Fetal Pig Dissection Packet Answers

 

1. Do you have space with a sink? Pigs are a lot more involved than frogs and the preservatives will need to be drained and pigs rinsed. This is not a good dissection for classrooms that do not have sinks.

2. Have your students completed the frog dissection? The pig is more advanced, students should have a basic understanding of dissection protocols.

3. Pigs will need to be ordered from a biological supply company. If they are not injected, the circulartory system is very difficult to view. Generally, 1 pig for two students is a good match, but you could get away with 3-4 students per pig.

4. Safety: Goggles are required for all dissections. Latex gloves are optional, though generally preferred. Students should always wash hands even if they wore gloves. Many chemicals will seep through the latex. I have switched to nitrile gloves because it provides more of a barrier from harsh chemicals, but they are slightly more expensive.

5. Assessment. I take a grade on the completion of this lab guide. But as worksheets go, you do want the students to work out the answers together and ask for help when needed. Generally I use a quick and easy method to grade it. Each section is worth 5 pts. If its completed and looks mostly right, then they get the full 5 pts. Reduce pts if there are blanks or incorrect answers.

The biggest part of their grade comes from the LAB PRACTICAL. This is where pigs are set up at stations with numbered or colored tags in the structures. Students have 1 minute at each station to identify the structure and write it on their answer sheet. This is done in complete silence with no working together. Depending on the class, I may or may not allow them a word bank. Honors classes do not get a word bank usually unless I have an IEP or student that needs differentiation. The sheets below can be printed for the practical, they are numbered 1-50, though you don’t need to use all of the blanks. Just make sure your practical contains enough stations to keep students busy. If you have 30 students, you can have 25 stations with questions, and 5 “rest stations” interspersed.

Lab Practical Blanks | .doc file

Also print out the fetal pig lab guide – this just lists all of the structures they need to find with a checkbox. It makes for a good reference and study guide.

Fetal Pig Dissection: External Anatomy

External Anatomy

1. Determine the sex of your pig by looking for the urogenital opening. On females, this opening is located near the anus. On males, the opening is located near the umbilical cord. Check the bags and packaging, they are often labeled with the pig’s sex. Make sure you mix them up within the classroom.

If your pig is female, you should also note that urogenital papilla is present near the genital opening. Males do not have urogenital papilla.

Both males and females have rows of nipples, and the umbilical cord will be present in both.
What sex is your pig? _________

2. Make sure you are familiar with terms of reference: anterior, posterior, dorsal, ventral. In addition, you’ll need to know the following terms

Medial: toward the midline or middle of the body
Lateral: toward the outside of the body
Proximal: close to a point of reference
Distal: farther from a point of reference

*label the sides on the pig picture above. On the pig picture, they should just labe the anterior, posterior, dorsal, ventral.

3. Open the pig’s mouth and locate the hard and soft palate on the roof of the mouth. Can you feel your own hard and soft palates with your tongue?

Note the taste buds (also known as sensory papillae) on the side of the tongue. Locate the esophagus at the back of the mouth. Feel the edge of the mouth for teeth. Does the fetal pig have teeth? _yes
Are humans born with teeth? ___ no _

Locate the epiglottis, a cone-shaped structure at the back of the mouth, a flap of skin helps to close this opening when a pig swallows. The pharynx is the cavity in the back of the mouth – it is the junction for food (esophagus) and air (trachea). To find the epiglottis, you will need to make deep cuts at the edges of the mouth, I also place a lot of pressure on the jaw to break it and to get the mouth to fully open. Students will often be too gentle opening the mouth.

4. Gestation for the fetal pig is 112-115 days. The length of the fetal pig can give you a rough estimate of its age.
11mm – 21 days | 17 mm – 35 days | 2.8 cm – 49 days
4 cm – 56 days | 22 cm – 100 days | 30 cm — birth

5. Observe the toes of the pig. How many toes are on the feet? _________________
Do they have an odd or even number of toes? ______odd toed – artiodactyls_________________

6. Observe the eyes of the pig, carefully remove the eyelid so that you can view the eye underneath. Does it seem well developed? Do you think pigs are born with their eyes open or shut? _____________eyes developed, they usually open their eyes within first day__________________

7. Carefully lay the pig on one side in your dissecting pan and cut away the skin from the side of the face and upper neck to expose the masseter muscle that works the jaw, lymph nodes, and salivary glands. The salivary glands kind of look like chewing gum, and are often lost if you cut too deeply. Salivary glands are usually in the same spot, near the cheek and jaw. Lymph nodes can be in different spots and be difficult to locate.

**Make sure you know the locations of all the bold words on this handout**

The Anatomy of the Fetal Pig (internal)

pig diagramIdentify the structures on the diagram.

1. _____esophagus___

2. _____liver_______________

3. _____gall bladder_________

4. ____bile duct___________

5. ____stomach____________

6. ____duodenum_____________

7. ____pancreas______________

8. ____small intestine_______

9. ____spleen_____________

10. ___cecum_____________

11. ___large intestine__________

12. ___rectum_____________

13. ___umbilical arteries__________

Identify the organ (or structure)

14. _____pyloric sphincter valve____ Opening (valve) between stomach and small intestine.
15. _____gall bladder_____________ Stores bile, lies underneath the liver.
16. ___________cecum___________ A branch of the large intestine, a dead end.
17. ___________diaphragm________ Separates the thoracic and abdominal cavity; aids breathing.
18. ________mesentery___________ Membrane that holds the coils of the small intestine.
19. ________duodenum___________ The straight part of the small intestine just after the stomach.
20. _______bile duct______________ Empties bile into the duodenum from the gall bladder.
21. _______rectum_______________ The last stretch of the large intestine before it exits at the anus.
22. _______pancreas_____________ Bumpy structure under the stomach; makes insulin
23. _______bladder_______________ Lies between the two umbilical vessels.

Urinary and Reproductive Systems

1. Locate the kidneys; the tubes leading from the kidneys that carry urine are the ureters. The ureters carry urine to the urinary bladder – located between the umbilical vessels. To find the ureters, expose the kidney and wiggle it, the ureter is attached and you’ll see it move.
2. Lift the bladder to locate the urethra, the tube that carries urine out of the body. checkbox
3. Note the vessels that attach to the kidney – these are the renal vessels checkbox

Male
1. Find the scrotal sacs at the posterior end of the pig (between the legs), testis are located in each sac. Open the scrotal sac to locate the testis. checkbox
2. On each teste, find the coiled epididymis. Sperm cells produces in the teste pass through the epididymis and into a tube called the vas deferens (in humans, a vasectomy involves cutting this tube).checkbox
3. The penis can be located by cutting away the skin on the flap near the umbilical cord. This tube-like structure eventually exits out the urogenital opening, also known as the urethra. checkbox The penis of the fetal pig is actually pretty difficult to find because it is internal (this can lead for much hilarity in the lab as students try to locate the structure. A simple technique I use to find it is to find the area just behind the urethral opening and roll this area (its also where the umbilical arteries are attached) between the thumb and forefinger. You should feel a solid tube like structure just under the skin – this is the penis.
Female
4. In the female pig, locate two bean shaped ovaries located just posterior to the kidneys and connected to the curly oviducts. checkbox
5. Trace the oviducts toward the posterior to find that they merge at the uterus. Trace the uterus to the vagina. The vagina will actually will appear as a continuation of the uterus. checkbox

LABEL THE DIAGRAMS
urinary system femalemale

Dissection of the Thoracic Cavity

Identify by number:
Aorta __2__ Dorsal Aorta _9__Pulmonary Trunk _1_
Common carotid _4__ Left & Right Carotid _7,8__
Coronary vessels _6__ Left Subclavian__5__
Right Subclavian __10__ Right Brachiocephalic _3___
Right Atrium __12__ Left Atrium _13__
Intercostal __11___ Ventricle __14_
Identify the structure.

1. _______pericardium________ Membrane over the heart.
2. _____trachea________ Airway from mouth to lungs
3. ____carotids_______ Blood supply to head
4. ____ventricles_________ Lower heart chambers
5. ____dorsal aorta_______ Blood supply to lower body
6. ____diaphragm_______ Muscle to aid breathing
7. ____vena cava_______ Returns blood to heart
8. ____aorta (or pulmonary)____ Large vessel at top of heart
9. ____larynx______ Used to make noises
10. ___coronary______ Arteries on heart surface.

Fetal Pig – Dissection of the Lower Arteries

I often do this part as an “optional section”. Some students will work very fast and will need something to do while others catch up. I have also offered extra credit to students who can expose these arteries to view (cleanly), which gives them extra incentive to work on it. The problem is, if you are spending time with groups that are farther behind, then you don’t have a lot of time to help students with the arteries. Giving them extra credit encourages them to try, but also requires them to work on their own.

1. Trace the abominal aorta (also called the dorsal aorta) to the lower part of the body, careful tweezing of the tissue will reveal several places where it branches, though some of the arteries may have been cut when you removed organs of the digestive system. checkbox
2. The hepatic artery leads to the liver. (may not be visible) checkbox
3. The splenic artery leads to the spleen (may not be visible) checkbox
4. The renal arteries lead to the kidney. checkbox
5. The mesenteric artery leads to the mesentery and branches into many smaller vessels. Look in the small intestine to find this artery. checkbox
6. Trace the abominal aorta and note where it joins the umbilical arteries. You will need to cut the muscle in the leg to trace the next vessels. Use a pin to carefully tease away the surrounding muscle and tissue. checkbox
7. The abominal aorta splits into two large vessels that lead to each leg – the external iliac arteries will turn into the femoral arteries as they enter the leg checkbox
8. Follow the umbilical artery toward the pig, you’ll find that it branches and a small artery stretches toward the posterior of the pig – this is the ilio-lumbar artery. checkbox
9. Follow the external iliac into the leg (carefully tease away muscle),it will branch into two arteries: the femoral (toward the outside of the leg) and the deep femoral (toward the back of the leg) checkbox

lower arteries

flow chart
Publisher: Biologycorner.com;
Creative Commons LicenseThis work is licensed under a Creative Commons Attribution-NonCommercial 3.0 Unported License.

Mastering Biology Chapter 16

Activity: DNA and RNA Structure

In the accompanying image, a nucleotide is indicated by the letter _____.

screen-shot-2017-02-17-at-9-22-35-am

Which of these is a difference between a DNA and an RNA molecule?

screen-shot-2017-02-17-at-9-23-51-am

This is an image of a(n) _____.

The figure shows a chemical compound, which consists of one phosphate group, ribose, and nitrogenous base.

screen-shot-2017-02-17-at-9-24-36-am

The letter A indicates a _____.

The figure shows a chemical compound, which consists of three parts. The first, marked A, contains a phosphorus atom with four oxygen atoms. Two oxygen atoms have negative charge. The second, marked B, represented by the five-membered ring, of which one carbon is replaced with an oxygen atom. Letter C marks the six-membered ring, which contains nitrogen atoms.

screen-shot-2017-02-17-at-9-25-40-am

A nitrogenous base is indicated by the letter _____.

The figure shows a chemical compound, which consists of three parts. The first, marked A, contains a phosphorus atom with four oxygen atoms. Two oxygen atoms have negative charge. The second, marked B, represented by the five-membered ring, of which one carbon is replaced with an oxygen atom. Letter C marks the six-membered ring, which contains nitrogen atoms.

screen-shot-2017-02-17-at-9-26-24-am

You can tell that this is an image of a DNA nucleotide and not an RNA nucleotide because you see a _____.

The figure shows a chemical compound, which consists of three parts. The first, marked A, contains a phosphorus atom with four oxygen atoms. Two oxygen atoms have negative charge. Letter B marks a pentagon with an O at its top vertex and a CH2OH attached above its left vertex. One can numerate carbon atoms clockwise starting from next to O. Then OH group is attached below to third carbon atom. The hydroxyl of the fifth carbon connects with the part A. The second carbon atom connects with the part C. Letter C marks the six-membered ring, in which the second and the fourth carbons are replaced with N atoms with –NH2 group attached to the first carbon, with an O atom attached to the third carbon by a double bond, and with double bonds between the first and the second, the fifth and the sixth atoms in the ring.

screen-shot-2017-02-17-at-9-27-19-am

Which of these nitrogenous bases is found in DNA but not in RNA?

screen-shot-2017-02-17-at-9-28-03-am

Which of these is(are) pyrimidines?

The figure shows two types of nitrogen bases. The first type contains two bases: guanine, marked A, and adenine, marked B. The second contains three other: C, D, E. The bases A and B consist of a six-membered ring attached to  five-membered ring. The bases C, D, and E have only six-membered ring in their structure.

screen-shot-2017-02-17-at-9-28-42-am

In a nucleotide, the nitrogenous base is attached to the sugar’s _____ carbon and the phosphate group is attached to the sugar’s _____ carbon.

screen-shot-2017-02-17-at-9-29-16-am

Nucleic acids are assembled in the _____ direction.

screen-shot-2017-02-17-at-9-29-57-am

In a DNA double helix an adenine of one strand always pairs with a(n) _____ of the complementary strand, and a guanine of one strand always pairs with a(n) _____ of the complementary strand.

screen-shot-2017-02-17-at-2-00-11-pm


Activity: The Hershey-Chase Experiment

This is an image of a _____.

The figure shows a particle, which consists of a hexagonal head, collar, tail and several tail fibers at the end.

screen-shot-2017-02-17-at-2-01-30-pm

Who demonstrated that DNA is the genetic material of the T2 phage?

screen-shot-2017-02-17-at-2-02-07-pm

The radioactive isotope 32P labels the T2 phage’s _____.

screen-shot-2017-02-17-at-2-02-40-pm

Hershey and Chase used _____ to radioactively label the T2 phage’s proteins.

screen-shot-2017-02-17-at-2-03-20-pm

After allowing phages grown with bacteria in a medium that contained 32P and 35S, Hershey and Chase used a centrifuge to separate the phage ghosts from the infected cell. They then examined the infected cells and found that they contained _____, which demonstrated that _____ is the phage’s genetic material.

screen-shot-2017-02-17-at-2-03-46-pm


DNA Replication (1 of 2): DNA Structure and Replication Machinery (BioFlix tutorial)

Part A – The chemical structure of DNA and its nucleotides

The DNA double helix is composed of two strands of DNA; each strand is a polymer of DNA nucleotides. Each nucleotide consists of a sugar, a phosphate group, and one of four nitrogenous bases. The structure and orientation of the two strands are important to understanding DNA replication.
Drag the labels to their appropriate locations on the diagram below. Pink labels can be used more than once.
screen-shot-2017-02-17-at-2-05-17-pm

Part B – The role of DNA polymerase III

In DNA replication in bacteria, the enzyme DNA polymerase III (abbreviated DNA pol III) adds nucleotides to a template strand of DNA. But DNA pol III cannot start a new strand from scratch. Instead, a primer must pair with the template strand, and DNA pol III then adds nucleotides to the primer, complementary to the template strand. Each of the four images below shows a strand of template DNA (dark blue) with an RNA primer (red) to which DNA pol III will add nucleotides.
In which image will adenine (A) be the next nucleotide to be added to the primer?
 screen-shot-2017-02-17-at-2-06-28-pm

Part C – The replication bubble and antiparallel elongation

DNA replication always begins at an origin of replication. In bacteria, there is a single origin of replication on the circular chromosome, as shown in the image here. Beginning at the origin of replication, the two parental strands (dark blue) separate, forming a replication bubble. At each end of the replication bubble is a replication fork where the parental strands are unwound and new daughter strands (light blue) are synthesized. Movement of the replication forks away from the origin expands the replication bubble until two identical chromosomes are ultimately produced.

Diagram showing DNA replication in a circular chromosome. For simplicity, the double-stranded DNA is shown as two concentric circles. There is one origin of replication, where the two parental strands (shown in dark blue) separate, forming a replication bubble. At each end of the replication bubble is a replication fork (indicated by a pink arrow) where the parental strands are unwound and new daughter strands (shown in light blue) are synthesized. The replication forks move away from the origin and expand the replication bubble. As the light blue strands elongate, the two double-stranded circles that are forming start to peel off from each other. The end result is two separate, identical daughter DNA molecules, each composed of one parental strand (dark blue) and one new strand (light blue).

In this activity, you will demonstrate your understanding of antiparallel elongation at the replication forks. Keep in mind that the two strands in a double helix are oriented in opposite directions, that is, they are antiparallel.

Drag the arrows onto the diagram below to indicate the direction that DNA polymerase III moves along the parental (template) DNA strands at each of the two replication forks. Arrows can be used once, more than once, or not at all.
screen-shot-2017-02-17-at-2-07-40-pm

Part D – Unwinding the DNA

As DNA replication continues and the replication bubble expands, the parental double helix is unwound and separated into its two component strands. This unwinding and separating of the DNA requires three different types of proteins: helicase, topoisomerase, and single-strand binding proteins.
Sort the phrases into the appropriate bins depending on which protein they describe.
screen-shot-2017-02-17-at-2-08-51-pm

DNA Replication (2 of 2): Synthesis of the Leading and Lagging Strands (BioFlix tutorial)

Part A – Comparing the leading and lagging strands

As the two parental (template) DNA strands separate at a replication fork, each of the strands is separately copied by a DNA polymerase III (orange), producing two new daughter strands (light blue), each complementary to its respective parental strand. Because the two parental strands are antiparallel, the two new strands (the leading and lagging strands) cannot be synthesized in the same way.

Diagram showing a replication bubble with replication forks on the left and the right of the origin of replication. The top parental DNA strand goes 5' to 3' and the bottom parental DNA strand goes 3' to 5'. At each replication fork, two DNA pol IIIs are synthesizing two new daughter strands. At the left replication fork, the top strand is the leading strand and the bottom strand is the lagging strand. At the right replication fork, the top strand is the lagging strand and the bottom strand is the leading strand. Replication proceeds to the left and to the right.

Drag each phrase to the appropriate bin depending on whether it describes the synthesis of the leading strand, the synthesis of the lagging strand, or the synthesis of both strands.
screen-shot-2017-02-17-at-2-10-32-pm

Part B – RNA primers on the leading and lagging strands

The diagram below shows a replication bubble with synthesis of the leading and lagging strands on both sides of the bubble. The parental DNA is shown in dark blue, the newly synthesized DNA is light blue, and the RNA primers associated with each strand are red. The origin of replication is indicated by the black dots on the parental strands.

Diagram showing a replication bubble. The top parental DNA strand goes 5' to 3' and the bottom parental DNA strand goes 3' to 5'. Four segments of new DNA are being synthesized inside the bubble--two as leading strands and two as lagging strands. One leading strand is in the top left, and its single primer is labeled (a) near the origin of replication. The other leading strand is in the bottom right, and its single primer is labeled (h) near the origin of replication. One lagging strand is in the top right, and its three primers are labeled left to right (b), (c), and (d). The other lagging strand is in the bottom left, and its three primers are labeled left to right (e), (f), and (g).

Rank the primers in the order they were produced. If two primers were produced at the same time, overlap them.
screen-shot-2017-02-17-at-2-11-59-pm

Part C – Synthesis of the lagging strand

In contrast to the leading strand, the lagging strand is synthesized as a series of segments called Okazaki fragments. The diagram below illustrates a lagging strand with the replication fork off-screen to the right. Fragment A is the most recently synthesized Okazaki fragment. Fragment B will be synthesized next in the space between primers A and B.

Diagram showing a lagging strand being synthesized on a dark blue parental strand running 5' to 3'. The replication fork is off to the right. Fragment A (light blue) is on the far left and is the most recently synthesized Okazaki fragment. Primer A (red) sets just to the right of fragment A. Then there is a space for fragment B. Then there is primer B (red) at the right end of the diagram, before the replication fork.

Drag the labels to their appropriate locations in the flowchart below, indicating the sequence of events in the production of fragment B. (Note that pol I stands for DNA polymerase I, and pol III stands for DNA polymerase III.)
screen-shot-2017-02-17-at-2-13-32-pm

Activity: DNA Packing

Part A

The letter A indicates _____.

The figure shows the complex of globules containing eight subunits. Around each of these globules filamentary structure forms about two turns. Letter A marks the filamentary structure. Letter B marks one of the globules. Letter C marks one subunit of the globule.

screen-shot-2017-02-17-at-2-14-30-pm

Part B

Where would RNA polymerase attach?

The figure shows the complex of globules containing eight subunits. Around each of these globules filamentary structure forms about two turns. Letter A marks the filamentary structure. Letter B marks one of the globules. Letter C marks one subunit of the globule.

screen-shot-2017-02-18-at-5-20-43-pm

Part C

The letter C indicates _____.

The figure shows the complex of globules containing eight subunits. Around each of these globules filamentary structure forms about two turns. Letter A marks the filamentary structure. Letter B marks one of the globules. Letter C marks one subunit of the globule.

screen-shot-2017-02-18-at-5-21-15-pm

Part D

What is this an image of?

The figure shows a cylinder, which is formed by spirally stranded fiber. The cylinder is hollow inside.

screen-shot-2017-02-18-at-5-21-49-pm

Part E

What is this an image of?

The figure shows a thread. It is undulating. From this undulating thread, larger waves are formed.

screen-shot-2017-02-18-at-5-22-42-pm


Scientific Skills Exercise: Working With Data in a Table

Given the percentage composition of one nucleotide in a genome, can we predict the percentages of the other three nucleotides?

Even before the structure of DNA was elucidated, Erwin Chargaff and his coworkers noticed a pattern in the base composition of nucleotides from different organisms: the number of adenine (A) bases roughly equaled the number of thymine (T) bases, and the number of cytosine (C) bases roughly equaled the number of guanine (G) bases. Further, each species they studied had a different balance of A/T and C/G bases. We now know that these consistent ratios are due to complementary base pairing between A and T and between C and G in the DNA double helix, and interspecies differences are due to the unique sequences of bases along a DNA strand. In this exercise, you will apply Chargaff’s rules to predict the composition of nucleotide bases in a genome.

In Chargaff’s experiments, DNA was extracted from the given organism, denatured, and hydrolyzed to break apart the individual nucleotides before analyzing them chemically. These experiments provided approximate values for each type of nucleotide. Today, the availability of whole-genome sequencing has allowed base composition analysis to be done more precisely directly from the sequence data.

Part A – Analyzing the data

Tables like the one shown here are useful for organizing sets of data representing a common set of values (in this case, percentages of A, G, C, and T) for a number of different samples (in this case, species).

Source of DNA Adenine Guanine Cytosine Thymine
Sea urchin 32.8% 17.7% 17.3% 32.1%
Salmon 29.7 20.8 20.4 29.1
Data from several papers by Chargaff: for example, E. Chargaff et al., Composition
of the desoxypentose nucleic acids of four genera of sea-urchin, Journal of Biological
Chemistry
195: 155-160 (1952).
Does the distribution of bases in sea urchin DNA and salmon DNA follow Chargaff’s rules?
screen-shot-2017-02-18-at-5-24-03-pm

Part B – Calculating missing data

You can use Chargaff’s rules to predict the percentage of one or more bases in the DNA of a species if at least one value is known.

Source of DNA Adenine Guanine Cytosine Thymine
Sea urchin 32.8% 17.7% 17.3% 32.1%
Salmon 29.7 20.8 20.4 29.1
Wheat 28.1 21.8 22.7
Data from several papers by Chargaff: for example, E. Chargaff et al., Composition
of the desoxypentose nucleic acids of four genera of sea-urchin, Journal of Biological
Chemistry
195: 155-160 (1952).
What is the %T in wheat DNA?
screen-shot-2017-02-18-at-5-24-41-pm

Part C

Use Chargaff’s rules to predict the missing values for E. coli, human, and ox DNA. Round your answers to the nearest whole number.
screen-shot-2017-02-18-at-5-28-41-pm

Part D – Evaluating a hypothesis

If Chargaff’s equivalence rule is valid, then hypothetically we could extrapolate this to the combined genomes of all species on Earth (as if there were one huge Earth genome). In other words, the total amount of A in every genome on Earth should equal the total amount of T in every genome on Earth. Likewise, the total amount of G in every genome on Earth should equal the total amount of C in every genome on Earth.
Calculate the average percentage for each base in your completed table. Do Chargaff’s equivalence rules still hold true when you consider those six species together?
screen-shot-2017-02-20-at-6-07-55-pm

Misconception Question 79

Part A

Meselson and Stahl cultured E. coli for several generations in a medium with a heavy isotope of nitrogen, 15N. They transferred the bacteria to a medium with a light isotope of nitrogen, 14 N. After two rounds of DNA replication, half the DNA molecules were light (both strands had 14N) and half were hybrids (15N-14N). What did the researchers conclude from these results?
screen-shot-2017-02-20-at-4-41-12-pm

Misconception Question 78

Part A

The figure shows a diagram of a nucleoside triphosphate. It consists of three phosphate groups joined successively to five-membered sugar, which is joined to a nitrogen base labeled T.Nucleotides are added to a growing DNA strand as nucleoside triphosphates. What is the significance of this fact?
screen-shot-2017-02-20-at-4-42-16-pm

Misconception Question 80

Part A

Select the most accurate statement describing DNA replication complexes.
screen-shot-2017-02-20-at-4-42-53-pm

Misconception Question 77

Part A

During DNA replication, the leading strand is synthesized continuously, whereas the lagging strand is synthesized as Okazaki fragments. Why is this so?
screen-shot-2017-02-20-at-4-43-18-pm

Misconception Question 76

Part A

DNA is a self-replicating molecule. What accounts for this important property of DNA?
screen-shot-2017-02-20-at-4-43-43-pm

DNA Replication

When cells divide or multiply, the DNA must be copied, or replicated. All known organisms replicate their DNA by a method known as semi-conservative replication, in which each of the two strands of the parent DNA becomes part of the daughter DNA.

Part A

Complete the following vocabulary exercise related to DNA replication.
Match the words in the left-hand column with the appropriate blank in the sentences in the right-hand column.
screen-shot-2017-02-20-at-4-49-13-pm

AP Exam Prep Question 30

Part A

In E. coli replication the enzyme primase is used to attach a 5 to 10 base ribonucleotide strand complementary to the parental DNA strand. The RNA strand serves as a starting point for the DNA polymerase that replicates the DNA. If a mutation occurred in the primase gene, which of the following would you expect?screen-shot-2017-02-20-at-4-50-39-pm


AP Exam Prep Question 31

Part A

Hershey and Chase used a DNA-based virus for their work. What would the results have been if they had used an RNA virus?

screen-shot-2017-02-20-at-4-51-22-pm


AP Exam Prep Question 32

Part A

The lagging strand is characterized by a series of short segments of DNA (Okazaki fragments) that will be joined together to form a finished lagging strand. The experiments that led to the discovery of Okazaki fragments gave evidence for which of the following ideas?
screen-shot-2017-02-20-at-4-52-38-pm

Chapter 16 Question 1

Part A

In his work with pneumonia-causing bacteria and mice, Griffith found that
screen-shot-2017-02-20-at-4-53-09-pm

Chapter 16 Question 2

Part A

What is the basis for the difference in how the leading and lagging strands of DNA molecules are synthesized?
screen-shot-2017-02-20-at-4-53-34-pm

Chapter 16 Question 3

Part A

In analyzing the number of different bases in a DNA sample, which result would be consistent with the base-pairing rules?
screen-shot-2017-02-20-at-4-54-03-pm

Chapter 16 Question 4

Part A

The elongation of the leading strand during DNA synthesis
 screen-shot-2016-02-20-at-4-54-32-pm

Chapter 16 Question 5

Part A

In a nucleosome, the DNA is wrapped around
screen-shot-2017-02-20-at-4-55-51-pm

Chapter 16 Question 6

Part A

E. coli cells grown on 15N medium are transferred to 14N medium and allowed to grow for two more generations (two rounds of DNA replication). DNA extracted from these cells is centrifuged. What density distribution of DNA would you expect in this experiment?
screen-shot-2017-02-20-at-4-56-27-pm

Chapter 16 Question 7

Part A

A biochemist isolates, purifies, and combines in a test tube a variety of molecules needed for DNA replication. When she adds some DNA to the mixture, replication occurs, but each DNA molecule consists of a normal strand paired with numerous segments of DNA a few hundred nucleotides long. What has she probably left out of the mixture?
screen-shot-2017-02-20-at-4-56-54-pm

Chapter 16 Question 8

Part A

The spontaneous loss of amino groups from adenine in DNA results in hypoxanthine, an uncommon base, opposite thymine. What combination of proteins could repair such damage?
screen-shot-2017-02-20-at-4-57-22-pm

Activity: DNA Double Helix

Part A

In a DNA double helix an adenine of one strand always pairs with a(n) _____ of the complementary strand, and a guanine of one strand always pairs with a(n) _____ of the complementary strand.
screen-shot-2017-02-20-at-4-58-18-pm

Chapter 16 Pre-Test Question 3

Part A

Who conducted the X-ray diffraction studies that were key to the discovery of the structure of DNA?
screen-shot-2017-02-20-at-4-58-40-pm

Chapter 16 Pre-Test Question 2

Part A

In the Hershey and Chase experiment that helped confirm that DNA, not protein, was the hereditary material, what was the key finding?
screen-shot-2017-02-20-at-4-59-33-pm

Chapter 16 Pre-Test Question 1

Part A

Griffith’s experiments with S. pneumoniae were significant because they showed that traits could be transferred from one organism to another. What else did he find that was significant?
screen-shot-2017-02-20-at-5-00-23-pm

Activity: DNA Replication: A Review

Part A

Short segments of newly synthesized DNA are joined into a continuous strand by _____.
 screen-shot-2017-02-20-at-5-00-56-pm

Part B

After DNA replication is completed, _____.
 screen-shot-2017-02-20-at-5-01-18-pm

Part C

The first step in the replication of DNA is catalyzed by _____.
 screen-shot-2017-02-20-at-5-01-45-pm

Part D

The action of helicase creates _____.
screen-shot-2017-02-20-at-5-02-05-pm

Part E

Why is the new DNA strand complementary to the 3′ to 5′ strands assembled in short segments?
screen-shot-2017-02-20-at-5-02-36-pm

Part F

The synthesis of a new strand begins with the synthesis of a(n) _____.
screen-shot-2017-02-20-at-5-03-22-pm

Part G

Which of these is responsible for catalyzing the formation of an RNA primer?

The figure shows the replication of the DNA. Each letter marks the definite structure. Letter A marks the part of a double-stranded DNA. Letter B marks the enzyme, which cuts the bonds between two strands of DNA. Letter C marks the structures, which prevent the linking of the DNA strands. Letter D marks the substance, which binds to DNA strand and initiates the synthesis. Letter E marks the component of the DNA strand.

screen-shot-2017-02-20-at-5-54-18-pm

Part H

An old DNA strand is used as a _____ for the assembly of a new DNA strand.
screen-shot-2017-02-20-at-5-54-50-pm

Activity: DNA Replication: A Closer Look

Part A

In a DNA double helix an adenine of one strand always pairs with a(n) _____ of the complementary strand, and a guanine of one strand always pairs with a(n) _____ of the complementary strand.
screen-shot-2017-02-20-at-9-11-26-pm

Part B

After DNA replication is completed, _____.
screen-shot-2017-02-20-at-9-11-47-pm

Part C

The first step in the replication of DNA is catalyzed by _____.
screen-shot-2017-02-20-at-9-12-08-pm

Part D

The action of helicase creates _____.
screen-shot-2017-02-20-at-9-12-36-pm

Part E

Why is the new DNA strand complementary to the 3′ to 5′ strands assembled in short segments?
screen-shot-2017-02-20-at-9-12-54-pm

Part F

The synthesis of a new strand begins with the synthesis of a(n) _____.
screen-shot-2017-02-20-at-9-13-15-pm

Part G

An old DNA strand is used as a _____ for the assembly of a new DNA strand.
screen-shot-2017-02-20-at-9-13-51-pm

Activity: DNA Replication: An Overview

Part A

Short segments of newly synthesized DNA are joined into a continuous strand by _____.
 screen-shot-2017-02-20-at-9-14-19-pm

Part B

After DNA replication is completed, _____.
screen-shot-2017-02-20-at-9-14-45-pm

Part C

The action of helicase creates _____.
screen-shot-2017-02-20-at-9-15-07-pm

Part D

Why is the new DNA strand complementary to the 3′ to 5′ strands assembled in short segments?
screen-shot-2017-02-20-at-9-15-27-pm

Part E

An old DNA strand is used as a _____ for the assembly of a new DNA strand.
screen-shot-2017-02-20-at-9-15-44-pm

Activity: DNA Synthesis

Part A

What catalyzes DNA synthesis?
screen-shot-2017-02-20-at-9-16-21-pm

Part B

Which of the following statements about DNA synthesis is true?
screen-shot-2017-02-20-at-9-16-39-pm

Part C

Which part of a deoxynucleoside triphosphate (dNTP) molecule provides the energy for DNA synthesis?
screen-shot-2017-02-20-at-9-17-17-pm

Part D

Which of the following enzymes creates a primer for DNA polymerase?
screen-shot-2017-02-20-at-9-17-52-pm

Part E

Which of the following statements about Okazaki fragments in E. coli is true?
screen-shot-2017-02-20-at-9-18-08-pm

Part F

Which of the following enzymes is important for relieving the tension in a helix as it unwinds during DNA synthesis?
screen-shot-2017-02-20-at-9-18-27-pm

Part G

True or false? Single-stranded DNA molecules are said to be antiparallel when they are lined up next to each other but oriented in opposite directions.
screen-shot-2017-02-20-at-9-18-45-pm

DNA Replication: Mechanism and Proteins

DNA replication is the process by which DNA is copied. It is highly accurate in both bacteria and eukaryotes and requires a variety of DNA polymerases and other accessory proteins. In this tutorial you will learn how DNA is replicated and understand the roles of the proteins involved in the process.

Part A – The mechanism of DNA replication

The diagram below shows a double-stranded DNA molecule (parental DNA).
Drag the correct labels to the appropriate locations in the diagram to show the composition of the daughter DNA molecules after one and two cycles of DNA replication. In the labels, the original parental DNA is blue and the DNA synthesized during replication is red.
screen-shot-2017-02-20-at-9-22-02-pm

Part B – Processes occurring at a bacterial replication fork

The diagram below shows a bacterial replication fork and its principal proteins.
Drag the labels to their appropriate locations in the diagram to describe the name or function of each structure. Use pink labels for the pink targets and blue labels for the blue targets.
screen-shot-2017-02-20-at-9-22-29-pm

Chapter 16 Pre-Test Question 9

Part A

What are chromosomes made of?
screen-shot-2017-02-20-at-9-29-15-pm

Chapter 16 Pre-Test Question 10

Part A

Which of the following is true of DNA during interphase?
screen-shot-2017-02-20-at-9-29-34-pm

Mastering Biology Chapter 15

Chapter 15 Pre-Test Question 9

How are human mitochondria inherited?

Screen Shot 2017-02-01 at 11.13.41 AM.png


Chapter 15 Pre-Test Question 2

What name is given to the most common phenotype in a natural population?
 Screen Shot 2017-02-01 at 1.24.49 PM.png

Chapter 15 Pre-Test Question 3

A white-eyed female Drosophila is crossed with a red-eyed male Drosophila. Which statement below correctly describes the results?
Screen Shot 2017-02-01 at 1.25.58 PM.png

Chapter 15 Question 1

When Thomas Hunt Morgan crossed his red-eyed F1 generation flies to each other, the F2 generation included both red- and white-eyed flies. Remarkably, all the white-eyed flies were male. What was the explanation for this result?

Screen Shot 2017-02-01 at 1.26.40 PM.png


Chapter 15 Pre-Test Question 4

 In humans, what determines the sex of offspring and why?
screen-shot-2017-02-01-at-1-29-19-pm

Chapter 15 Pre-Test Question 10

Which of the following is true of an X-linked gene, but not of a Y-linked gene?
screen-shot-2017-02-01-at-1-30-50-pm

Chapter 15 Question 3

Males are more often affected by sex-linked traits than females because _____.
screen-shot-2017-02-01-at-1-31-47-pm

Chapter 15 Question 25

What does a frequency of recombination of 50% indicate?

Screen Shot 2017-02-01 at 1.21.27 PM.png


Chapter 15 Question 26

What is the definition of one map unit?
 screen-shot-2017-02-01-at-1-22-55-pm

Chapter 15 Pre-Test Question 5

In general, the frequency with which crossing over occurs between two linked genes depends on what?

Screen Shot 2017-02-01 at 1.34.09 PM.png


Chapter 15 Pre-Test Question 7

What results if a fragment of a chromosome breaks off and then reattaches to the original chromosome at the same place but in the reverse direction?
 Screen Shot 2017-02-01 at 1.35.29 PM.png

Chapter 15 Question 40

If cell X enters meiosis, and nondisjunction of one chromosome occurs in one of its daughter cells during meiosis II, what will be the result at the completion of meiosis?

Screen Shot 2017-02-01 at 1.37.28 PM.png

Chapter 15 Pre-Test Question 8

What phenomenon occurs when a particular allele will either be expressed or silenced, depending on whether it is inherited from a male or a female?

Screen Shot 2017-02-01 at 1.38.00 PM.png


AP Exam Prep Question 29

 Inheritance patterns cannot always be explained by Mendel’s models of inheritance. If a pair of homologous chromosomes fails to separate during meiosis I, select the choice that shows the chromosome number of the four resulting gametes with respect to the normal haploid number (n)?Screen Shot 2017-02-01 at 1.39.58 PM.png

Misconception Question 72

Which of these descriptions of the behavior of chromosomes during meiosis explains Mendel’s law of independent assortment?

Screen Shot 2017-02-01 at 1.42.10 PM.png


Chapter 15 Question 43

Of the following human aneuploidies, which is the one that generally has the most severe impact on the health of the individual?

Screen Shot 2017-02-01 at 1.44.05 PM.png


Chapter 15 Question 23

Which of the following statements is true of linkage?

Screen Shot 2017-02-01 at 1.48.35 PM.png


Chapter 15 Pre-Test Question 6

Which of the following results in a situation in which the chromosome number is either 2n+1 or 2n-1 ?

Screen Shot 2017-02-01 at 1.49.39 PM.png


Chapter 15 Question 41

One possible result of chromosomal breakage is for a fragment to join a nonhomologous chromosome. What is this alteration called?

Screen Shot 2017-02-01 at 8.15.50 PM.png


Chapter 15 Question 42

A nonreciprocal crossover causes which of the following products?

screen-shot-2017-02-01-at-8-17-00-pm


Chapter 15 Question 27

Recombination between linked genes comes about for what reason?

screen-shot-2017-02-01-at-8-18-07-pm


Chapter 15 Question 24

How would one explain a testcross involving F1 dihybrid flies in which more parental-type offspring than recombinant-type offspring are produced?

Screen Shot 2017-02-01 at 8.19.50 PM.png


Chapter 15 Question 31

In a series of mapping experiments, the recombination frequencies for four different linked genes of Drosophila were determined as shown in the figure above. What is the order of these genes on a chromosome map?

screen-shot-2017-02-01-at-8-23-39-pm


Chapter 15 Question 32

Use the following information to answer the question(s) below.

A plantlike organism on the planet Pandora can have three recessive genetic traits: bluish leaves, due to an allele (a) of gene A; a feathered stem, due to an allele (b) of gene B; and hollow roots due to an allele (c) of gene C. The three genes are linked and recombine as follows:

A geneticist did a testcross with an organism that had been found to be heterozygous for the three recessive traits and she was able to identify progeny of the following phenotypic distribution (+ = wild type):

Phenotypes Leaves Stems Roots Number
1 a + + 14
2 a + c 0
3 a b + 32
4 a b c 440
5 + b + 0
6 + b c 16
7 + + c 28
8 + + + 470
Total 1000

Which of the following are the phenotypes of the parents in this cross?

screen-shot-2017-02-01-at-8-26-41-pm


Sex Linkage

The inheritance of a skin condition in humans

Consider the following family history:

  • Bob has a genetic condition that affects his skin.
  • Bob’s wife, Eleanor, has normal skin. No one in Eleanor’s family has ever had the skin condition.
  • Bob and Eleanor have a large family. Of their eleven children, all six of their sons have normal skin, but all five of their daughters have the same skin condition as Bob.

Based on Bob and Eleanor’s family history, what inheritance pattern does the skin condition most likely follow?

screen-shot-2017-02-01-at-8-28-01-pm

Part B – A sex-linked gene for eye color in Drosophila

The inheritance of eye color in Drosophila is controlled by genes on each of the fly’s four chromosome pairs. One eye-color gene is on the fly’s X chromosome, so the trait is inherited in a sex-linked manner. For this sex-linked trait, the wild-type (brick red) allele is dominant over the mutant vermilion (bright red) allele.

A homozygous wild-type female fly is mated with a vermilion male fly.
X+X+×XvY

Predict the eye colors of F1 and F2 generations. (Assume that the F1 flies are allowed to interbreed to produce the F2 generation.)

Drag the correct label to the appropriate location in the table. Labels can be used once, more than once, or not at all.
screen-shot-2017-02-01-at-8-34-14-pm

Part C – The inheritance of both a sex-linked trait and an autosomal trait in humans

Red-green color blindness is due to an X-linked recessive allele in humans. A widow’s peak (a hairline that comes to a peak in the middle of the forehead) is due to an autosomal dominant allele.Consider the following family history:

  • A man with a widow’s peak and normal color vision marries a color-blind woman with a straight hairline.
  • The man’s father had a straight hairline, as did both of the woman’s parents.

Use the family history to make predictions about the couple’s children.

Drag the correct label to the appropriate location in the table. Not all labels will be used.
screen-shot-2017-02-01-at-8-35-23-pm

AP Exam Prep Question 28

 During meiosis, a defect occurs in a cell that results in the failure of microtubules, spindle fibers, to bind at the kinetochores, a protein structure on chromatids where the spindle fibers attach during cell division to pull sister chromatids apart. Which of the following is the most likely result of such a defect?

screen-shot-2017-02-01-at-8-38-50-pm


Experimental Inquiry: What Is the Inheritance Pattern of Sex-Linked Traits?

The study of modern genetics began with the work of Thomas Hunt Morgan in the early 1900s. Morgan’s experiments with the fruit fly Drosophila melanogaster were the first to demonstrate that genes are located on chromosomes and are the basis of heredity.One of Morgan’s first challenges was to find fruit flies with mutant phenotypes. After nearly two years of tedious work, he discovered a single male fly with white eyes, instead of the usual (wild-type) red eye color. Through his genetic experiments with the white-eyed fly, Morgan deduced that a fruit fly’s eye color was somehow linked to its sex.

Part A – Experimental technique: Reciprocal crosses

When Gregor Mendel conducted his genetic experiments with pea plants, he observed that a trait’s inheritance pattern was the same regardless of whether the trait was inherited from the maternal or paternal parent. Mendel made these observations by carrying out reciprocal crosses: For example, he first crossed a female plant homozygous for yellow seeds with a male plant homozygous for green seeds, and then crossed a female plant homozygous for green seeds with a male plant homozygous for yellow seeds.Unlike Mendel, however, Morgan obtained very different results when he carried out reciprocal crosses involving eye color in his fruit flies. The diagram below shows Morgan’s reciprocal cross: He first crossed a homozygous red-eyed female with a white-eyed male, and then crossed a homozygous white-eyed female with a red-eyed male.

Drag the labels to their appropriate locations to complete the Punnett squares for Morgan’s reciprocal cross.

  • Drag blue labels onto the blue targets to indicate the genotypes of the parents and offspring.
  • Drag pink labels onto the pink targets to indicate the genetic makeup of the gametes (sperm and egg).

Labels can be used once, more than once, or not at all.

screen-shot-2017-02-01-at-8-41-22-pm

Part B – Experimental results: The F2 generation

In one of Morgan’s experiments, he crossed his newly discovered white-eyed male with a red-eyed female. (Note that all of the females at that time were homozygous for red eyes because the allele for white eyes had not yet propagated through Morgan’s flies.) All of the F1 flies produced by this cross (both males and females) had red eyes.

Next, Morgan crossed the red-eyed F1 males with the red-eyed F1 females to produce an F2 generation. The Punnett square below shows Morgan’s cross of the F1 males with the F1 females.

Drag the labels to their appropriate locations to complete the Punnett square for Morgan’s F1 x F1 cross.

  • Drag pink labels onto the pink targets to indicate the alleles carried by the gametes (sperm and egg).
  • Drag blue labels onto the blue targets to indicate the possible genotypes of the offspring.

Labels can be used once, more than once, or not at all.

screen-shot-2017-02-01-at-8-43-32-pm

Part C – Experimental prediction: Comparing autosomal and sex-linked inheritance

You now know that inheritance of eye color in fruit flies is sex-linked: The gene encoding eye color is located on the X chromosome, and there is no corresponding gene on the Y chromosome.How would the inheritance pattern differ if the gene for eye color were instead located on an autosome (a non-sex chromosome)? Recall that for autosomes, both chromosomes of a homologous pair carry the same genes in the same locations.
Suppose that a geneticist crossed a large number of white-eyed females with red-eyed males.
Consider two separate cases:

  • Case 1: Eye color exhibits sex-linked inheritance.
  • Case 2: Eye color exhibits autosomal (non-sex-linked) inheritance. (Note: In this case, assume that the red-eyed males are homozygous.)
For each case, predict how many of the male and female offspring would have red eyes and white eyes.Drag the correct numbers on the left to complete the sentences on the right. Numbers can be used once, more than once, or not at all.
 screen-shot-2017-02-01-at-8-45-52-pm

Linked Genes and Linkage Mapping

If two genes are found on different chromosomes, or if they are far enough apart on the same chromosome that the chance of a crossover between them is very high, the genes are considered to be unlinked. Unlinked genes follow Mendel’s law of independent assortment.If, however, two genes tend to “travel together” because they are near one another on the same chromosome, they are said to be linked. Linked genes do not follow Mendel’s law of independent assortment.

In this tutorial, you will compare the inheritance patterns of unlinked and linked genes.

Part A – Independent assortment of three genes

A wild-type tomato plant (Plant 1) is homozygous dominant for three traits: solid leaves (MM), normal height (DD), and smooth skin (PP).Another tomato plant (Plant 2) is homozygous recessive for the same three traits: mottled leaves (mm), dwarf height (dd), and peach skin (pp).

In a cross between these two plants (MMDDPP x mmddpp), all offspring in the F1 generation are wild type and heterozygous for all three traits (MmDdPp).

Now suppose you perform a testcross on one of the F1 plants (MmDdPp x mmddpp). The F2 generation can include plants with these eight possible phenotypes:

  • solid, normal, smooth
  • solid, normal, peach
  • solid, dwarf, smooth
  • solid, dwarf, peach
  • mottled, normal, smooth
  • mottled, normal, peach
  • mottled, dwarf, smooth
  • mottled, dwarf, peach
Assuming that the three genes undergo independent assortment, predict the phenotypic ratio of the offspring in the F2 generation.
screen-shot-2017-02-01-at-8-48-42-pm

Part B – Gene linkage and phenotypic ratios

Now, suppose that the three tomato genes from Part A did not assort independently, but instead were linked to one another on the same chromosome. Would you expect the phenotypic ratio in the offspring to change? If so, how?
Which statement best predicts the results of the cross MmDdPp x mmddpp assuming that all three genes are linked?
Screen Shot 2017-02-01 at 8.50.03 PM.png

Part C – Building a linkage map

Suppose that you perform the cross discussed in Part B: MmDdPp x mmddpp. You plant 1000 tomato seeds resulting from the cross, and get the following results:

Use the data to complete the linkage map below.

Drag the labels onto the chromosome diagram to identify the locations of and distances between the genes. Use the blue labels and blue targets for the genes; use the white labels and white targets for the distances. Gene m has already been placed on the linkage map.
screen-shot-2017-02-01-at-8-54-49-pm

Scientific Skills Exercise: Using the Chi-Square Test

Are two genes linked or unlinked?

Genes that are in close proximity on the same chromosome will result in the linked alleles being inherited together more often than not. But how can you tell if certain alleles are inherited together due to linkage or due to chance?

If genes are unlinked and therefore assort independently, the phenotypic ratio of offspring from an F1 testcross is expected to be 1:1:1:1. If the two genes are linked, however, the observed phenotypic ratio of the offspring will not match the expected ratio.

Given random fluctuations in the data, how much must the observed numbers deviate from the expected numbers for us to conclude that the genes are not assorting independently but may instead be linked? To answer this question, scientists use a statistical test called a chi-square (χ2) test. This test compares an observed data set to an expected data set predicted by a hypothesis (here, that the genes are unlinked) and measures the discrepancy between the two, thus determining the “goodness of fit.”

If the difference between the observed and expected data sets is so large that it is unlikely to have occurred by random fluctuation, we say there is statistically significant evidence against the hypothesis (or, more specifically, evidence for the genes being linked). If the difference is small, then our observations are well explained by random variation alone. In this case, we say the observed data are consistent with our hypothesis, or that the difference is statistically insignificant. Note, however, that consistency with our hypothesis is not the same as proof of our hypothesis.

Part A – Calculating the expected number of each phenotype

In cosmos plants, purple stem (A) is dominant to green stem (a), and short petals (B) is dominant to long petals (b). In a simulated cross, AABB plants were crossed with aabb plants to generate F1 dihybrids (AaBb), which were then test crossed (AaBb X aabb). 900 offspring plants were scored for stem color and flower petal length. The hypothesis that the two genes are unlinked predicts the offspring phenotypic ratio will be 1:1:1:1.

Using the ratio of 1:1:1:1, calculate the expected number of each phenotype out of the 900 total offspring. Drag the correct values onto the data table. Labels may be used once, more than once, or not at all.
screen-shot-2017-02-01-at-8-56-43-pm

Part B – Calculating the χ2 statistic

The goodness of fit is measured by χ2. This statistic measures the amounts by which the observed values differ from their respective predictions to indicate how closely the two sets of values match.

The formula for calculating this value is

χ2=(oe)2e

where o = observed and e = expected.

The expected and observed data have been entered into the table below. Carry out the operations indicated in the top row. In the last column, enter your answers to two decimal places. Then add up the entries in the last column to find the χ2 value.
screen-shot-2017-02-01-at-9-05-37-pm

Part C – Interpreting the data

The χ2 value means nothing on its own–it is used to find the probability that, assuming the hypothesis is true, the observed data set could have resulted from random fluctuations. A low probability suggests the observed data is not consistent with the hypothesis, and thus the hypothesis should be rejected.
What is the hypothesis that you are testing?
screen-shot-2017-02-01-at-9-08-42-pm

Part D

A standard cut-off point biologists use is a probability of 0.05 (5%). If the probability corresponding to the χ2 value is 0.05 or less, the differences between observed and expected values are considered statistically significant and the hypothesis should be rejected. If the probability is above 0.05, the results are not statistically significant; the observed data is consistent with the hypothesis.

To find the probability, locate your χ2 value (2.14) in the χ2 distribution table below. The “degrees of freedom” (df) of your data set is the number of categories (here, 4 phenotypes) minus 1, so df = 3.

Chi-square distribution table

Between which values on the df = 3 line does your calculated χ2 value lie?
screen-shot-2017-02-01-at-9-35-18-pm

Part E

The column headings show the probability range for your χ2 number.

Chi-square distribution table

What is the probability range that your data fit the expected 1:1:1:1 ratio of phenotypes?

screen-shot-2017-02-01-at-9-36-14-pm

Part F

Based on whether there are non-significant differences (p > 0.05) or significant differences (p ≤ 0.05) between the observed and expected values, you can determine if the data are consistent with the hypothesis that the two genes are unlinked and assorting independently.
Do your results support the hypothesis that the stem color and petal length genes are unlinked and assorting independently, or do the observed values differ from the expected values enough to reject this hypothesis?
screen-shot-2017-02-01-at-9-56-35-pm

Chromosomal Mutations

Chromosomal mutations are changes in the normal structure or number of chromosomes.

  • Changes in chromosome structure can result from errors in meiosis or from exposure to radiation or other damaging agents.
  • Certain changes in chromosome number can result from nondisjunction during either meiosis or mitosis.

Both structural mutations and nondisjunction can play a role in trisomy 21, commonly known as Down syndrome.

Part A – Changes in chromosome structure

The diagram below shows two normal chromosomes in a cell. Letters represent major segments of the chromosomes.

The following table illustrates some structural mutations that involve one or both of these chromosomes. Identify the type of mutation that has led to each result shown.

Drag one label into the space to the right of each chromosome or pair of chromosomes. You can use a label once, more than once, or not at all.
screen-shot-2017-02-01-at-9-13-44-pm

Part B – Nondisjunction

Suppose a diploid cell with three pairs of homologous chromosomes (2n = 6) enters meiosis.

How many chromosomes will the resulting gametes have in each of the following cases?

Drag one label into each space at the right of the table. Labels can be used once, more than once, or not at all.
screen-shot-2017-02-01-at-9-16-02-pm

Part C – Trisomy 21

Down syndrome is caused by trisomy 21, the presence of three copies of chromosome 21. The extra copy usually results from nondisjunction during meiosis.In some cases, however, the extra copy results from a translocation of most of chromosome 21 onto chromosome 14. A person who has had such a translocation in his or her gamete-producing cells is a carrier of familial Down syndrome. The carrier is normal because he or she still has two copies of all the essential genes on chromosome 21, despite the translocation. However, the same may not be true for the carrier’s offspring.The diagram shows the six possible gametes that a carrier of familial Down syndrome could produce.

Suppose that a carrier of familial Down syndrome mated with a person with a normal karyotype. Which gamete from the carrier parent could fuse with a gamete from the normal parent to produce a trisomy-21 zygote?

Drag one of the white cells (representing gametes) to the white target in the diagram. Drag one of the pink cells (representing zygotes) to the pink target.
screen-shot-2017-02-01-at-9-40-03-pm

Make Connections: Chromosomal Inheritance and Independent Assortment of Alleles

Mendel’s law of independent assortment (click on the figure on the left) tells us that the alleles for one character segregate into gametes independently of the alleles for a different character. Independent assortment occurs when the genes for two characters are found on different chromosomes. Why this is the case is explained by the way gametes inherit chromosomes during meiosis (click on the figure on the right).

Part A – Reviewing independent assortment of alleles

At the time of Mendel’s pea plant experiments, no one knew how organisms formed gametes. As Mendel studied the inheritance of two different characters, he wondered how the alleles for the two characters segregated into gametes. Mendel had two hypotheses for how this might work.

  • Under the hypothesis of dependent assortment, the alleles inherited from the parental generation should always be transmitted to the next generation in the same combinations.
  • Under the hypothesis of independent assortment, alleles for different characters should segregate independently of each other, meaning that alleles should be packaged into gametes in all possible combinations, as long as each gamete has one allele for each gene.

The figure below shows the experiment that Mendel used to distinguish between these two hypotheses. The results of the experiment confirmed that the alleles for these characters undergo independent assortment.

Drag the terms to the appropriate blanks to complete the sentences below. Not all terms will be used.
screen-shot-2017-02-01-at-9-43-23-pm

Part B – Chromosomal inheritance during meiosis

In the figure below, you can see Mendel’s experiment again, this time superimposed on the events of meiosis and fertilization. How does chromosomal inheritance during meiosis explain Mendel’s law of independent assortment?

Drag the labels to their appropriate locations in the table below.

  • The number at the top of each column corresponds to the same number in the image above. Each column describes what happens at that numbered stage.
  • Use only white labels for white targets, blue labels for blue targets, and pink labels for pink targets.

screen-shot-2017-02-01-at-9-45-00-pm

Part C – Do the alleles for different characters always sort independently?

In biology lab, you conduct a breeding experiment to test Mendel’s law of independent assortment. You study two characters in a new plant species recently discovered on campus:

  • Flower color, which can be blue (BB) or purple (bb)
  • Petal shape, which can be pointy (PP) or rounded (pp)

You use the following procedure.

  • In the parental generation, you breed a plant that you know to be homozygous for blue-pointy flowers (BBPP) with a plant that you know to be homozygous for purple-rounded flowers (bbpp).
  • In the F1 generation, all your plants have blue-pointy flowers (BbPp).
  • You then allow the F1 plants to self-pollinate to produce F2 offspring. In the F2 generation, you obtain 80 plants with the following phenotypes. Note that an underscore “_” in the genotype indicates that the second allele for that gene could be either dominant or recessive:
Phenotype Number of individuals
Blue flower/pointy petal (B_P_) 59
Blue flower/rounded petal (B_pp) 1
Purple flower/pointy petal (bbP_) 0
Purple flower/rounded petal (bbpp) 20

To try to explain this unusual data, you come up with two alternate hypotheses in addition to your original hypothesis of independent assortment.

Hypothesis 1: The alleles for flower color and petal shape are found on different chromosomes. (This is independent assortment as observed by Mendel with the characters of seed color and shape.)

Hypothesis 2: The alleles for flower color and petal shape are found on different chromosomes, but the blue-rounded (B_pp) and purple-pointy (bbP_) phenotypes typically do not survive, for a reason that has yet to be determined.

Hypothesis 3: The alleles for flower color and petal shape are found close to each other on the same chromosome.

Drag the labels to their appropriate locations in the table below. Labels may be used more than once. (Hint: First, figure out the predicted F1 gametes for each hypothesis; then construct a Punnett square to help you fill in the rest of the table.)
screen-shot-2017-02-01-at-9-47-41-pm

Activity: Mistakes in Meiosis

screen-shot-2017-02-01-at-9-51-35-pmscreen-shot-2017-02-01-at-9-51-43-pmscreen-shot-2017-02-01-at-9-51-51-pmscreen-shot-2017-02-01-at-9-51-58-pm


Activity: Polyploid Plants

screen-shot-2017-02-01-at-9-54-41-pmscreen-shot-2017-02-01-at-9-54-49-pmscreen-shot-2017-02-01-at-9-54-57-pmscreen-shot-2017-02-01-at-9-55-03-pm


 

Misconception Question 71

Which of these descriptions of the behavior of chromosomes during meiosis explains Mendel’s law of segregation?

screen-shot-2017-02-01-at-9-59-01-pm


Misconception Question 73

Select the correct statement(s) about sex determination in animals.
Select all that apply.
 screen-shot-2017-02-01-at-10-00-12-pm

Misconception Question 74

Imagine a human disorder that is inherited as a dominant, X-linked trait. How would the frequency of this disorder vary between males and females?

screen-shot-2017-02-01-at-10-00-55-pm

Chapter 14 Homework – Mastering Biology

***

Use control+F (Command+F for mac) to search this page for the answer. User either Chapter 14 Question __ or some part of the question to easily find the answer.

***

Chapter 14 Question 19

Albinism is an autosomal (not sex-linked) recessive trait. A man and woman are both of normal pigmentation and have one child out of three who is albino (without melanin pigmentation). What are the genotypes of the albino’s parents?

screen-shot-2017-01-09-at-5-08-25-pm


Chapter 14 Question 20

A black guinea pig crossed with an albino guinea pig produced twelve black offspring. When the albino was crossed with a second black animal, six blacks and six albinos were obtained. What is the best explanation for this genetic situation?

Screen Shot 2017-01-09 at 5.10.53 PM.png


Chapter 14 Question 31

Phenylketonuria is an inherited disease caused by a recessive autosomal allele. If a woman and her husband are both carriers, what is the probability that their first child will be a phenotypically normal girl?

Screen Shot 2017-01-09 at 5.12.03 PM.png


Chapter 14 Question 32

Assuming independent assortment for all gene pairs, what is the probability that the following parents, AABbCc × AaBbCc, will produce an AaBbCc offspring?

Screen Shot 2017-01-09 at 5.12.55 PM.png


Chapter 14 Question 51

In humans, ABO blood types refer to glycoproteins in the membranes of red blood cells. There are three alleles for this autosomal gene: IA, IB, and i. The IA allele codes for the A glycoprotein, The IB allele codes for the B glycoprotein, and the i allele doesn’t code for any membrane glycoprotein. IA and IB are codominant, and i is recessive to both IA and IB. People with type A blood have the genotypes IAIA or IAi, people with type B blood are IBIB or IBi, people with type AB blood are IAIB, and people with type O blood are ii. If a woman with type AB blood marries a man with type O blood, which of the following blood types could their children possibly have?

Screen Shot 2017-01-09 at 5.13.48 PM.png


Chapter 14 Question 36

Which of the following is an example of polygenic inheritance?

screen-shot-2017-01-09-at-5-15-10-pm


Chapter 14 Question 38

Which of the following provides an example of epistasis?
 Screen Shot 2017-01-09 at 5.16.17 PM.png

Chapter 14 Pre-Test Question 2

Which of the following is true about a plant with the genotype AABbcc?

screen-shot-2017-01-09-at-5-17-01-pm


Chapter 14 Pre-Test Question 4

Consider pea plants with the genotypes GgTt and ggtt . These plants can each produce how many type(s) of gametes?

screen-shot-2017-01-09-at-5-17-55-pm


Chapter 14 Pre-Test Question 5

Two organisms with genotype AaBbCcDdEE mate. These loci are all independent. What fraction of the offspring will have the same genotype as the parents?

screen-shot-2017-01-09-at-5-18-51-pm


Chapter 14 Pre-Test Question 8

Folk singer Woody Guthrie died of Huntington’s disease, an autosomal dominant disorder. Which statement below must be true?

screen-shot-2017-01-09-at-5-21-33-pm


Chapter 14 Question 2

A man with type A blood marries a woman with type B blood. Their child has type O blood.

screen-shot-2017-01-09-at-5-25-25-pm

Screen Shot 2017-01-09 at 7.27.32 PM.png


Determining Genotype: Pea Pod Color

A botanist has acquired a group of sweet pea plants. All of the plants have yellow pea pods (the recessive trait), except for one, which has green pea pods (the dominant trait). Pea pod color is a trait caused by a single gene. In this tutorial, you will determine how the botanist can identify the genotype of the green pea pod, and how this relates to Mendel’s laws and meiosis.

screen-shot-2017-01-09-at-5-27-44-pm

screen-shot-2017-01-09-at-5-30-11-pm

Screen Shot 2017-01-09 at 5.30.44 PM.png

Screen Shot 2017-01-09 at 5.38.49 PM.png


Activity: The Principle of Independent Assortment

Click to launch activity

View this animation. Then answer the questions.

Screen Shot 2017-01-09 at 5.41.20 PM.png

screen-shot-2017-01-09-at-5-42-00-pm

screen-shot-2017-01-09-at-5-43-04-pm


Misconception Question 68

Each chromosome in this homologous pair possesses a different allele for flower color. Which statement about this homologous pair of chromosomes is correct?

screen-shot-2017-01-09-at-5-48-18-pm


Misconception Question 67

Look at the Punnett square, which shows the predicted offspring of the F2 generation from a cross between a plant with yellow-round seeds (YYRR) and a plant with green-wrinkled seeds (yyrr). Select the correct statement about wrinkled yellow seeds in the F2 generation.

Screen Shot 2017-01-09 at 5.50.18 PM.png


Chapter 14 Question 60

The figure below shows the pedigree for a family. Dark-shaded symbols represent individuals with one of the two major types of colon cancer. Numbers under the symbols are the individual’s age at the time of diagnosis. Males are represented by squares, females by circles.

From this pedigree, this trait seems to be inherited _____.

Screen Shot 2017-01-09 at 5.52.52 PM.png


Chapter 14 Question 49

The following question refer to the pedigree chart in the figure below for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle.

What is the probability that individual III-1 is Ww?

Screen Shot 2017-01-09 at 5.54.40 PM.png


Incomplete Dominance and Codominance

You decide to conduct a genetic analysis of these mutant lines by crossing each with a pure wild-type line. The numbers in the F2 indicate the number of progeny in each phenotypic class.

From these results, determine the relationship between the mutant allele and its corresponding wild-type allele in each line.

Label each mutant line with the best statement from the list below. Labels may be used once, more than once, or not at all.
Screen Shot 2017-01-09 at 6.57.21 PM.png
You continue your genetic analysis by crossing the forked and pale mutant lines with each other. The leaves of the F1 are light green (intermediate between pale and wild-type leaves) and forked. The F2 has six phenotypic classes, as shown below.You designate the forked mutant allele as F (wild type = f+ ) and the pale mutant allele as p (wild type = P).

  1. Consider the alleles for leaf color first. Drag the white labels to the white targets to identify the genotype of each F2 class. Remember that p (the pale mutant allele) and P (the wild-type allele) are incompletely dominant to each other.
  2. Consider the alleles for leaf shape next. Drag the blue labels to the blue targets to identify the genotype of each F2 class. Remember that F (the forked mutant allele) is dominant to f +(the wild-type allele).

Labels may be used once, more than once, or not at all. For help getting started, see the hints.

Screen Shot 2017-01-09 at 7.00.26 PM.png
You continue your analysis by crossing the forked and twist lines. Your results are as follows:

Which of the following statements best explains the outcome of this cross?

Screen Shot 2017-01-09 at 7.05.11 PM.png
You decide to designate the twist allele as FT to distinguish it from the forked allele F.
Using the following allele symbols, identify the genotypes of the three F2 classes in Part C by dragging one label to each class. Labels may be used once, more than once, or not at all.
screen-shot-2017-01-09-at-7-08-55-pm

Chapter 14 Question 54

Phenylketonuria (PKU) is a recessive human disorder in which an individual cannot appropriately metabolize the amino acid phenylalanine. This amino acid is not naturally produced by humans. Therefore, the most efficient and effective treatment is which of the following?

Screen Shot 2017-01-09 at 7.10.01 PM.png


Chapter 14 Question 57

The following question refer to the pedigree chart in the figure below for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle.

What is the genotype of individual II-5?

Screen Shot 2017-01-09 at 7.13.40 PM.png


Chapter 14 Question 56

One of two major forms of a human condition called neurofibromatosis (NF 1) is inherited as a dominant gene, although it may range from mildly to very severely expressed. Which of the following is the best explanation for why a young, affected child is the first in her family to be diagnosed?

screen-shot-2017-01-09-at-7-14-39-pm


Activity: Mendel’s Experiments

Click to launch activity

Click here to view this animation.

Then answer the questions.

Screen Shot 2017-01-09 at 7.16.32 PM.png

Screen Shot 2017-01-09 at 7.18.21 PM.png

Screen Shot 2017-01-09 at 7.18.45 PM.png

screen-shot-2017-01-09-at-7-19-10-pm


Chapter 14 Pre-Test Question 1

In his breeding experiments, Mendel first crossed true-breeding plants to produce a second generation, which were then allowed to self-pollinate to generate the offspring. How do we name these three generations?

Screen Shot 2017-01-09 at 7.22.29 PM.png


Pedigree Analysis: Dominant and Recessive Autosomal Conditions

Geneticists analyze pedigrees to follow the inheritance of genetically controlled conditions. Three things must be determined in a pedigree analysis:

  1. The mode of inheritance of the condition. In this tutorial, this means deciding if a condition is caused by an autosomal dominant or autosomal recessive allele.
  2. The genotypes of individuals in the pedigree (as far as can be known) based on their phenotype and the phenotypes of their parents or children.
  3. The probability that certain individuals will have the condition. This requires assigning probability values to some individuals whose genotypes cannot be determined with certainty. It also requires an understanding of how and when to apply the multiplication and addition rules.

Part A – Determining the mode of inheritance

The pedigrees below show the inheritance of three separate, rare autosomal conditions in different families. For each pedigree, decide if the condition is better explained as recessive or dominant.
Drag the correct label to the appropriate location. Labels can be used once, more than once, or not at all.
Screen Shot 2017-01-09 at 7.32.37 PM.png

Part B – Determining genotypes in autosomal dominant pedigrees

Pedigree 2 from Part A is shown below. Recall that this pedigree shows the inheritance of a rare, autosomal dominant condition.
Fill in the genotypes for the indicated individuals in the pedigree by dragging the best label to the appropriate location. Labels can be used once, more than once, or not at all.
Screen Shot 2017-01-09 at 7.33.07 PM.png

Part C – Determining genotypes in autosomal recessive pedigrees

Pedigree 3 from Part A is shown below. Recall that this pedigree shows the inheritance of a rare, autosomal recessive condition.

Note that individual II-3 has no family history of this rare condition.

Fill in the genotypes for the indicated individuals in the pedigree by dragging the best label to the appropriate location. Labels can be used once, more than once, or not at all.
Screen Shot 2017-01-09 at 7.35.30 PM.png

Part D – Calculating probabilities in pedigrees

The pedigree from Part C is shown below.

Individuals III-3 and III-4 are expecting their first child when they become aware that they both have a family history of this recessive condition. As their genetic counselor, you can calculate the probability that they are carriers and that their child will be affected with the condition.

Complete each statement by dragging the correct label to the appropriate location. Labels can be used once, more than once, or not at all.
Screen Shot 2017-01-09 at 7.36.39 PM.png