Ex.

∙ Independent Assortment

o Alleles on different chromosomes will separate independently of  each other during meiosis.

▪ Ex.

16.) How can animal pedigrees, Al Sire Linage and Generations, and Punnett  Squares help us understand the segregation of alleles?

∙ A pedigree can give us information on the genotype or performance of the  ancestors of an individual.

∙ By looking at sire lineage we can see which sires had really good traits and  thus, were bred to pass the same traits down to the next generation.

∙ Punnett Squares help us understand segregation of alleles by helping us  calculate the different possible outcomes of a progeny based on the  parent’s genotype.  

17.) What information is needed in order to construct a Punnett square?

∙ Are we looking at single loci or multiple loci.

∙ Is there a linkage (haplotype) present or it just a regular chromosome. ∙ Are we looking at one chromosome or multiple chromosomes. ∙ Information (genotype) of the sire and dam.

18.) Compare and contrast homozygosity and heterozygosity.

∙ Homozygosity

o When the alleles are the same.

▪ Ex. BB or bb

⮚ If an individual possess the genotype BB they are said  

to be homozygous dominant for that trait.

⮚ If an individual possess the genotype bb they are said  

to be homozygous recessive for that trait.

∙ Heterozygosity

o When the alleles are different.

▪ Ex. Bb

⮚ If an individual has the genotype Bb they a said to be  

heterozygous for that trait.

o The number of heterozygous outcomes can be calculated by using  the following equation: 2n, where n equals the number of  

heterozygous loci.

▪ Ex. BbxCc

⮚ There are two different alleles, so the equation is 22  

which gives you 4 different out comes. (BC, Bc, bC,  

bc).

Week 7 Review (3/1/16-3/3/16) 

1.) Define the following vocab terms:

∙ Autosomal Chromosome

o Any other chromosome besides the sex chromosomes.

∙ Sex Chromosome

o Chromosomes that determine the gender of a species, where  

female has two X chromosomes and a male has one X  

chromosome and one Y chromosome.

∙ Locus

o The specific location of a gene on a chromosome.

∙ Genotype

o One’s genetic makeup.

∙ Homozygous

o When the alleles are the same.

▪ Ex. BB or bb

⮚ If an individual possess the genotype BB they are said  

to be homozygous dominant for that trait.

⮚ If an individual possess the genotype bb they are said  

to be homozygous recessive for that trait.

∙ Heterozygous

o When the alleles are different.

▪ Ex. Bb

⮚ If an individual has the genotype Bb they a said to be  

heterozygous for that trait.

∙ Allele

o A variation of a gene.

∙ Bi-allelic

o When you are dealing with just two alleles.

▪ Ex. BB, Bb, and bb are bi-allelic for they each contain two  

alleles.

∙ Multi-Allelic

o When dealing with three or more alleles.

▪ Ex. AaBbCc  

∙ Deleterious Allele

o A mutation within a gene that can harm the animal if passed down  from parents to progeny.

▪ Most deleterious alleles are homozygous recessive, meaning  

the offspring must acquire one recessive allele from each  

parent in order to have the condition.

∙ No Dominance

o When an animal is heterozygous for a trait, but both are expressed  half-way between the two homozygotes.

▪ Additive

⮚ When each allele has an independent effect on a  

trait.

o Ex.  

▪ Holstein

▪ Chromosome 14

▪ G allele (alanine advantageous form effect = 100lbs for milk  yield).

⮚ If an animal has two GG alleles it will yield an extra  

200lbs of milk.

⮚ If an animal has the GC alleles it will yield only 100lbs of  

extra milk.

⮚ If an animal has the CC alleles it will yield 0lbs of extra  

milk.

∙ Complete Dominance

o When an animal is homozygous or heterozygous for a trait. ▪ Ex.

⮚ Polled vs. Horned

❖ PP????An animal with these allele will have  

polled horns (no horns).

❖ Pp????An animal with these alleles with have  

polled horns (no horns).

❖ pp????An animal with these alleles will have  

horns.

∙ Incomplete or Partial Dominance

o When an animal is heterozygous for a trait, but portrays the traits of  the homozygous alleles.

▪ Ex.

⮚ HYPP in horses.

⮚ Impressive Syndrome miss-sense mutation.

∙ Over Dominance

o When an animal is heterozygous for a trait, but has a heterozygous  advantage and thus have a higher fitness than homozygous  individuals.  

▪ Ex.

⮚ Callipyge in sheep.

❖ A mutation in sheep in that allows them to  

have more muscular buttocks.  

✔ It is a heterozygous condition and the  

big C can only be inherited from the  

mother.

∙ Co-Dominance

o When animal possess a heterozygote trait and both the dominant  and recessive allele appear together in the animal’s phenotype. ▪ Ex.

⮚ Roan Coat Color in Short Horn Cattle

❖ RR????Red Coat

❖ RW????Red and White Coat

❖ WW????White Coat

∙ Epistasis

o The interaction of genes at different loci. The expression of genes at  one locus depends on the alleles on another locus.

▪ Ex.

⮚ Coat Color In Labrador Retrievers  

❖ A locus on chromosome 11 (Bb) controls the  

color of the coat.

❖ A locus on chromosome 5 (Ee) controls the  

extension of pigment.

✔ bb????Chocolate coat color

✔ Bb????Black coat color

✔ E_B_ ???? Black Coat

✔ E_bb ???? Chocolate Coat

✔ ee__ ????Yellow Coat

∙ Sex Linked

o When a specific trait is controlled on the loci of a sex chromosome  (XY vs. XX).

▪ The X is inactivated in the male; although calico and male  tortoiseshell cats are possible, they are rare and born sterile  (meaning they can not breed).

▪ Ex. Coat Color in Cats

⮚ XOXO????Female With an Orange Coat

⮚ XOXo???? Female Tortoiseshell Cat (Has Orange and Black  

Coat)

⮚ XoXo???? Female With a Black Coat

⮚ XOY????Male With an Orange Coat

∙ Sex Limited

o When the phenotypic expression of a trait is limited to one sex (also  known as sexual dimorphism).

▪ Ex.

⮚ Beard growth

⮚ Male Patterned baldness

⮚ Milk Production (milk epd)

⮚ Hen (h-h+) vs. Cock (hh) for colored feathers.

∙ Sex Influenced

o When the phenotypic expression of a trait is different between  males and females (one allele may be dominant in the male and  recessive in the female or vice versa.)

▪ Ex. Scurs in Cattle

⮚ Sn????Normal Dominant in Female

⮚ SC????Scur Dominant in Males

∙ Filial

o F1, F2, F3, etc. the generations of the family and progeny.

▪ The first generation of a progeny is known as the f1 generation. ∙ Inter-Se Mate

o The process of random mating.

∙ Repeated Back Crossing

o The crossing of a hybrid to one of its parents or someone that is  genetically similar to the parent, in order to get offspring that are  genetically similar to the parent.

∙ Grading Up

o Cross breeding a breed with another breed.  

▪ Upon breeding back 7/8 of your original breed, that animal is  considered purebred.

∙ Phenotypic Ratio

o The ratio of how many individual will display those particular traits. ∙ Genotypic Ratio

o The ratio of how many different genetic combinations there are.

2.) What is the foundational information needed to understand inheritance?  

∙ Need to know the genotypes of the parents.

∙ Need to know how genes interact with each other.

∙ Mendel’s principles and how they work.

∙ How genotype and allele frequency work.

3.) What the difference between sex and autosomal chromosomes?

∙ Autosomal Chromosome

o Any other chromosome besides the sex chromosomes.

∙ Sex Chromosome

o Chromosomes that determine the gender of a species, where  female has two X chromosomes and a male has one X  

chromosome and one Y chromosome.

4.) What the different nomenclature of genotypes and alleles?

∙ Letters  

o Ex. BB and Dd

∙ Abbreviations

o Ex. N????Normal  

o Ex. HYPP

∙ Super and Subscripts

o Ex. XOXo or XO/Xo 

∙ +/-

∙ Descriptor (size) or a PCR Product

o Ex. AT AT AT ????200 Base Pair Repeat

 AT AT AT AT????201 Base Pair Repeat

▪ Can be summarized as 200/201  

∙ Sequence (AGCT, SNP vs. indel vs. CNV)

5.) What are the different types of autosomal modes of inheritance?

∙ No Dominance

o When an animal is heterozygous for a trait, but both are expressed  half-way between the two homozygotes.

▪ Additive

⮚ When each allele has an independent effect on a  

trait.

o Ex.  

▪ Holstein

▪ Chromosome 14

▪ G allele (alanine advantageous form effect = 100lbs for milk  

yield).

⮚ If an animal has two GG alleles it will yield an extra  

200lbs of milk.

⮚ If an animal has the GC alleles it will yield only 100lbs of  

extra milk.

⮚ If an animal has the CC alleles it will yield 0lbs of extra  

milk.

∙ Complete Dominance

o When an animal is homozygous or heterozygous for a trait.

▪ Ex.

⮚ Polled vs. Horned

❖ PP????An animal with these allele will have  

polled horns (no horns).

❖ Pp????An animal with these alleles with have  

polled horns (no horns).

❖ pp????An animal with these alleles will have  

horns.

∙ Incomplete or Partial Dominance

o When an animal is heterozygous for a trait, but portrays the traits of  the homozygous alleles.

▪ Ex.

⮚ HYPP in horses.

⮚ Impressive Syndrome miss-sense mutation.

∙ Over Dominance

o When an animal is heterozygous for a trait, but has a heterozygous  advantage and thus have a higher fitness than homozygous  

individuals.  

▪ Ex.

⮚ Callipyge in sheep.

❖ A mutation in sheep in that allows them to  

have more muscular buttocks.  

✔ It is a heterozygous condition and the  

big C can only be inherited from the  

mother.

∙ Co-Dominance

o When animal possess a heterozygote trait and both the dominant  and recessive allele appear together in the animal’s phenotype. ▪ Ex.

⮚ Roan Coat Color in Short Horn Cattle

❖ RR????Red Coat

❖ RW????Red and White Coat

❖ WW????White Coat

∙ Epistasis

o The interaction of genes at different loci. The expression of genes at  one locus depends on the alleles on another locus.

▪ Ex.

⮚ Coat Color In Labrador Retrievers  

❖ A locus on chromosome 11 (Bb) controls the  

color of the coat.

❖ A locus on chromosome 5 (Ee) controls the  

extension of pigment.

✔ bb????Chocolate coat color

✔ Bb????Black coat color

✔ E_B_ ???? Black Coat

✔ E_bb ???? Chocolate Coat

✔ ee__ ????Yellow Coat

6.) What are the different sex-related modes of inheritance?

∙ Sex Linked

o When a specific trait is controlled on the loci of a sex chromosome  (XY vs. XX).

▪ The X is inactivated in the male; although calico and male  

tortoiseshell cats are possible, they are rare and born sterile  

(meaning they can not breed).

▪ Ex. Coat Color in Cats

⮚ XOXO????Female With an Orange Coat

⮚ XOXo???? Female Tortoiseshell Cat (Has Orange and Black  

Coat)

⮚ XoXo???? Female With a Black Coat

⮚ XOY????Male With an Orange Coat

∙ Sex Limited

o When the phenotypic expression of a trait is limited to one sex (also  known as sexual dimorphism).

▪ Ex.

⮚ Beard growth

⮚ Male Patterned baldness

⮚ Milk Production (milk epd)

⮚ Hen (h-h+) vs. Cock (hh) for colored feathers.

∙ Sex Influenced

o When the phenotypic expression of a trait is different between  males and females (one allele may be dominant in the male and  recessive in the female or vice versa.)

▪ Ex. Scurs in Cattle

⮚ Sn????Normal Dominant in Female

⮚ SC????Scur Dominant in Males

7.) What are some exceptions to Mendelian inheritance?

∙ Deleterious or lethal alleles

∙ Incomplete or co-dominance

∙ Silent or null alleles

∙ Epistasis

∙ Pleiotropy

∙ Epigenetics

∙ Variable expressivity

o Multiple variations in phenotype between the individuals with the  same genotype.  

▪ Ex.  

⮚ Piebald Spotting

❖ S=Solid Color

❖ SP=Pie-bald spotting (large areas of white).

❖ 10 different possible phenotypes, 1 genotype.

∙ Incomplete penetrance

o When individuals are carrying a variation of a gene (allele or  

genotype) that expresses an associated trait (phenotype).

▪ Ex. Huntington Disease and Neurodegenerative  

8.) Know how to draw and use a Punnett square.

9.) Find the genotypic and phenotypic ratio when you cross a male with the alleles  BB and a female with the alleles bb. Where BB equals a black coat and bb  equals a red coat.

b b  

B Bb Bb

B Bb Bb

⮚ Phenotypic Ratio: 100% Black

⮚ Genotypic Ratio: 0 BB, 4Bb, 0bb (100% Heterozygous)

10.) Find the genotypic and phenotypic ratio when you cross a heterozygous male  (Bb) and a heterozygous female (Bb) from the previous generation.

B b

B BB Bb

b Bb bb

⮚ Phenotypic Ratio: 3 Black: 1 Red (75% Black : 25% Red)

⮚ Genotypic Ratio: 1 BB: 2Bb: 1bb (1:2:1)

11.) Find the genotypic and phenotypic ratio when you cross a female that has RW  alleles and a male that has RW alleles. Where RR codes for a red coat, RW codes  for a roan coat, and WW codes for a white coat.

R W

R RR RW

W RW WW

⮚ Phenotypic Ratio: 1 Red: 2 Roan: 1 White

⮚ Genotypic Ratio: 1 RR: 2 RW: 1WW

12.) Find the genotypic and phenotypic ratio when you cross a female with the  following alleles BBPP and a male with the following alleles bbpp. Where the  allele b represents coat color and the p allele represents the presence of horns.

bp bp bp bp

BP BbPp BbPp BbPp BbPp

BbPp

BbPp

BP BbPp BbPp BbPp

BP BbPp BbPp BbPp

BP BbPp BbPp BbPp BbPp

⮚ Phenotypic Ratio: 100% Black and 100% Polled

⮚ Genotypic Ratio: 100% BbPp

13.) Find the genotypic and phenotypic ratio when you cross a male with the alleles  BbPp and a female with the alleles BbPp from the previous generation. Where the  allele b represents coat color and the p allele represents the presence of horns.  (Remember PP and Pp mean polled and pp means horned).

 BP Bp bP bp

BP BBPP BBPp BbPP BbPp

BBpp

BbPp

Bp BBPp BbPp Bbpp

bP BbPP bbPP bbPp

bp BbPp Bbpp bbPp bbpp

⮚ Phenotypic Ratio: 9 black and polled, 3 black and  

horned, 3 red and polled, 1 red and horned.

⮚ Genotypic Ratio: 1 BBPP: 2BBPp: 2BbPP: 4BbPp: 1BBpp:  

2Bbpp: 1bbPP: 2bbPp: 1bbpp

14.) Find the genotypic and phenotypic ratio when you cross a female black lab  with the alleles BBEE and a male yellow lab with the alleles bbee.

BE BE BE BE

be BbEe BbEe BbEe BbEe

BbEe

BbEe

be BbEe BbEe BbEe

be BbEe BbEe BbEe

be BbEe BbEe BbEe BbEe

⮚ Phenotypic Ratio: 100% Black  

⮚ Genotypic Ratio: 100% Heterozygous

15.) Find the genotypic and phenotypic ratio when you cross a male with the alleles  BbEe and a female with the alleles BbEe from the previous generation.

⮚ Phenotypic Ratio: 9 Black, 4 Yellow, 3 Chocolate  

16.) For more Punnett square practice review the Punnett Square practice we did in  class on 3/8/16. (The worksheet along with the answer key can be found on  Canvas in the modules section under supplementary information.) Also  reviewing the homework problems is a good idea.

Week 8 Review (3/8/16-3/10/16) 

1.) Define the following terms:

∙ Test Cross

o Determines if an animal is carrying specific alleles by breeding that  particular animal.

∙ DNA Test

o Testing the animal’s DNA in order to get a better understanding of  its genome.

∙ Introgression

o The movement of genes from one species to another through  

repeated back crossing.

∙ Precision Breeding

o Using DNA technology to remove or keep a specific gene from an  embryo in order to retain or remove unwanted traits.

∙ Simple Trait

o When a gene is controlled by 1 or a few loci.

▪ Can be qualitative or categorical.

⮚ Ex. The coat color of a horse is only controlled by a few  

loci.

∙ Agouti Gene

o Affects 18 species of animals.

o The agouti gene is involved in determining if an animal’s coat is  

banded (agouti) or a solid color (non-agouti).

2.) Know the different characteristics of various simple traits in domestic animals. ∙ Cattle

o Coat Color In Cattle

Animal

Bovine (Cattle)

Trait

Coat Color Expression

Phenotype(s)

Coat Color is Black or Red

Involved Chromosomes

18

Involved Genes

Melanocortin 1 receptor or MSHR

Biology/Physiology

A mutation on the 99th amino acid  in which leucine changes to  proline.

Mutation(s) (Genotype)

Complete Dominance

E+e and ee = Red Coat

EE and E+E = Black Coat

Can It Be Tested For  

Through A DNA Test?

Yes

Cost To Perform DNA Test

$15-$25

o Curly Calf Syndrome

Animal

Bovine (Cattle)

Trait

Curly Calf Syndrome  

(Arthrogryposis Multiplex  

Congenita or AM)

Phenotype(s)

Can give rise to a healthy born  calf or a still-born calf

Involved Chromosomes

N/A

Involved Genes

N/A

Biology/Physiology

Neurologic

The spine is bent and twisted in  affected calves.

Mutation(s) (Genotype)

Homozygous Recessive

AMF=Normal

AMC=Carrier

Can It Be Tested For  Through A DNA Test?

Yes

Cost To Perform DNA  Test

$15-$25

∙ Horse

o Hyperkalemic Periodic Paralysis (HYPP)

Animal

Equine (Horses)

Trait

Impressive Syndrome or HYPP

Phenotype(s)

Muscle Spams, Paralysis, and or  Weakness

Involved Chromosomes

11

Involved Genes

SCN4A

Biology/Physiology

Caused by a miss-sense  

mutation, it causes sodium  channels within the skeletal  muscles to function improperly.

Mutation(s) (Genotype)

Partial Dominance

Miss-Sense Mutation

NN=Normal

NH= Normal

HH= Have Disease

Can It Be Tested For  Through A DNA Test?

Yes

Cost To Perform DNA  Test

~$30

o Agouti In Horses

Animal

Equine (Horses)

Trait

Coat Color

Phenotype(s)

Bay, plus many, many more.

Involved Chromosomes

22

Involved Genes

Agouti Signaling Protein (ASIP)

Biology/Physiology

A frameshift mutation causes the  ASIP to malfunction.

Mutation(s) (Genotype)

Deletion in the extension of  pigment.

Can It Be Tested For  Through A DNA Test?

Yes

Cost To Perform DNA  Test

$40

∙ Sheep

o Scrapies

Animal

Ovine (Sheep)

Trait

Scrapie (TSE)

Phenotype(s)

Spongiform encephalopathy,  wasting

Involved Chromosomes

13

Involved Genes

PRNP or PrP

Biology/Physiology

In which one miss-folded prion  causes other prions to miss-fold in  the brain.

Mutation(s) (Genotype)

171 SNP, RR, RQ, QQ (R=arginine  vs. Q= glutamine).

Can It Be Tested For  Through A DNA Test?

Yes

Cost To Perform DNA  Test

$11-$29

∙ Swine

o Porcine Stress Syndrome

Animal

Swine (Pigs)

Trait

Porcine Stress Syndrome (PSS) Pale Soft Exudated Pork (PSE)

Phenotype(s)

Oxidative Metabolism,  

Halothane (HAL), Anesthesia  Response and Collapse

Involved Chromosomes

6

Involved Genes

Ryanodine Receptor (RyR1)

Biology/Physiology

A defective ryanodine receptor  causes a huge calcium influx,  which leads to muscle  

contracture and increase in  metabolism.

Mutation(s) (Genotype)

Autosomal Recessive Disorder NN=No Disease

Nn=No Disease But a Carrier nn=Has Disease

Can It Be Tested For  Through A DNA Test?

Yes

Cost To Perform DNA  Test

$22

∙ Dog

o Coat Color In Labrador Retrievers

Animal

Canine (Dogs)

Trait

Coat Color

Phenotype(s)

Black, Chocolate, Yellow

Involved Chromosomes

5

11

Involved Genes

MC1R

Tyrp1

Biology/Physiology

Responsible  

for extending  color pigment  in the fur.

Responsible for  the color of the  fur

Mutation(s) (Genotype)

E vs. e

B vs. b

Can It Be Tested For  Through A DNA Test?

Yes

Cost To Perform DNA  Test

$55

∙ Cat

o Feline Coat Length

Animal

Feline (Cats)

Trait

Hair Length (FGF5)

Phenotype(s)

Short, Medium, or Long Hair

Involved Chromosomes

1

Involved Genes

Fibroblast growth factor (FGF5)

Biology/Physiology

The fibroblast growth factor  controls the length of hair grown.

Mutation(s) (Genotype)

NN=Short Hair

N/M= 1, 2, 3, and 4, where 4  equals long hair.

Can It Be Tested For  Through A DNA Test?

Yes

Cost To Perform DNA  Test

$35

Week 10 Review (3/22/16-3/24/16) 

1.) Define the following terms:

∙ Pedigree

o A record of an animal’s descendants.

∙ Inbreeding

o When two genetically related animals breed.

▪ Inbred Pedigree

⮚ When a common ancestor is on the top and bottom of  

a pedigree.

❖ Crossing in an inbred line gives you hybrid  

vigor.

✔ Ex. AAbbCC x aaBBcc

F1 Generation: AaBbCc

∙ Line Breeding

o A breeding system designed to maintain a substantial degree of  pedigree relationship to a highly regarded ancestor, or group of  

ancestors, without causing high levels of inbreeding.

2.) Know the structures and components of a pedigree.

3.) Know how to read a Mendelian flow chart.

https://www.wikipedia.org/ 

4.) Know how to create a Mendelian flow chart from a pedigree.

∙ Read the pedigree keeping in mind who has the disease, who is a carrier of  the disease, who is disease free, and what mode of inheritance the  disease is.

∙ Once you have all these components you can easily transfer the  information from a pedigree to a Mendelian flow chart.

o Here is an example with Grid Maker and Curly Calf Syndrome.

5.) What is the foundational information needed to understand inheritance?

∙ Need to know the genotypes of the parents.

∙ Need to know how genes interact with each other.

∙ Mendel’s principles and how they work.

∙ How genotype and allele frequency work.

6.) Probability is ranked on a scale of zero too one. Thus, when an event has a score  of zero is has 0% chance of occurring, where when an event has a score of one then it has 100% chance of occurring.

7.) What is the rule of multiplication and when do I use it?

∙ How do I use the rules of multiplication?

o Calculate the probability of each independent event, then multiply  the individual probabilities together in order to obtain the final  

probability that these events will occur together.

▪ Ex. A heterozygous polled cow (Pp) mates with a  

heterozygous pulled bull (Pp). What is the probability that  

they with produce a horned calf (pp)?

⮚ From the cow whose alleles are Pp, she has 50%  

chance of passing the small p down her offspring. The  

bull whose alleles are also Pp, he too has a 50%  

chance of passing down the small p. So since each  

parent has a 50 % or ½ chance of passing down the  

small, you multiply ½ times ½ and you get ¼. Thus, the  

calf has a 25% or ¼ chance of being horned. You can  

double check your work with a Punnett square.

∙ When do I use the rules of multiplication?

o When you want to know the probability that two or more  

independent events will occur in a specific combination.

8.) How do you use the rules of multiplication when solving a di-hybrid problem? ∙ Allows us to predict the probability of the F1 generation without the need to  construct a 16-part Punnett square.

o Ex. When mating two heterozygous parents with the same  

genotypes of BbPp what is the probability of producing offspring  with the gamete BP?

▪ Since each parent has a 50% of producing a big B and big P,  

then you multiply ½ and ½ and you get ¼. Meaning the  

offspring will have a ¼ or 4/16 chance in inheriting the BP  

genotype.

o Ex. When mating two heterozygous parents with the same  

genotypes of BbPp what is the probability of producing offspring  with the gamete BBPP?

∙ Since the dam has a 50% chance of producing a big B and big P, then you  multiply ½ and ½ and you get ¼. Since the sire also has a 50% chance of  producing a big B and big P, then you multiply ½ and ½ and you get ¼. When you multiply ¼ and ¼ you get 1/16, meaning the offspring will have  a 1/16 chance of having the BBPP gamete.

9.) What is the rule of addition and when do I use it?

∙ Use the rules of addition when the probability of an event can occur in two  of more different ways.

o Ex. Probability of a heterozygote.

▪ There is 50% chance that the dominant allele could come  

from the sperm, and a 50% chance the recessive allele can  

come from the egg. Therefore, multiplying ½ and ½ gets you  

¼.

OR

▪ There is 50% chance that the dominant allele could come  

from the egg, and a 50% chance the recessive allele can  

come from the sperm. Therefore, multiplying ½ and ½ gets  

you ¼.

⮚ So the probability of a heterozygote is the addition of  

½ from the first event possibility and ½ from the  

second event possibility. ½ + ½= ¼  

10.) How do I use the rules of multiplication and addition to solve complex  problems?

∙ First use the rule of multiplication to calculate the probability for each of the  genotypes and then use the rule of addition to calculate the probabilities  of the recessive traits.

▪ Ex. Determine the probability of producing an offspring with

the genotypes of bbppCc, Bbppcc, bbPpc, bbPPcc, and  

bbppcc. Then, find the probability of two recessive  

phenotypes for at least two of three resulting from a tri

hybrid cross in cattle that are BpPpCc (dam) and bbPpcc

(sire).

⮚ An example on how to find the probability of one of  

the above genotypes. What is the probability of  

producing an offspring with the genotype of bbppCc  

with parents that have the following genotypes  

BpPpCc (dam) and bbPpcc (sire)?

❖ The probability of producing bb is a ½ (50%)  

chance from the dam and 1 (100%) chance  

from the sire. So by multiplying ½ and 1 you get  

½.

❖ The probability of producing pp is a ½ (50%)  

chance from the dam and a ½ (50%) chance  

from the sire. So by multiplying ½ and ½ you  

get ¼.

❖ The probability of producing Cc is a ½ (50%)  

chance from the dam and 1 (100%) chance  

from the sire. So by multiplying ½ and 1 you get  

¼.

✔ Thus, the probability of producing an offspring  

with the genotype bbppCc is ½ x ¼ x ½  

which equals 1/16. Therefore there is a 1/16  

chance in that the offspring will have the  

bbppCc genotype.

⮚ After calculating the probability of the remaining  

genotypes (bbppCc, Bbppcc, bbPpc, bbPPcc, and  

bbppcc), find the probability of two recessive  

phenotypes for at least two of three resulting from a  

tri-hybrid cross in cattle that are BpPpCc (dam) and  

bbPpcc (sire).

❖ From the calculation above the probability of  producing an offspring with the genotype  bbppCc is ½ x ¼ x ½ which equals 1/16.

❖ The probability of an offspring having the  genotype Bbppcc is ½ x ¼ x ½ which equals  1/16.