1. Reporting category #2- Mechanisms of genetics
This reporting category builds on Reporting Category #1- Cell Structure & Function.
This reporting category will have 11 questions on the EOC Biology exam.

2. Se 6- the mechanisms of genetics, including the role of nucleic acids & the principles of Mendelian Genetics

3. Phosphate= Nucleotide
DNA
A with T
C with G
Deoxyribose sugar +
Nitrogen base +
Phosphate= Nucleotide
RS 6A- identify components of DNA & describe how information for specifying the traits of an organism is carried in the DNA.
This is the same slide that was used in RC#1.
Review the building blocks of nucleic acids nucleotides. Also review that the sugar phosphate backbone of DNA is held together by covalent (relatively strong) bonds, while the nitrogenous bases are held together by hydrogen (very weak) bonds.

4. Chromosomes contain genes
RS 6A- identify components of DNA & describe how information for specifying the traits of an organism is carried in the DNA.
Remind the students that DNA is found in the nucleus of the cell- normally as uncondensed chromatin, but when the cell is getting ready to divide, the chromatin is tightly wound into chromosomes that are easier to separate into the daughter cells.
Human have 23 pairs of chromosomes- 23 from their mother & 23 from their father.
Genes are sections on the DNA molecule that contain the code for a particular protein particular trait.

5. Chromosomes contain genes
RS 6A- identify components of DNA & describe how information for specifying the traits of an organism is carried in the DNA.
Another visual to reinforce the fact that DNA is the genetic code which is translated into a chain of amino acids  protein.

6. genetic code- determined by the order of the nitrogen bases
RS 6A- identify components of DNA & describe how information for specifying the traits of an organism is carried in the DNA.
Show how the DNA of each organism is unique- the order of the nitrogen bases in each organism determines that organism’s traits.
Different proteins = different traits.

7. DNA specifies traits Central dogma
RS 6A- identify components of DNA & describe how information for specifying the traits of an organism is carried in the DNA; & SS 6C-
explain the purpose & process of transcription & translation using models of DNA & RNA.
This graphic shows the Central Dogma of Molecular Biology- a nice way to tie together how DNA controls what proteins are produced in the cell; therefore DNA controls cell activity.
Review that DNA replication (making copies of DNA) & transcription (rewriting DNA as an RNA strand) occur in the nucleus; translation or protein synthesis occur at the ribosome in the cell cytoplasm.

8. Sample question:
Question #1:
The sugar & phosphate portions of a DNA molecule function mainly to do which of the following?
A. store & transmit information about an organism’s traits
B. attract the units that are used to create new DNA molecules
C. provide the structural support for the sequence of nitrogenous base pairs
D. provide the energy that the molecule needs to carry out its function
Question #2:
DNA affects the organism’s traits by doing which of the following?
A. directing the assembly of amino acids to form proteins
B. determining which traits will be most beneficial to the organism
C. establishing the rules by which all of the elements in the body will form compounds
D. identifying which genetic material should be expressed & which material discarded
RS 6A- identify components of DNA & describe how information for specifying the traits of an organism is carried in the DNA.

9. The Genetic code is common to all organisms
SS 6B- recognize that components that make up the genetic code are common to all organisms.
This is a graphic that students may see on the STAAR Exam- shows the genetic code= the order of the three nitrogen bases in a three letter codon determines what amino acid is brought to the ribosome during protein synthesis to make the protein.
The genetic code is read using the first base of codon, then second base of codon, & then the third base of codon- look to see where these intersect to determine which amino acid is coded for or what punctuation is used in the genetic message (i.e. start or stop).

10. Sample question: What do these three statements describe?
I. Instructions for translating information into proteins II. Alignment & sequence of genes on a chromosome III. Composed of nucleotide triplets
A. protein synthesis
B. the genetic code
C. cell differentiation
D. DNA fingerprinting
SS 6B- recognize that components that make up the genetic code are common to all organisms.

11. Sample question:
Question #2:
Refer to the chart. The triplet codes found in this chart apply to:
A. humans only.
B. plants & animals.
C. prokaryotes only.
D. all living things.
SS 6B- recognize that components that make up the genetic code are common to all organisms.

12. Single strand U instead of T
RNA
Single strand
U instead of T
SS 6C- explain the purpose & process of transcription & translation using models of DNA & RNA.
Review the differences between DNA & RNA: ribose sugar instead of deoxyribose; the nitrogen base uracil replaces thymine (there’s no T in RNA!); single-stranded molecule instead of double stranded.
Also remind them that DNA cannot exist outside the cell nucleus, so it has to be rewritten as RNA to travel to the place in the cell where proteins are made.

13. Transcription DNA making RNA Animation link Video link
SS 6C- explain the purpose & process of transcription & translation using models of DNA & RNA.
Animation link is of RNA transcription.
Video link by DNA Learning Center shows a computer animation of transcription (it is too cool!).
Key Points: in the cell nucleus, the DNA molecule is opened by an enzyme that pairs RNA nucleotides that are complementary to the DNA nucleotides in one direction. This process occurs very quickly with the DNA molecule re-zipping once the enzyme has passed along the DNA strands.
Point out how the RNA nucleotides are complementary to the DNA nucleotides they paired with: U with A; A with T; C with G; G with C.

14. Translation Animation link Video link
SS 6C- explain the purpose & process of transcription & translation using models of DNA & RNA.
Animation link is of translation.
The video link shows a computer animation of the same process.
Key Points: the mRNA (messenger) molecule travels out of the nucleus (through nuclear membrane pores) & to the ribosome complex in the cell cytoplasm. It pairs with the ribosome briefly while tRNA (transfer or taxi-cab) molecules bring the amino acids that correspond to the mRNA codon to the complex. The tRNA anticodon pairs with the mRNA codon as the amino acids are linked together to form a protein (polypeptide). The process occurs in one direction & stops when the mRNA stop codon is reached.

15. Sample question:
Question #1:
The diagram shows how a section of protein containing the amino acid sequence “QYWQ” is formed. What is the nucleotide sequence of the mRNA section shown?
A. 5'UCGGAUACUACU3‘ B. 5'CCACCACCACCA3‘
C. 5'ACUUCGGAUACU3‘ D. 5'CAGUAUUGGCAA3'
SS 6C- explain the purpose & process of transcription & translation using models of DNA & RNA.

16. Sample question:
Question #2:
The illustration shows the transcription process. What is the main purpose of the structure labeled W?
A. carrying instructions for protein synthesis
B. transforming into a protein
C. replacing damaged DNA
D. passing traits to offspring
SS 6C- explain the purpose & process of transcription & translation using models of DNA & RNA.

17. gene expression is a regulated process
SS 6D- recognize that gene expression is a regulated process.
Animation Link shows the process involved in the tryptophan operon.
Key Points: genes (esp. in bacteria & viruses) can be turned on or off. An operon is a genetic regulatory system on a DNA molecule that consists of a cluster of genes that code for one or more proteins.
Review these definitions with the students:
Promoter= short DNA sequence to which the RNA polymerase attaches.
Operator= another short DNA sequence near the promoter where a regulatory protein may attach.
Repressor= a regulatory protein that blocks transcription by stopping the production of RNA.
Animation link

18. gene expression is a regulated process
SS 6D- recognize that gene expression is a regulated process.
This graphic review the idea that the genetic message, as RNA, is processed cut apart & spliced together to create a mature mRNA molecule.
Review these terms:
intron= intervening sequence noncoding regions of RNA that are transcribed, but then removed before the final mRNA is produced.
exon= expressed sequence DNA sequences that code for either an RNA or a protein.

19. Sample question: Every cell contains DNA. The main purpose of DNA
is to store the cell’s genetic information. How does
DNA control the cell?
A. DNA activates nerve signals
B. DNA protects the cell from invaders
C. DNA speeds up chemical reactions
D. DNA determines what proteins are made
Question #2:
The base sequence that remains after RNA splicing, & can be translated into a protein is a/an:
A. exon.
B. deletion.
C. insertion.
D. intron.
SS 6D- recognize that gene expression is a regulated process.

20. MUTATION
RS 6E- identify & illustrate changes in DNA & evaluate the significance of these changes.
Scientists have been able to genetically engineer these organisms in the laboratory- top right= the zebra-striped tree frog; bottom left= the kangarion; & middle= the tigbit (NOT).
Just funny pictures…

21. MUTATION= change in dna
RS 6E- identify & illustrate changes in DNA & evaluate the significance of these changes.
A mutation is defined as any change in the structure/sequence of a DNA molecule or in a cell’s chromosome. They may arise when a single nucleotide or multiple nucleotides are inserted, deleted, or exchanged in the genetic material.
The illustrations above show- at left= a single change from A to C; at right= from C to T. This may have little/no or a great effect on the protein that will be produced from this DNA sequence.
Many biology teachers use the sentence, “thecatatethefatrat” to illustrate both a point mutation (change one letter of the sentence) & a frameshift mutation (insert or delete one or two letters from the sentence). Point mutations are generally harmless (one exception is sickle cell anemia), while frameshift mutations are generally more damaging since they significantly alter the DNA reading frame.
thecatatethefatrat

22. MUTATION= change in a chromosome
RS 6E- identify & illustrate changes in DNA & evaluate the significance of these changes.
These illustrations show how a variety of types of mutations can affect chromosomes. Remind students that homologous chromosomes line up next to each other (gene to gene) during prophase I of meiosis & undergo crossing over (an exchange of genetic material).
Discuss each circumstance using the illustration above:
A deletion occurs when a chromosome section is lost
A duplication occurs when a chromosome section is copied multiple times
An inversion occurs when a chromosome section is inserted with the associated genes in the wrong order
An insertion occurs when a section from one chromosome is joined to another chromosome
A translocation occurs when one or more chromosome fragments break & are rejoined together

23. MUTATION & significance
Beneficial or detrimental?
RS 6E- identify & illustrate changes in DNA & evaluate the significance of these changes.
This slide illustrates the results of some various mutations.
Discuss each one:
Top left shows albinism (a recessive genetic mutation) results from the body’s inability to produce an enzyme needed to produce the pigment melanin affected individuals are white & have difficulty [(ike this white tail deer) hiding from predators.
Bottom left shows sickle cell anemia (result of a point mutation in the DNA strand that produces the oxygen carrying protein hemoglobin) results in red blood cells taking on an abnormal sickle shape which clumps together easily & can block blood circulation to some body parts.
Top right shows hemophilia (this form is a genetic mutation on the X chromosome) results in individuals (typically males since they only have one X chromosome) that lack the ability to clot their blood. Half shaded females in the graphic can be assumed to be carriers on this disorder.
Bottom right shows red green color blindness (another genetic disorder on the X chromosome) results in individuals that have a difficult time differentiating between shades or red &/or green. Individuals with this disorder will not be able to see the number 45 in the left circle or 7 in the right circle.

24. Sample question:
Question #1:
A scientist is searching for a chemical that can alter the characteristics a pea plant will pass on to its offspring. The scientist needs a chemical that can affect pea plants in which way?
A. by altering the sequence of nitrogenous bases in the plant’s DNA
B. by reducing the total number of copies of the plant’s genetic material
C. by changing the spiral shape of the plant’s DNA molecules to wheels or horseshoes
D. by adding extra deoxyribose molecules to each strand of the plant’s genetic material
RS 6E- identify & illustrate changes in DNA & evaluate the significance of these changes.

25. Sample question:
Question #2:
Mutated DNA in somatic cells occurs frequently, but this modified DNA cannot be passed along to offspring because:
A. modified DNA in somatic cells is recessive.
B. offspring typically reject modified DNA.
C. only gamete cells carry genetic material to offspring.
D. offspring mutations happen only when somatic cells are diseased.
RS 6E- identify & illustrate changes in DNA & evaluate the significance of these changes.

26. Animation link
RS 6F- predict possible outcomes of various genetic combinations such as monohybrid crosses, dihybrid crosses & non-Mendelian inheritance.
Remind students that they will be asked to work problems using what they learned in the genetics unit. The easiest way to solve a genetic problem is using a Punnett Square where the alleles from one parent are placed along the left side of the square while the alleles from the other parent are placed along the top of the same square. The genetic make up or genotype of the offspring are predicted by filling in the boxes in the square using one allele from each parent (dominant or upper case allele always goes first). Once the genotypes of the offspring are shown, you can determine their appearance or phenotype.
Also remind them that we use upper case letters to represent dominant genes & lower case letters to represent recessive genes.
Animation link is from Cold Spring Harbor- it does a nice job reviewing how to set up & interpret a Punnett Square!

27. heterozygous & HOMOZYGOUS
RS 6F- predict possible outcomes of various genetic combinations such as monohybrid crosses, dihybrid crosses & non-Mendelian inheritance.
This slide reviews the meanings of two important terms. Ask the students to define them from the illustrations.
Heterozygous= having two different alleles (i.e. R r)  both parents & the two blue children
Homozygous= having two of the same alleles (i.e. R R or r r)  the white child & the pink child
You may want to point out that genes are not really “faulty” though blue hair would be considered an unusual mutation!

28. Monohybrid cross
RS 6F- predict possible outcomes of various genetic combinations such as monohybrid crosses, dihybrid crosses & non-Mendelian inheritance.
This graphic illustrates one of Mendel’s classic crosses. A true-breeding tall plant (homozygous dominant or TT) is crossed with a true-breeding dwarf plant (homozygous recessive or tt) & all the first generation showed the tall trait. When you show this in a Punnett Square, you end up with all F1 offspring having the genotype Tt which shows the phenotype tall.
When you take these F1 offspring & cross them (all are Tt); you get a 3 tall to 1 dwarf phenotypic ration in the F2 generation. Work this Punnett Square for the class or ask a student to work it. What is the genotypic ratio of the F2 generation? (1TT, 2Tt, 1tt).

29. dihybrid cross
RS 6F- predict possible outcomes of various genetic combinations such as monohybrid crosses, dihybrid crosses & non-Mendelian inheritance.
Here is a graphic that shows the outcome of a dihybrid cross involving two round, wrinkled plants (RrYy) when yellow (Y) is dominant over green (y) & round (Y) is dominant over wrinkled (r). Use the graphic to show the relationship between round (R) & wrinkled (r); & the relationship between yellow (Y) & green (y). Each trait shows a 3 to 1 phenotypic ratio as expected. Work your way from the top left corner to the bottom right corner as you illustrate each genotype to remind students of the expected pattern in a dihybrid cross of two heterozygous parents.

30. Unusual patterns of inheritance
RS 6F- predict possible outcomes of various genetic combinations such as monohybrid crosses, dihybrid crosses & non-Mendelian inheritance.
Remind the students that not all traits show simple dominance & recessiveness. Some show an unusual pattern.
The left graphic shows incomplete dominance in snapdragon flowers where the phenotype of the heterozygous individual is intermediate or a blending of the dominant & recessive trait (i.e. red X white  pink). This also occurs in other organisms like cattle & chickens.
The right graphic shows codominance where the heterozygous individual with both dominant alleles shows both the dominant & recessive trait. Blood groups in humans is a great example an individual with the dominant allele for type A antigen & type B antigen can show in an individual with type AB blood. There are 4 possible combinations of alleles for type A, type B, type AB, & type O blood.
Incomplete dominance
Codominance

31. Nom-mendelian inheritance
RS 6F- predict possible outcomes of various genetic combinations such as monohybrid crosses, dihybrid crosses & non-Mendelian inheritance.
Here are some additional examples of non-Mendelian patterns.
The left graphic shows how leaf color can result from the presence of either green or colorless chloroplasts a result of the genetic contribution of the DNA in plant chloroplasts. The chloroplast exerts an influence on the phenotype of the plant extranuclear inheritance.
The right graphic shows how, since female cats have two X chromosomes, one X chromosome becomes inactive (forms a Barr body) while the other X chromosome shows the fur pigmentation gene. Depending on which copy of the X chromosome each cell chooses to leave active either an orange or black coat color results. Therefore, the female cat shows a calico or tortoiseshell fur pigmentation pattern.
X-inactivation
Extranuclear inheritance

32. Sample question: Look at the Punnett square shown below.
In pea plants, having round peas (R) is dominant over wrinkled peas (r). Two plants with round (Rr) peas were crossed. If 4 offspring are produced, how many offspring are expected to have round peas?
A. 1
B. 2
C. 3
D. 4
RS 6F- predict possible outcomes of various genetic combinations such as monohybrid crosses, dihybrid crosses & non-Mendelian inheritance.

33. Sample question:
Question #2:
Refer to the diagram. What are the genotypes of the parents in this cross?
A. TTRr & ttRr
B. TtRr & TtRr
C. TTrr & TtRr
D. TTRr & TtRr
RS 6F- predict possible outcomes of various genetic combinations such as monohybrid crosses, dihybrid crosses & non-Mendelian inheritance.

34. MEIOSIS allows sexual reproduction
Video link
Occurs in sex cells
4 cells result
½ # of chromosomes
SS 6G- recognize the significance of meiosis to sexual reproduction.
The video link is from McGraw-Hill.
The graphic shows the meiosis I & II. Review the stages with the students & remind them of the importance of meiosis to the organism allows production of haploid sex cells allows sexual reproduction since the chromosome number is divided in half.
Also remind the students of two events that increase genetic variation in organisms that undergo meiosis crossing over & independent assortment.
Crossing over occurs in prophase I of meiosis, where homologous chromosomes line up- gene to gene- & switch genetic material. The graphic shows this with the two blue colored chromosomes.
Independent assortment results by the way the homologous chromosomes (which just underwent crossing over) line up on the cell equator during metaphase I of meiosis. This is illustrated in the graphic. Point out how the daughter cells at telophase II are very different from the chromosomes in prophase I.
Without meiosis, the chromosome number of organisms would double each time they sexually reproduce!
Crossing over + Independent assortment

35. Sample question:
Question #1:
A rat has 42 chromosomes in each somatic cell. How many chromosomes are in each gamete?
A. 18
B. 21
C. 40
D. 42
Question #2:
The diagram shows the process of meiosis. The chromosome separation that occurs during meiosis results in a:
A. single fertilized egg cell.
B. group of genetically identical cells.
C. single sperm cell.
D. reduction in the number of chromosomes per cell.
SS 6G- recognize the significance of meiosis to sexual reproduction.

36. (can you find the hidden words?)
How to study a genome
Karyotype analysis
lqvrdlmnqvtthequickababcmfxlqbrownfoxjulrvsmpedoverthelazyyyzplfdogjjiurttiythedoglayhhbeldquietlydreaminghwwiqldnsofdinnerplwosiucnd
SS 6H- describe how techniques such as DNA fingerprinting, genetic modifications, & chromosomal analysis are used to study the genomes of organisms.
These graphics show how scientists effectively use different techniques to study an organism’s genetic makeup.
The left graphic shows DNA fingerprinting extracting & treating the DNA from an organism with an enzyme that cuts the DNA into fragments of various sizes. These fragments are run through gel electrophoresis to form bands (according to their size smaller move farther than larger), the DNA fragments are transferred to a membrane, then the gel is treated with a chemical that binds to the fragments the fragments can be visualized using X-ray film to produce a picture of their banding pattern.
During karyotype analysis- a white blood cell is chemically treated & cultured & all of the chromosomes are photographed, then arranged from largest to smallest to determine if a chromosome mutation has occurred. See if the students can determine the disorder shown in the top right graphic (trisomy 21).
The bottom right graphic simply shows how the genetic message results from useful & nonuseful words in the DNA molecule. Have the students attempt to find the hidden sentence in the four lines of letters, then go forward to the next page to show the answer.
DNA fingerprinting
Genome sequencing
(can you find the hidden words?)

37. (did you find the hidden words?)
How to study a genome
Karyotype analysis
lqvrdlmnqvtthequickababcmfxlqbrownfoxjulrvsmpedoverthelazyyyzplfdogjjiurttiythedoglayhhbeldquietlydreaminghwwiqldnsofdinnerplwosiucnd
SS 6H- describe how techniques such as DNA fingerprinting, genetic modifications, & chromosomal analysis are used to study the genomes of organisms.
DNA fingerprinting
Genome sequencing
(did you find the hidden words?)

38. Sample question:
Question #1:
Which of the following is an inappropriate place to gather information for a karyotype?
A. a sex cell
B. a skin cell
C. a nerve cell
D. a muscle cell
Question #2:
Most species have a standard karyotype. It is common, however, that there is a difference between males & females within a species. How does the human karyotype differ between males & females?
A. males only have 45 chromosomes & females have 46
B. the sex chromosomes in males are XY & in females they are XX
C. scientists cannot tell the difference but know there is one
D. males have two satellites attached to their sex chromosomes
SS 6H- describe how techniques such as DNA fingerprinting, genetic modifications, & chromosomal analysis are used to study the genomes of organisms.