Answers to Problem Set #2

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Answers to Problem Set #2

1)Genotype: The actual genetic makeup of the cells of an individual with respect to a given trait.

Phenotype: The observable properties of an organism with regard to a specific trait.

2)a. 9/16 F-P-, 3/16 F-pp, 3/16 ffP-, 1/16 ffpp b. 1/4 BbTt, 1/4 Bbtt, 1/4 bbTt, 1/4 bbtt

3)a. 2x2x2x2=16 b. 1x2x1x2=4 c. 2x3x2x4=48 d. 2x2x1x3=12 e. ½ x ¼ x ½ x ¼ = 1/64 f. ½ x ¾ x 1 x ¾ = 9/32

4) II-6, 2/3; IV-1, 1

5)Primary Sex Ratio: ratio of males to females at conception.

Secondary Sex Ratio: ratio of males to females at birth

Tertiary Sex Ratio: ratio of males to females at any time after birth

6)Female cats have two chromosomes and male cats have one X chromosome. Since the gene/locus for coat color in cats is on the X chromosome, male cats have only one coat color gene and thus are one solid color. Since female cats have two X chromosomes they can be heterozygous for coat color. Depending on which X is active in a given cell (the other X being an inactive Barr body) coat color patches are possible. A male with a calico coat color pattern would have to be XXY, and would be sterile.

7)Difference: Spermatogenesis yields four functional sperm produced from one primary spermatocyte; Oogenesis produces only one egg from one primary Oocyte along with three non-functional polar bodies.

8)1 – Meiosis produces gametes containing the haploid numbers of chromosomes. 2 –

Meiosis makes genetic recombination possible through crossing over and independent segregation.

9)Meiosis I is where crossing over between homologous chromosomes occurs, as well as reduction in chromosomes # from 46 to 23. Meiosis II is similar to Mitosis in that sister chromatids separate and move to opposite poles of the cell.

10) An antigen is a substance that is capable of stimulating the production of a specific antibody. An antibody is a specific complex protein made in response to the entry and recognition of a “foreign” antigen.

Test #2 Answer Key

March 19, 1999

1a. "The Stonger Sex" (genetically/chromosomally)

- Females, since they have two X-chromosomes (although there are more males than females at birth and most likely conception, too) and males only have one. Many detrimental conditions are due to X-linked recessive genes. Hemizygous recessive males are affected, heterozygous females are not. Females live longer prenatally and postnatally.

1b. Hemizygous Recessive – a condition where a recessive trait is expressed because a gene is only located on the X chromosome. Males who carry a recessive trait on their X chromosome do not have another X to counter the trait due to the fact that they are XY or hemizygous (carrying only one copy of an X chromosome).

1c. Mutagen - Any agent which increases the likelihood that DNA or chromosomal change/damage will occur e.g., radiation or cigarette smoke exposure alters our DNA.

1d. Primary Sex Ratio - The ratio of males to females at conception. No one knows what it is. In theory we say it is 1 to 1, but some evidence, e.g., from sex ratio at birth and in miscarried fetuses, indicate that there may be more XY than XX zygotes.

1e. The Pleiotropic Effects Found in Cystic

Fibrosis – Sticky secretions in the lungs, pancreas, gonads, etc. compromising respiration, digestion, reproduction . . . i.e., several effects throughout the body all due to one gene/genotype.

2.

1) 1 x 1 x 2 x 2 = 4

2) 2 x 1 x 2 x 2 = 8

3) 2 x 1 x 3 x 4 = 24

4) 2 x 1 x 2 x 2 = 8

5) 1/2 x 1 x 3/4 x 1/2 = 3/16

6) 1/2 x 1 x 3/4 x 1/4 = 3/32

Show Punnett Squares

3a. A

3b. D

3c. B

3d. D

3e. D

4a. F - XO

4b. T

4c. F - limited

4d. T

4e. F - labia major, labia minor, and clitoris or penis and scrotum

4f. F - 1/20

4g. 1 1/2 for F

5a. OO

5b. BO

5c. A,B,O

6a. No

6b. Yes

7a. Diagrams and indicate

7b. It signals the beginning and the direction of sexual development . . . at 7-8 weeks prenatal and in a male like direction.

8a. HY

8b. Hh

8c. 1/4

8d. 0 (zero)

9a. She had Testicular Feminization (Androgen

Insensitivity or male pseudohermaphroditism). She was XY, but had an X-linked mutation (transmitted from mom) which failed to activate her testosterone, thus her secondary sex characteristics were female in spite of having abdominally located testes.

9b. In Marilyn's case, it helped her understand herself . . . that there was a natural, logical explanation for her intersexual development. It helped her appreciate her "normalness", that she was not a "freak" like she had been led to believe. In addition, in other cases, often surgery and/or hormone therapy can aid in their development as well as minimize risks of cancer, etc.

10a. D

10b. B

10c. C

10d. E

10e. F

10f. A

11a. B'B' and BB

11b. B'B'and B'B

11c. 1/2 (1/2 male x 1 B'B)

11d. 1/4 (1/2 female x 1/2 B'B')

12. Up to 5 points for Class Attendance

13. Signature

October 27, 1999

1a. Severe Combined Immune Deficiency Syndrome

(SCIDS) - An inherited inability to recognize foreign antigens and to make antibodies against them. David "the bubble boy" had one of these

Mendelianly inherited conditions. He survived for more than 12 years in a relatively germ free environment.

1b. Multiple alleles; n/2 (n+1) - Some genes, e.g., ABO have more than 2 alleles. In ABO there are 3: A, B, and O. N/2 (n+1) is the formula to determine the number of genotypes possible where n =# of alleles. So, for ABO: (3/2)(3+1) = 6

1c. Testicular Feminization - XY chromosome, but more so female than male. These intersexual individuals have abdominally located tests. However, the testosterone produced in their testes and released into the blood stream fails to be activated.

This failure is due to an X-linked recessive allele. Thus hemizygous recessive (tY) are affected. Failing to activate/stimulate male secondary sex characteristic shifts sexual development in a female direction.

1d. Y-linked characteristic - SRY or TDF are gene(s) on the Y chromosome which cause the ovotestes to become testes. Also, "hairy ears" may be due to a gene on the Y. Only males have Y-linked characteristics. Not many examples, i.e., the Y chromosome does not seem to have many useful genes.

1e. Syngamy is not fertilization - Syngamy is when the egg and sperm nuclei unite . . .which occurs several hours after the sperm enters the egg . . . the latter being fertilization. Fertilization triggers the egg to go through Meiosis II extruding a second polar body. After that, syngamy can take place.

2.

2 x 1 x 1 x 2 = 4

2 x 2 x 1 x 1 = 8

3 x 2 x 1 x 4 = 24

2 x 2 x 1 x 4 = 16

3/4 x 1/2 x 1 x 1/2 = 3/16

3/4 x 1/2 x 1 x 1/4 = 3/32

Show your work with Punnett Squares.

3a. A (1 pt) and C (1 pt) (Full credit for either

A or C only)

3b. D

3c. A

4a. Yes

4b. 4 (2 x 2)

4c. Yes

4d. AbCd or AbcD or aBCd or aBcD - (Any of the 4 which do not involve crossing over)

5a. F - 1/50 x 1/50 = 1/2500

5b. F - ovaries or testes

5c. T

5d. F - limited

5e. F - and/or

5f. F - XXY

5g. 1/2 for F; 1 pt for correcting statement

6a. Cutting DNA with a specific DNA cutting enzyme; i.e., DNA from different individuals will be cut into a different number of pieces and different sizes of pieces. This is VERY VERY accurate in distinguishing between different people . . . in making accurate identification.

6b. Paternity Testing

Military

Accident reconstruction

Murder, rape . . . forensics

Others?

7a. aa (homozygous recessive)

7b. Aa (heterozygous)

7c. A- (-1 pt for AA or Aa . . . because you can't tell which!)

7d. 1/2

7e. 1/4 x 1/2 = 1/8 (-1 pt if 1/4 or 1/2)

8a - 8b. Up to 5 points for each of two well communicat$

examples, situations, etc. from one/both videos. Must be put in a

Genetics in Human Affairs context.

9a. HY (hemizygous dominant)

9b. Hh (heterozygous)

9c. 1/4 (-1 pt if 1/2)

9d. 0

10a. A

10b. E

10c. C

10d. F

11a. No, Mom must be Rh- and fetus Rh+ for there to be a potential problem.

11b. Yes. The B allele in the child can be transmitted from an

AB dad. The Rh+ (D allele) could have been transmitted from the

Rh+ mom.

12. Up to 5 points for Class Attendance Record

13. Signature

March 27, 2000

1a. Substitution Point Mutation in “THEFATCATATETHERAT”

One “base” is replaced by another, e.g. R replaces C to now read

“THEFATRATATETHERAT”

1b. “The Stronger Sex” (Genetically/Chromosomally) – Females, since they have two

X-chromosomes (although there are more males than females at birth and most likely conception, too) and males only have one. Many detrimental conditions are due to Xlinked recessive genes. Hemizygous recessive males are affected, heterozygous females are not. Females live longer prenatally and postnatally.

1c. Unlinked Loci – Diagram will be given in Help Sessions. “A” and “B” are on different chromosome pairs, thus are not linked. Instead, the two loci segregate independently.

1d.

David, “The Bubble Boy” – A boy with a homozygous recessive genotype

(inherited from his heterozygous parents) leaving his immunogenetic system unable to make specific antibodies. Lived in a germ-free bubble to +/- age 14.

1e. Syngamy is Not Fertilization – Fertilization is when a sperm penetrates the egg.

Syngamy (several hours later) is when the egg and sperm nuclei unite. During the time between Fertilization and syngamy the egg undergoes Meiosis II.

1f. Radiation Exposure . . . Background Sources – About ½ of our radiation exposure comes from background . . . earth’s crust and cosmic sources. These environmental exposures are mutagenic.

2.

1)1 x 2 x 1 x 2 = 4

2)2 x 2 x 1 x 2 = 8

3)2 x 3 x 1 x 4 = 24

4)2 x 2 x 1 x 4 = 16

5)½ x ½ x 1 x ¼ = 1/16

6)female

7)½ x ¾ x 1 x ½ = 3/16

Show your work with Punnett Squares (diagrams will be given in Problem Session)

3a. F

3b. D

3c. C

3d. E

3e. B

3f. A

4a. T

4b. F; XY sex chromosome constitution

4c. F; (1/100)^2 = 1/10,000

4d. T

4e. F; sex-limited

4f. F; genital tubercle

5. Cutting DNA into specific numbers and sizes of pieces. Two samples from the same person yield identical results; while two from different persons have different results. A powerful means of identifying where/who cells came from.

5a-5b. Examples:

- Rape case – compare semen with cells of suspects

- Other forensic/ID applications such as murder, breaking and entering, since cells are left at the scene of the crime.

- Missing in action, etc., military ID applications

- Accident reconstruction . . . who was driving, etc.

- Immigration

- Mix-ups at hospitals, etc.

- Others?

6a. Bee stings: First sting caused antibodies to be made against foreign antigens. The stings several weeks later were much more painful due to “clash” of antibodies (keys) now available to attack foreign antigens (locks).

6b. Polio virus: Schoolmates contracted polio. It was likely that all the students exposed to the polio virus. Some died/handicapped; others developed antibodies, fought off the virus, didn’t get sick. Probably didn’t need the vaccine when it became available.

6c. RhoGAM: Second daughter was Rh+ (although the mother was Rh-). RhoGAM was injected into the mother to intercept and destroy Rh+ cells from fetus before Mom was stimulated to make anti-Rh antibodies, thus subsequent pregnancies would not have been at risk.

7a. Hh (heterozygous)

7b. HY (hemizygous dominant)

7c.

½

7d. 1 or 100%

8a. Aa (heterozygous)

8b. albino (albinism)

8c. ½ female x ¼ aa = 1/8

8d. Several answers are possible:

- II-4 is NOT the father

- A new mutation in sperm, egg, zygote, etc. made III-3 Aa.

-Gene interaction . . . two different genetic forms of albinism in this family

9a - 9b. Up to 4 points for each of two different, well communicated examples, situations, etc. from the video. Must be put in a “Genetics in Human Affairs” context.

10a. No. Woman (Type B) makes anti-A antibodies which would/could attack a type A or AB fetuses cells. She and her partner’s kids cannot be those blood types.

10b. Yes. Their pregnancies can involve an Rh+ fetus. Cells from such a fetus can stimulate Rh- Mom to make Anti-Rh antibodies. Usually a 1 st

pregnancy is not a problem. Subsequent one could be if Mom does not receive a RhoGAM injection after each Rh+ baby is born.

11.Up to 5 points extra credit for Class Attendance Record.

12.Signature

October 27, 2000

1a.

Frameshift Point Mutation in “THEFATCATATETHERAT” – e.g., delete (or add) a letter, in the case of deleting the first A, the new sequence would be 3-letter nonsense: THE FTC ATA TET HER AT

1b. Background Radiation Sources – a) Earth’s crust . . . radioactive ores/minerals/elements. b) Cosmic/outerspace. These two categories (or is it one?) contribute to ~ ½ of our “natural” radiation exposure; medical use being the other “big player”.

1c. Sex-linked Characteristic – Colorblindness, muscular dystrophy, and hemophilia are classic examples of “traits” due to genes in the X chromosome; males more often than females are affected.

1d. Variable Expressivity – Some inherited conditions are considerably more strongly/severely revealed in some affected individuals than in others, e.g., varying degrees of diabetes, hearing impairment, etc.

1e. Nondirective Counseling – Genetic counselors give information, calculate risks, discuss treatments, supply perspective, offer alternatives, etc., but do NOT tell clients what to do.

1f.

“The Stronger Sex” (Genetically/Chromosomally) – Females, not males, due to females having two X chromosomes, males only one. As a result, the incidence of Xlinked recessive conditions (some are detrimental) affect males more frequently. Females outlive males.

2a. Tt (heterozygous)

2b. tt (homozygous recessive)

2c.

¾

2d.

½ x ½ = ¼

3a. cY (hemizygous recessive)

3b. Cc (heterozygous)

3c. ¼ (show Punnett Squares)

3d.

½ (show Punnett Squares)

4a – 4b. Up to 4 points for each of two different, well communicated, detailed, and complete examples, situations, etc. from the video. Must be put in a “Genetics in Human

Affairs” context.

5.

1)1 x 2 x 1 x 2 = 4

2)2 x 2 x 2 x 2 = 16

3)2 x 3 x 2 x 4 = 48

4)1 x 2 x 2 x 4 = 16

5)½ x ½ x ½ x ¼ = 1/32

6)1 x ¾ x ½ x ¼ = 3/32

Show work with Punnett Squares.

6a. A

6b. D

6c. B

6d. C

6e. B

7a. F; ovotestes

7b. T

7c. T

7d. F; male sexual development

7e. F; (1/200)^2 = 1/40,000

7f. T

8.Up to 10 points for EXCELLENT, complete, well communicated response.

9.Up to 10 points for a careful, thoughtful, well communicated response.

10.Up to 5 points for Class Attendance Record.

11.Signature.

March 23, 2001

1a. Linked Loci – Two or more genes on the same chromosome, e.g., “A” and “B” are linked. Linked loci do not segregate independently in gamete formation as do loci on different pairs of chromosomes.

1b. Mutagen – Any agent that causes DNA/Genes/Chromosomes to be altered/changed/damaged/mutated. Examples: radiation, cigarette smoke, formaldehyde.

1c. Pleiotropism – A gene with multiple phenotypic effects, e.g., skeletal, visual, and circulatory problems in M- individuals with Marfan Syndrome.

1d. Sex-limited Characteristic – One due to gene(s) (either autosomal or X-linked) which can be present in both male and female, but where expression phenotypically is in one sex only. For example: testicular cancer, fraternal twinning.

1e. Intersexual individual. XY chromosomes, testes, but other sexual development is predominantly female such as breast development, vagina, psychosexual orientation.

2

1)1 x 2 x 2 x 2 = 8

2)2 x 2 x 1 x 2 = 8

3)2 x 3 x 2 x 4 = 48

4)1 x 2 x 2 x 4 = 16

5)1 x ¾ x ½ x ¼ = 3/32

Show your work with Punnett Squares.

3a. A

3b. B

4a. C

4b. G

4c. F

4d. B

4e. A

4f. E

4g. D

4h. H

5a. influenced (or conditioned)

5b. Bald

5c.

B’B’

5d.

½ (50%)

5e. 1 (100%)

6a - 6b. Up to 4 points for each of two different, well communicated, detailed, and complete examples, situations, etc. from the video. Must be put in a “Genetics in Human

Affairs” context.

7a. Cc (heterozygous), AO

7b. cY (hemizygous recessive), BO

7c. cc (homozygous recessive), OO

7d. cc (homozygous recessive), AB

7e. No

7f. No

7g.

¼ CY x ¼ AO = 1/16 (Show Punnett Squares)

8a. Bee stings: First sting caused antibodies to be made against foreign antigens. The stings several weeks later were much more painful due to “clash” of antibodies (keys) now available to attack foreign antigens (locks).

8b. Polio virus: Schoolmates contracted polio. It was likely that all the students exposed to the polio virus. Some died/handicapped; others developed antibodies, fought off the virus, didn’t get sick. Probably didn’t need the vaccine when it became available.

8c. RhoGAM: Second daughter was Rh+ (although the mother was Rh-). RhoGAM was injected into the mother to intercept and destroy Rh+ cells from fetus before Mom was stimulated to make anti-Rh antibodies, thus subsequent pregnancies would not have been at risk.

9a. F; 1/2500, hemophilic (hh)

9b. T

9c. T

9d. T

9e. F; ovotestes

9f. T

10a. Cutting DNA into pieces, separating the pieces to determine number and size of pieces. DNA samples from two different people will be cut into uniquely different numbers and sizes. A means of precise identification.

10b. - Matching cells/DNA left at a crime scene, accident, etc. to those of suspect.

- Military identification

- Others?

11a. Yes. Rh- women whose partner is Rh+ could have Rh+ babies. During pregnancy or at delivery Rh+ cells from baby could enter Mom’s system. If so, Mom would make antibodies which could attack subsequent Rh+ fetuses.

11b. Yes. Type O women make anti-A and anti-B antibodies. Her pregnancies with an

AB partner would be A or B, either of which would be at risk should Mom’s antibodies traverse the placenta.

12.Up to 5 points for Class Attendance Record.

13.Signature

October 26, 2001

1. a.

Synapsis - The "pairing", i.e., close alignment of the two members of each chromosome pair during prophase of Meiosis I. Crossing over happens at this time in the early stages of gamete formation./li> b.

Mutagen - Any agent (chemical or physical) which increases the chance that a mutation . . . DNA/chromosome alteration will occur. Examples include radiation, formaldehyde, etc. c.

Testicular Feminization - Example of intersexual or pseudohermaphrodite individual. Although they appear to be female, they have XY sex chromosomes, and testes located in their abdomen. Faulty gene is in the X chromosome, resulting in sex developmental confusion. d.

Patches, the Calico Cat - Dr. McKenzie's cat. XX, normal mosaic coat color pattern due to heterozygosity for X linked coat color gene and inactivation of one

X or the other in various regions of the skin/hair. e.

Syngamy - Fusion of egg nucleus and sperm nucleus several hours after the sperm entered the egg (fertilization). During those several hours the egg completes

Meiosis II. f.

TDF or SRY - Gene in the Y chromosome which activates other genes, i.e., sets in motion sex development . . . male when present, female when absent. Causes ovotestes to become testes.

2.

2 x 1 x 1 x 1 = 2

2 x 2 x 1 x 2 = 8

3 x 2 x 1 x 2 = 12

2 x 1 x 1 x 2 = 4

3/4 x 1 x 1 x 1/2 = 3/8

Show your work with Punnett Squares.

3. a.

C b.

F c.

E d.

D e.

A f.

B

4. a.

D b.

A,B

c.

E, -1 for A or C d.

C e.

B f.

B

5. a.

No b.

16 c.

No d.

ABCD - Several correct answers, but each correct one must 1) be haploid, i.e., have one dose of "A","B","C","D" . . . meaning big or little of each letter and 2) show that crossing over had happened involving loci "C" and "B".

6. a.

T b.

F, Turner Syndrome c.

F, either the ovaries or testes d.

F, 2 x 2 x 2 x 2 x 2 = 32 different haploid gametes e.

F, X-limited f.

F, either Mom or Dad

7. a.

aa, homozygous recessive b.

Aa, heterozygous c.

A-, can't tell so -1 for AA or Aa d.

1/2 e.

1/4 aa x 1/2 female = 1/8

8. a.

Hh b.

HY c.

1/2 d.

1 or 100%

9a-b. Up to 4.5 points for each of two different examples, conditions, situations which convincingly communicate the essence of the video in question.

10. Up to 4 points for Class Attendance Record.

11. Signature

March 22, 2002

1.

a.

Crossing Over - The exchange of DNA/chromosome segments of the two members of a chromosome pair during the early stages of Meiosis I/gamete formation. Contributes greatly to the genetic uniqueness of each gamete, and hence the genetic uniqueness of ea ch zygote. b.

Spontaneous Mutation - DNA or chromosome changes that occur without known cause, i.e., without knowledge of exposure to a mutagen. Occurs rarely, only 1 in

100,000 per gene per cell division. c.

Pseudohermaphrodite - False hermaphroditism. Such individuals have features common to both male and female, e.g., testes but no penis, instead of a clitoris and vagina. . .an example is Testicular Feminization. d.

Tertiary Sex Ratio - Ratio of male to female at any time after birth, e.g., et age 18

= 100:100; at age 50 = 87:100; at age 100 = 20:100. Males throughout post natal life die younger than females. This is likely due to both genetic and environmental f actors. e.

Variable Expressivity - Concept that different individuals with the same diagnosed condition have varied severities of it, e.g., some with Cystic Fibrosis have more serious lung problems than others.

2.

2 x 1 x 1 x 2 = 4

1 x 2 x 1 x 2 = 4

2 x 2 x 1 x 4 = 16

1 x 2 x 1 x 4 = 8

1 x 1/2 x 1 x 1/4 = 1/8

1/2 x 1/2 x 0 x 1/4 = 0

Show your work with Punnett Squares.

3. a.

D b.

C c.

F d.

E e.

B f.

A

4. a.

A b.

E c.

A d.

A,B e.

B f.

D

5.

a.

T b.

T c.

F, either the ovaries or testes d.

F, 8 different haploid e.

F, Turner Syndrome f.

F, 1/10,000

6. a.

Left-handed b.

Rr c.

R- (RR or Rr) d.

1/4 rr x 1/2 female = 1/8

7. a.

F, WalMart b.

T c.

T d.

F, 17 years old and will never (according to Mom) e.

F, Rachel goes to a regular high school f.

F, two children, one of which has

8. a.

Horseback Riding b.

Acting c.

Cooking d.

Gardening e.

Sports, e.g. bowling, softball, gymnastics, weight lifting f.

TV g.

Others?

9. a.

C b.

B c.

B d.

A

10. Up to 4.5 points for each of two well thought out examples, conditions, situations that draw from the "Birth, Sex, and Death" video.

11. Up to 4 points for Class Attendance Record.

13.Signature

October 25, 2002

1. a.

Tasting PTC is dominant, nontasting is recessive. A nontaster woman and a nontaster man have a taster child. GIVE two different explanations as to how that could have happened. -

1. A new mutation resulting in a Tt child.

2. The man is not the father; a taster guy is.

3. Two different loci are involved, e.g. AAtt x aaTT -> AaTt./li> b.

Syngamy - Syngamy is when the egg and sperm nuclei unite . . .which occurs several hours after the sperm enters the egg causing fertilization. Fertilization triggers the egg to go through Meiosis II extruding a second polar body. After that, syngamy can take place. c.

Frameshift Mutation - Due to addition of deletion of one or more DNA base pairs.

The triplet reading frame is shiftd such that the codons are improperly read, e.g.

The FAT CAT ATE THE RAT becomes THE FAT ATA TET HER AT if the seventh letter is omitted . d.

Pleiotropism - A gene with multiple phenotypic effects, e.g., M-individuals with

Marfan syndrome have very long bones, aneurisms, and visual distortion. e.

"Patches," the Calico Cat - XX female heterozygous for X-linked coat color gene, i.e., Bb. Due to Barr body inactivation of one X or the other some cells express B, other b, leading to Black and Yellow patches. Dr. McKenzie's pet cat.

2.

1 x 2 x 1 x 2 = 4

1 x 2 x 2 x 2 = 8

2 x 3 x 2 x 4 = 24

1 x 2 x 2 x 4 = 16

1 x 3/4 x 1/2 x 1/4 = 3/32

0 x 1/2 x 1/2 x 1/4 = 0

Show your work with Punnett Squares.

3. a.

G b.

E c.

C d.

F e.

H f.

A

4. a.

B b.

B c.

D d.

E e.

B

f.

A

5. a.

F. 1/400 (or 1/20 2 ) b.

T c.

F, equally to genes transmitted from mom and dad. d.

F, 2 5 or 32 different haploid e.

T f.

F, sex-limited

6. a. Hh (or heterozygous), aa (or homozygous recessive) b. hY (or hemizygous recessive), Aa (or heterozygous) c-d. hh (or homozygous recessive), Aa (or heterozygous); and hh (or homozygous recessive), aa (or homozygous recessive). e. No f. No g. 1/4 x 1/2 = 1/8

7. a.

influenced b.

Bald c.

B'B' d.

1/2 e.

1 or 100%

8. Up to 4.5 points for each of two different examples, conditions, situations from the video that clearly indicated they saw the video and gained from it.

9. Up to 5 points each for both of the Take Home Questions.

10. Signature

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