High school biology worksheets

There are six worksheets in total that needs to be done, I have linked it to this post with the data needed.

Biology200-29 Data Sheet

Messenger RNA Codons and Their Corresponding Amino Acids

First
Base

Second Base Third
Base U C A

G

U

UUU phenylalanine
UUC phenylalanine
UUA leucine
UUG leucine

UCU serine
UCC serine
UCA serine
UCG serine

UAU tyrosine
UAC tyrosine
UAA stop**
UAG stop**

UGU cysteine
UGC cysteine
UGA stop**
UGG tryptophan

U

C

A

G

C

CUU leucine
CUC leucine
CUA leucine
CUG leucine

CCU proline
CCC proline
CCA proline
CCG proline

CAU histidine
CAC histidine
CAA glutamine
CAG glutamine

CGU arginine
CGC arginine
CGA arginine
CGG arginine

U
C
A
G

A

AUU isoleucine
AUC isoleucine
AUA isoleucine
AUG methionine*

ACU threonine
ACC threonine
ACA threonine
ACG threonine

AAU asparagine
AAC asparagine
AAA lysine
AAG lysine

AGU serine
AGC serine
AGA arginine
AGG arginine

U
C
A
G

G

GUU valine
GUC valine
GUA valine
GUG valine

GCU alanine
GCC alanine
GCA alanine
GCG alanine

GAU aspartate
GAC aspartate
GAA glutamate
GAG glutamate

GGU glycine
GGC glycine
GGA glycine
GGG glycine

U
C
A
G

* Note: AUG is an initiator codon and also codes for the amino acid methionine.
** Note: UAA, UAG, and UGA are terminator codons.

Information About Nitrogen Bases

Nitrogen Base Classification Abbreviation
Adenine Purine A
Guanine Purine G
Cytosine Pyrimidine C
Thymine Pyrimidine T
Uracil Pyrimidine U

Biology30—Lab III.5 Page 4

PART C: Mutations and Base Sequence Errors

Not often are there errors in the process of forming proteins from the DNA code of instructions. An error in the process is a mutation and will result in formation of a different type of protein.

Hemoglobin is a protein in red blood cells. Hemoglobin results from the proper arrangement of almost 600 amino acids. Most humans have the correct type of hemoglobin. However, in some people the arrangement is incorrect. These people have a disease called sickle-cell anemia. Their red blood cells are sickle shaped rather than round. As a result, the red blood cells cannot transport oxygen as well.

The following amino acid sequence represents a portion of the normal hemoglobin molecule:

proline, glutamate, glutamate, lysine.

1. Use the Data table to translate the sequence of amino acids in normal hemoglobin into the following codes. Remember that the table shows mRNA codes (use first listing of the amino acid in the table)

a) mRNA base codes

b) tRNA base codes

c) DNA base codes

In sickle-cell anemia, the sequence of amino acids is slightly different. It is

proline, valine, glutamate, lysine.

2. Translate the sequence of amino acids in sickle-cell hemoglobin into (use third valine in tables)

a) mRNA base codes

b) tRNA base codes

c) DNA base codes

3. In terms of base nucleotides, explain the only difference between the DNA message for normal hemoglobin and the DNA message for sickle-cell hemoglobin.

A mutation, therefore, is a difference from what we consider to be the normal sequence of bases in a molecule of DNA. The differences or error does not have to be very great. As you have just determined, a base sequence of only one triplet (three bases) can cause the formation of the wrong type of hemoglobin. A change at only one base site of the triplet can cause mutation.

4. How are mutations passed on to offspring?

DNA and RNA

Sickle Cell Lab Page 6 of 7

Biology 30—Lab III.1 Page 4

BIOL200-27 Single Trait Lab Page 3

FINDING GENOTYPES AND PHENOTYPES FOR ONE TRAIT

In genetics, it is possible to calculate the results that should appear in offspring if the genotypes of both parents are known. These are called expected results. Expected results can be calculated by mathematics or use of Punnett squares. Thus, expected results are specific numbers and are not the result of random events. Observed results are those that appear in offspring in actual crossings. They are due to chance combinations of certain genes.

Expected and observed results may not always agree exactly, but there should be some agreement. Expected results are used to predict the results of a cross before the cross is done. If the expected results indicate that a certain type of offspring is likely, the cross can be carried out with some certainty that the type of offspring will appear in the observed results.

In this investigation, you will

a) substitute properly marked coins for gamete cells,

b) toss the marked coins 100 times to represent 100 offspring,

c) determine the expected numbers of genotypes for 100 offspring and compare them with the observed numbers of genotypes obtained through 100 coin tosses,

d) determine the numbers of expected phenotypes for a genetic cross, and compare them with the numbers of observed phenotypes obtained through coin tossing.

Part A: Determining Numbers of Expected Genotypes

How many of each genotype combination are expected in the offspring of a cross if both parents are Ss for a trait?

Use the Punnett square below to determine the genotypes. Record the number of each genotype Record the number of each genotype in column A of Table 1. If there are 100 offspring multiply each number in column A by 25. Record this number in column B of Table 1.

S

s

	S		s

Part B: Determining Numbers of Observed Genotypes

Cover both sides of two pennies with adhesive tape. Trim off any excess tape with scissors. Print an S on one side of each coin and an s on the other side of each coin.

Place both coins in cupped hands, shake, and then toss the coins onto your desk. Read and record the letter combination in column C (Toss Results) of Table 1. Make a slash in the proper row of column C to indicate the letter combination. Repeat this process until the coins have been tossed 100 times. Record the coin combinations for each toss in Table 1.

Record in column D the totals for each.

Table 1: Expected and Observed Genotypes
Gene Combination	(A) Expected Genotype For 4 Offspring	(B) Expected Genotype For 100 Offspring	(C) Toss Results	(D) Observed Genotype For 100 Offspring
SS
Ss or sS
ss

Part C: Determining Numbers of Expected Phenotypes

Assume that S represents the dominant gene for normal skin pigment. Assume that s represents a recessive condition called albinism, no skin pigment. From the Punnett square, list in column A of Table 2 the number of offspring expected to have normal skin colour (SS or Ss) and the number expected to be albino (ss).

Calculate the number expected to have each trait if there are 100 offspring. Do this by multiplying column A figures by 25. Record these numbers in column B of Table 2.

Part D: Determining Numbers of Observed Phenotypes

From your data in column D of Table 1, total and record in Column C of Table 2 the number of offspring who will have normal skin pigment (SS, Ss or sS) and those who will be albino (ss).

Table 2: Expected and Observed Phenotypes
Phenotype Possible	(A) Expected Phenotype For 4 Offspring	(B) Expected Phenotype For 100 Offspring	(C) Observed Phenotype For 100 Offspring
Normal Skin
Albino

Analysis

1. What does each side of each coin represent?

2. How does the chance of a coin landing on each side compare to the chance that a gamete cell will receive a particular gene at meiosis?

3. Why must two coins be used to determine the genotypes for the offspring?

4. What does the use of two coins compare to at fertilization?

5. Compare the expected genotypes of 100 offspring with the observed genotypes.

6. What is the advantage of comparing the 100 expect offspring with the 100 observed offspring rather than comparing only four expected offspring with four observed offspring?

7. Compare the expected phenotypes for 100 offspring with the observed phenotypes.

Genotype & Phenotype For One Trait

Genotype & Phenotype For One Trait Page 2 of 4

Biology

30—Lab III.5 Page 4

DNA

and

RNA

Deoxy

ribonucleic acid

(DNA) is a complex molecule found in all living organisms. DNA is the chemical of which genes are composed. An understanding of the organization of this molecule has answered many questions. Scientists now know how chromosomes can duplicate during cell division and transfer their genetic information to new chromosomes. Scientists also understand how chromosomes in the cell nucleus can direct the formation of specific proteins outside the nucleus.

In this investigation, you will

1. learn the names of the molecules which make up DNA.

2. use models to construct a molecule of DNA and show how it replicates.

3. learn the names of the molecules which make up RNA.

4. use models to show how the base sequence code in DNA is transcribed exactly to RNA.

PART A: Structure of DNA Nucleotides

Two important molecules which make up DNA are deoxyribose and phosphoric acid. Their structural formulas are:

deoxyribose	phosphoric acid

1. Give the molecular formula for

a) deoxyribose C___ H___ O___

b) phosphoric acid H___ P___ O___

Deoxyribose is a carbohydrate. In addition, there are four different molecules called nitrogen bases. Their structural formulas are:

guanine	cytosine
adenine	thymine

2. Of the four bases, which other base does

a) adenine most resemble in shape?

b) thymine most resemble in shape?

A molecule of deoxyribose joins with phosphoric acid and any one of the four bases to form a chemical compound called a nucleotide. A nucleotide is named after the nitrogen base that joins with the deoxyribose. For example, if thymine attaches to deoxyribose, the molecule is called a thymine nucleotide.

3. List four different nucleotides that can be made (4 marks).

4. a) How is each nucleotide alike (2 Marks)?

b) How does each nucleotide differ?

PART B: Structure of a DNA Molecule

A DNA molecule is “ladder like” in shape. Deoxyribose and phosphoric acid molecules join to form the sides or uprights of the ladder. Base molecules join to form the rungs of the ladder.

Print two copies of BIOL200-28 Lab Sheet 1 and one copy of BIOL200-28 Lab Sheet 2. Cut out the 24 nucleotide models (cut only on solid lines).

Fit six nucleotides together in puzzle like fashion to form a row in the following sequence from top to bottom: cytosine, thymine, guanine, adenine, guanine and cytosine. Let this arrangement represent the left half of a ladder molecule. It should consist of one side plus six half rungs.

5. If DNA is “ladder like”, which two molecules of a nucleotide form the sides, or the upright portion of the ladder (2 Marks)?

6. To which molecule does each base attach?

7. Name the molecule of each nucleotide that form part of the ladder’s rungs.

Complete the right side of the DNA ladder by matching the bases of other nucleotides to form complete rungs. It may be necessary to turn molecules upside down in order to join certain base combinations. NOTE: The ends of each base will allow only a specifically shaped matching new base to fit exactly.

Your completed model should look like a ladder with matched bases as the rungs. Besides being shaped like a ladder, a DNA molecules is twisted. It looks like a spiral staircase. However, your paper model cannot show this shape.

8. Is the order of half-rung bases exactly the same from top to bottom of each side of your model?

9. Only two combinations of base pairings are possible for the rungs. Name the molecule combinations or pairs (2 Marks).

10. If four guanine bases appear in a DNA model, how many cytosine bases should there be?

11. Your DNA model has four guanine bases.

a) Does the number of cytosine bases in your model agree with your prediction?

b) The following are the bases on the left side of a DNA molecule. List the bases that would make up the right side of a DNA molecule.

i) Thymine

ii) Adenine

iii) Guanine

iv) Guanine

v) Cytosine

PART C: DNA Replication

A chromosome contains DNA. Your DNA model represents only a short length of the DNA portion of a chromosome. An entire chromosome has thousands of rungs rather than only six. Although your model is only a small part of a chromosome, its replication is the same as that of an entire chromosome during mitosis and meiosis.

Open your DNA model along the point of attachment between base pairs (rungs) and separate the two ladder halves (A chromosome untwists and “unzips” in a similar way prior to replication).

Using the left half of your model as a pattern, add new nucleotide to form a new right side. Build a second DNA model by adding new nucleotides to the right half of the original model.

12. Do the two new molecules contain the same number of rungs?

13. Is the order from top to bottom of the base pairs (rungs) different or the same for each new DNA molecule?

14. How many molecules of adenine and thymine are in each DNA molecule?

15. Do the numbers agree?

16. Are the two DNA molecules exact copies of each other?

The specific order of bases in DNA serves as a code or language. When a chromosome replicates, the code (the order in which the bases occur) is carried over to the new chromosome.

17. What is the code of a chromosome?

PART D: RNA Structure

Besides ensuring the exact replication of chromosomes, the sequence (order) and pairings of bases are a genetic code of the instructions for the entire cell. How does a cell “read” the chemical message coded in its DNA in the form of specific base sequences? Part of the answer lies with a second molecule in the nucleus of cells called ribonucleic acid (RNA).

RNA is similar to DNA in that its molecules are also

formed from nucleotides

. However, deoxyribose and thymine are not found in RNA. Two other molecules, ribose and uracil, are present. Ribose replaces deoxyribose, and uracil replaces thymine. Looking at their structural formulas and models, you will see certain similarities between the molecules that they replace. Formulas and models are shown below:

ribose

uracil

18. a) Which base is replaced in RNA by uracil?

b) What chemical replaces deoxyribose in RNA?

19. To which base in DNA do the following RNA bases pair?

a) guanine

b) adenine

c) cytosine

d) uracil

Analysis

Complete the table by using “x” marks to indicate to which molecule each characteristic applies.

SIMILARITIES AND DIFFERENCES BETWEEN DNA AND RNA
DNA	RNA
deoxyribonucleic acid
ribonucleic acid
deoxy ribose present
ribose present
phosphoric acid present
adenine present
thymine present
uracil present
guanine present
cytosine present
contains a chemical message or code
formed from nucleotides
double stranded
single stranded
remains in nucleus
moves out of nucleus

DNA and RNA

DNA and RNA Lab Page 6 of 7

InsectClassification Lab

Six of the Orders of Class Insecta are given below. Do some research to find information about each Order to give a description and examples of organisms in them. Use your descriptions to determine which specimen below belongs to each Order. Record the letter of each specimen below.

A)	B)
C)	D)
E)	F)

All images from

https://pixabay.com/

Order Diptera

Description:

Example organism(s) in this Order:

Source(s) of Information:

Letter of specimen:

Order Lepidoptera

Description:

Example organism(s) in this Order:

Source(s) of Information:

Letter of specimen:

Order Coleoptera

Description:

Example organism(s) in this Order:

Source(s) of Information:

Letter of specimen:

Order Odonata

Description:

Example organism(s) in this Order:

Source(s) of Information:

Letter of specimen:

Order Orthoptera

Description:

Example organism(s) in this Order:

Source(s) of Information:

Letter of specimen:

Order Hymenoptera

Description:

Example organism(s) in this Order:

Source(s) of Information:

Letter of specimen:

Biology30—Lab III.2 Page 6

Pedigree Studies

Page 5

PEDIGREE STUDIES

Pedigrees are not reserved for show dogs and race horses. All living things, including humans, have pedigrees. A pedigree is a diagram that shows the occurrence and appearance, or phenotype, of a particular genetic trait from one generation to the next in a family. Genotypes for individuals in a pedigree usually can be determined with an understanding of inheritance and probability.

In this investigation, you will:

a)

learn the meaning of all symbols and lines that are used in a pedigree,

b)

calculate expected genotypes for all individuals shown in pedigrees.

Part A: Background Information

The pedigree in the diagram to the right shows the pattern of inheritance in a family for a specific trait. The trait being shown is ear lobe shape. Geneticists recognize two general ear lobe shapes, free lobes and attached lobes. The gene responsible for free lobes (E) is dominant over the gene for attached lobes (e).

Figure 1

In a pedigree, each generation is represented by a Roman numeral. Each person in a generation is numbered. Thus each person can be identified by a generation number and individual number. Males are represented by squares whereas females are represented by circles.

Part B: Reading A Pedigree

In Figure 1, persons I1 and I2 are the parents. The line which connects them is called a marriage line. Persons II1, 2, and 3 are their children. The line which extends down from the marriage line is the children line. The children are placed left to right in order of their births. That is, the oldest child is always on the left.

1. What sex is the oldest child?

2. What sex is the youngest child?

Using a different pedigree of the same family at a later time shows three generations.

Figure 2

shows a soninlaw as well as a grandchild. Generation I may now be called grandparents.

3. Which person is the soninlaw?

4. To whom is he married?

5. What sex is their child?

Figure 2	Figure 3

Part C: Determining Genotypes from a Pedigree

The value of a pedigree is that it can help predict the genes (genotype) of each person for a certain trait.

All shaded symbols on a pedigree represent individuals who are homozygous recessive for the trait being studied. Therefore, persons I1 and II2 have ee genotypes. They are the only two individuals who are homozygous recessive and show the recessive trait. They have attached ear lobes.

All unshaded symbols represent individuals who have at least one dominant gene. These persons show the dominant trait.

To predict the genotypes for each person in the pedigree, there are two rules you must follow.

Rule 1: Assign two recessive genes to any person on a pedigree whose symbol is shaded. (These persons show the recessive trait being studied.) Small letters are written below the person’s symbol.

Rule 2: Assign one dominant gene to any person on a pedigree whose symbol is unshaded. (These persons show the dominant trait being studied.) A capital letter is written below the person’s symbol.

These two rules allow one to predict some of the genes for the persons in a pedigree. Figure 3 shows the genes predicted by using these two rules.

To determine the second gene for persons who show the dominant trait, a Punnett square is used. In Figure 3, we already know that the grandfather (I1) is ee. If the grandmother (I2) were EE, could any ee children (like II2) be produced? A Punnett Square shows this combination to be impossible. Thus, the grandmother must be heterozygous or Ee.

6. Can an Ee parent and an ee parent have the results shown in generation II? Prove your answer by showing the results in a Punnett square.

e

e

e

E

7. Predict the second gene for person II3 and II4 (Read them from the Punnett square).

8. Could child II3 or II4 be EE ? Explain.

To predict the second gene for person II1, a different method must be used, since he could be either EE or Ee.

9. Can an EE person married to an ee person (II2) have children with free ear lobes?

10. Can an Ee person married to an ee person have children with free ear lobes?

11. Prove your answers by showing the results of these crosses in the Punnett squares below.

e

e

e

e

E

E

E

e

In this case, the second gene from person II1 cannot be predicted using Punnett squares. Either genotype Ee or EE may be correct. When this situation occurs, both genotypes are written under the symbol (Figure 4). Predicting the second gene for III1 results in her being heterozygous. Although her mother must provide her with one recessive gene, she has free lobes, so the second gene must be dominant (Figure 4).

Figure 4

At some time in the future, if II1 and II2 have many more children, one might be able to predict the father’s second gene. For example, if they have ten children and all show the dominant free lobes, one could safely conclude that he is EE. If, however, they have some children with attached ear lobes (ee), then he must be Ee.

E

e

E

e

E

E

E

E

E

E

e

E

12. Examine the pedigree on the right. Which Punnett square would best fit this family? Explain. B, Both parents must pass on a recessive gene.
a)	b)	c)

Analysis

1. Using the pedigrees below, predict the genotypes for these families.

a)

b)

c)

2. Examine the pedigree below.

a) How many generations are shown?

b) Identify by generation and number those persons with attached ear lobes.

c) Give the genotype for all persons have attached ear lobes.

3. Predict the genotypes for all persons in question 2.

Pedigree Studies

Pedigree Studies Page 1 of 6

Lesson 32 Climatogram Activity

Read the Climatogram Info to understand how they are useful, how to create them and how to interpret them.

1. Saskatoon is located at latitude 52.17° N. The table below presents average climate conditions in Saskatoon.

16.5

Month	Average Temperature (°C)	Average Precipitation (mm)
Jan	-17.8	18.2
Feb	-14.4	14.2
Mar	-7.6	16.9
Apr	3.6	20.1
May	11.0	38.3
Jun	15.7	62.5
Jul	18.6	57.4
Aug	17.3	42.8
Sep	11.2	35.0
Oct	4.9	19.4
Nov	-5.8		16.5
Dec	-14.2

a) Use the Climatographer to construct the climatogram.. Use the climatogram to answer the questions below.

b) Predict the biome in which Saskatoon is located using the map and biome descriptions in the textbook. Give reasons for your prediction.

c) Assume your prediction is correct. What kinds of plants and animals would you expect to find in natural areas around Saskatoon?

2. The figure below shows climatographs for two cities with similar latitudes: Manchester, United Kingdom, and Hamburg, Germany. Given that Hamburg is further inland than Manchester, deduce reasons for the differences and similarities in the climates of these two cities.

Manchester, UK, 53.35° N	Hamburg, Germany, 53.63° N

© NorthStar Academy BIOL200-32 Climatograph Activity Page 2 of 2

Climatograph

Average

Location – Saskatoon 52.17°N

Average

Month

Temp. (°C)

Ppt. (mm)

Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Sep

Oct

Nov

Dec

Ppt. (mm) Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Temp. (°C) Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Average Precipitation (mm)

Average Temperature (°C)

BIOL200-32Climatogram Info Page 1

Climatograms

Climatograms
In this exercise you will learn what climatograms are and how to interpret them. You
will also construct and interpret a number of climatograms.

Climatograms summarize monthly measurements of only two climatic factors:
temperature and precipitation. Other factors can also greatly affect climate but the
climatogram does give a reasonable indication of the climate for the location where
the data was gathered.

Since there is a distinct association between the climate and the biome found in any
one region, you can, through daily observation, become familiar with this association.
On a world-wide basis it would be quite time consuming to learn about this association
between climate and biome for all the biomes of the world. Consequently we must
rely on climatograms and descriptions of the biota to learn this association.

A climatogram is a graph that summarizes two variables, temperature and
precipitation. Precipitation is represented by a bar graph while temperature is
represented by a line graph. These two forms of graphing are used so that one
variable can easily be distinguished from the other. Look at the two examples:

BIOL200-32 Climatogram Info Page 2

Climatograms

Use the following procedures to prepare your own climatograms:
1. The monthly temperature averages are plotted on the climatogram from the

temperature data (in degrees Celsius) for each location. Use the scale shown on
the right hand side of the graph. The points representing each monthly average
should be at the mid-month position. Once you have plotted the points for all 12
months they should be joined by a line.

2. The monthly precipitation averages are plotted on the climatogram from the
precipitation data (in centimetres) for each location. Use the scale shown on the
left hand side of the graph to plot the point for each monthly average on the graph.
For each point, draw a line the width of the bar. Shade in all bars to make them
stand out.

3. Label your climatogram giving name of the location where the data was collected
and the biome.

BIOL200-32 Climatogram Info Page 3

Climatograms

Use the following data to interpret the climatograms you will prepare:
1. The hemisphere, northern or southern, in which an ecosystem is found can be

determined by interpreting the line on the climatogram that represents the
temperature. A line showing the highest temperatures in June and July and the
lowest temperatures in December and January would indicate a biome in the
northern hemisphere. If these temperatures were reversed for the stated times of
year the biome would be in the southern hemisphere.

2. The latitude of an ecosystem can also be determined by interpreting the
temperature line on a graph. Ecosystems that are equatorial show very little or
no seasonal fluctuation in temperature. Generally speaking, as you move farther
north or south of the equator, the difference between the summer high
temperature and the winter low temperature becomes more extreme. Also,
because the amount of solar radiation received as you move away from the
equator decreases, the average annual temperature decreases.

3. The biome can sometimes be determined by comparing temperature and
precipitation. Plants need a constant supply of water and temperatures above
freezing in order to be actively growing. Near the equator, average temperatures
are always high enough to maintain plant growth but water can act as a limiting
factor. High temperatures and low precipitation will lead to a water deficit due to
evaporation of water. When these conditions occur, plants will die or exist in a
dormant state. Thus at equatorial latitudes we can find biomes that range from
continuously wet tropical rain forests, to seasonally wet deciduous forests, to
fairly dry savanna, to very dry deserts.

At middle and northern latitudes, both precipitation and temperature can act as
limiting factors. High summer temperatures and low precipitation can lead to a
water deficit, thus affecting the kinds of plants and animals found. As well,
average minimum temperatures must be above freezing if plant growth is to occur.
Below this temperature, plants either die or become dormant. Consequently at
mid-latitudes we find middle latitude rain forests (coniferous), middle latitude
deciduous forests, taiga chaparral, grasslands and deserts.

4. Some of the factors that can influence the type of biome present in an area which
cannot be determined by reading a climatogram are:

a) proximity to large bodies of water (ocean, lakes)
b) direction of prevailing wind
c) altitude (mountains – complex zonation).