BIO 211 Exercise II-18

Genetics

Materials:

DNA/Chromosome Kits

Instructions:

The lab instructor will describe the genetic terms and explain how to do Punnet squares referring to the Objectives for Lab Quiz #5. Students will work the following genetics problems. Underline the lower case letter to prevent confusion (Ss and Ww and Cc). Additional instructions follow.

Single-gene human genetic traits and their alleles.

1. Give the genotypes for the listed phenotype. (D=dominant, R=recessive – to have the recessive phenotype you must have two copies of the recessive allele; example: Person with attached ear lobe has the genotype – aa)

Phenotype	Alleles	Genotype(s)

Bent pinky -D	B,b

Pattern Baldness-D	P,p

PTC tasters -D	T,t

Blue eyes -R	E,e

Red-green color blindness-R	C,c (x-linked)

Tongue rolling-D	R,r

Widow’s peak -D	W,w

Thumb crossing -D	C,c

Attached ear lobe- R	A,a

Hitchhikers’ thumb-R	H,h

Phenotype	Genotype(s)

Straight pinky

No Pattern Baldness

Non PTC tasters

colored eyes

color vision

nontongue rolling

straight hairline

Thumb crossing

(fold hands together right thumb over left)

Unattached ear lobe

Non-Hitchhikers’ thumb

(Cannot bend thumb back more than 30 degrees)

2. From the above list, check the characteristics you have (your phenotype). Circle your probable genotype(s).

3. Why is there a difference in the phenotype of an X-linked trait in males and females when each has only one recessive allele?

Genetics Problem Set

Work the problems before checking the solutions on page

1. In summer squashes, white fruit is dominant to yellow fruit. The factor for white fruit is represented by W and yellow by w. What gametes are produced by plants with the following genotypes?

Genotype Gametes

WW

Ww

ww

Do Punnett squares for the following crosses and give the percentage of the offspring with each fruit color (phenotype) for the offspring.

Ww X ww

WW X Ww

WW X ww

Ww X Ww

2. A. Hemophilia in humans is due to a sex-linked recessive gene, h. A normal woman with no history of hemophilia in her family marries a normal man who had a hemophilic father. What percentage of their children will have hemophilia?

B. What would be the results if the man had a hemophilic mother.

3. The determiner for brown eyes (B) is dominant to blue eyes (b). What are the probable genotypic and phenotypic percentages of children of a man with brown eyes, whose mother was blue-eyed, and a blue-eyed woman, whose father was brown-eyed..

4. If a brown-eyed man marries a blue-eyed woman and they have eight children, all brown eyed, what are the likely genotypes of each individual in this family?

Father:

Mother:

Children:

5. Blood types in humans are the result of a combination of two of three possible alleles IA, IB and i. The I allele is recessive and when an individual is homozygous for this allele, they have type O blood. Blood type A may be due to a homozygous condition or the the IA and i allele. Blood type AB is the result of having a copy of both the IA and IB alleles. Rh factor is another marker on human red blood cells and is either positive (dominant R) or negative (recessive r). Give all possible genotypes for each blood type regarding IA and IB and i alleles and R or r.

Blood Type – Phenotype Genotypes
A
A-
B+
B-
AB+
AB-
O+
O-

What are the percentages of offspring which each phenotype and genotype in a cross between a person with type O blood and a person with type AB blood?

6. In incomplete dominance the heterozygous condition results in some intermediate trait. In humans, carrying the sickle cell trait results is some intermediate signs of sickle cell anemia. Let S represent normal hemoglobin and s is the abnormal hemoglobin. What are the genotypes for each phenotype? If a father and mother are heterozygous what are the probable percentages of each phenotype and genotype of their children?

Phenotype	Genotype

Normal hemoglobin

Sickle cell anemia

Sickle cell trait

Dr. Bartholdi’s Rules for Setting-up Genetics Problems:

Suggestion – Feel Free to Use Your Own if You Choose

1.	a.	Rule #1: Always underline the lower case letter representing the recessive gene.

Example: b this also helps to determine lower case with letters s, c, w, v, z

b.	Rule #2: Dominant gene always written first regardless of donor (male/female). Example: Bb, Ww

c.	Rule #3: Female gamete X always written before Y gamete.

Example: XX (female) or XY (male)

d.	Rule #4: If sex is not indicated in a word problem (Rr x Rr), the first set of allele pairs can be designated as the male, the second allele pairs can be designated as the female.

e.	Rule #5: Male allele pairs go down the Punnet Square.

Genotype is represented by the off-springs in the square; Phenotype is the interpretation of the Genotype. R = rough coat; r = smooth coat

Female	RR x rr would be the written cross of both parents

Male	r	r
R	Rr	Rr
R	Rr	Rr

Male = RR

Female = rr

100% of all offspring will have rough coats = Phenotype

100% of all offspring will be heterozygous for R = Genotype

f.	Rule #6: Female allele pairs go across the Punnet Square (see “e” above).

g.	Rule #7: 1 single allele pair per parent would produce a 4 Punnet Square (see e above); RR x rr

h.	Rule #8: 2 different allele pairs per parent would produce a 16 Punnet Square (see #6).

Example: RRWW x RRWW produces 16 squares; would need to apply FOIL (see Rule 9).

i.	Rule #9: FOIL must be used to get the individual gametes when you cross 2 allele pairs .

Example: F (first); O (outer); I (inner); L (last)

j.	Rule #10: Heterozygous parents (RrWw x RrWw) produce a 16 Punnet Square = 9:3:3:1 for phenotypes of offspring; use FOIL to get the individual alleles for each parent; see #6 for complete demonstration of a 16 Punnet Square.

k.	Rule #11: You can represent off-spring Phenotype or Genotype as percent, % (50%), ratio (1:2) or fraction (½ ), just be consistent, do NOT mix.

2.	Dominant vs. Recessive

a.	A dominant gene will always be expressed phenotypically over a recessive gene in a perfect world: Mendalian genetics.

b.	Example would be the illustration above, Rule #5:

Genotype is represented by the off-springs in the square; Phenotype is the expression of the Genotype. R = rough coat; r = smooth coat

Female	RR x rr would be the written cross of both parents

Male	r	r
R	Rr	Rr
R	Rr	Rr

Male = RR

Female = rr

100% of all offspring will have rough coats = Phenotype

100% of all offspring will be heterozygous for R = Genotype

LAB QUIZ #5

3.	Exception #1: Co-Dominance

a.	Take the scenario of ABO blood groups; this is an example of incomplete dominance.

1)	There are 4 blood types that a person can state he/she is (Phenotype):

Phenotype	Anitgen(s) on RBC	Genotype
Group A	A antigens on RBC only	Genotype AA or AO
Group B	B antigens on RBC only	Genotype BB or BO
Group O	No antigens on RBC	Genotype OO only
Group AB	Both A&B antigens on RBC	Genotype AB only

2)	A person with AB blood MUST have both alleles: Genotype A, B

b.	Let’s now consider a Cross between AO x BO

Female	Off-spring would be:

Male	B	O
A	AB	AO
O	BO	OO

Phenotypically = 25% AB, 25% B, 25% A and 25%O.

Genotypically = 25% AB, 25% BO, 25%AO and 25% OO.

Note:	The offspring that is Phenotypically AB is the Genotypic A,B

example for co-dominance.

4.	Exception #2: Incomplete Dominance

a.	Take the scenario of Sickle Cell Disease vs. Sickle Cell Trait vs. Normal Hemoglobin (Hb).

1)	There are 3 ways that this can be expressed Phenotypically and Genotypically:

Person has Sickle Cell Disease: ss (American Association of Blood Banks nomenclature)

Person has Sickle Cell Trait:	Ss (AABB nomenclature)

Person has Normal Hb:	SS (AABB nomenclature)

2)	A person with Sickle Cell Disease MUST inherit s from both parents; a person with Sickle Cell Trait will inherit one s from a parent and one normal S from the other parent; a person with normal hemoglobin will NOT inherit s from either parent.

b.	Let’s consider a cross between Ss x Ss

Off-spring would be:

Female	Phenotypically = 25% Sickle Cell Ds; 50% Sickle Cell Trait; and

Male	S	s
S	SS	Ss
s	Ss	ss

25 % Normal hemoglobin.

Genotypically = 25% SS; 50% Ss, and 25% ss.

LAB QUIZ #5

5.	Exception #3: Sex Linked Inherited Disorders MUST Always Express Gene with X or Y

a.	Take the scenario of Hemophilia Disease.

1)	There are 3 ways that this can be expressed Phenotypically and Genotypically:

Male has Hemophilia Disease: XhY (allele is superscript of X gene)

Female is carrier for trait: XHXh (alleles are superscript of X genes)

Male nor Female has disease or trait: XHY or XHXH (alleles superscript of X genes)

NOTE: the Y chromosome/gene does NOT have any alleles (H or h) assigned, why???

2)	A male child with Hemophilia Disease MUST inherit Xh from the female parent; a female with Xh is considered a carrier for the Hemophilia gene.

3)	In this example, sex of the child is critical when calculating percent for disease or carrier; ONLY males can have the disease and ONLY females can be carriers.

b.	Female	Cross between XhY x XHXH

Male	XH	XH
Xh	XH Xh	XH Xh
Y	XH Y	XH Y

The offspring Phenotypically are 2 girls who are normal; 2 boys who are normal = no bleeding disorders.

The offspring Genotypically are 2 girls who are carriers,

XH Xh and 2 boys who are normal with XH Y.

Therefore, 100% of the male children (XH Y) will be normal, but 100% of all the female (XH Xh ) children will be carriers for Hemophilia gene.

c.	Female	Cross between XhY x XHXH; Work this problem for offspring

Male	XH	Xh
XH	XH XH	XH Xh
Y	XH Y	Xh Y

Genotype =

Phenotype =

6.	16 Square Punnet: TtYy x TtYy cross; follow the Rules #8, #9, #10 listed above

Tall = T	FOIL Rules apply to this genetic problem (why?)

Dwarf = t	TtYy = First alleles = TY	TtYy = Outer alleles = Ty

Yellow = Y

Green = y	TtYy = Inner alleles = tY	TtYy = Last alleles = ty

♂	♀TY	♀Ty	♀tY	♀ty
♂TY	TTYY	TTYy	TtYY	9 = hetero dom/dom Tall/Yellow 3 = hetero dom/recess Tall/Green 3 = homo rec/dom Short/Yellow 1 = homozygous recessive Short/Green TtYy
♂Ty	TTYy	TTyy	TtYy	Ttyy
♂tY	TtYY	TtYy	ttYY	ttYy
♂ty	TtYy	Ttyy	ttYy	ttyy

LAB QUIZ #5

16 Square Punnet: TtYy x ttyy cross; follow the rules listed above and work problem

Tall = T	FOIL Rules apply to this genetic problem but NOT the 9:3:3:1 ratio (why?)

Dwarf = t	♂ = TtYy = F = TY	♂ = TtYy = O = Ty

Green = y	♂ = TtYy = I = tY	♂ = TtYy = L = ty

Yellow = Y

♂	♀ty	♀ty	♀ty	♀ty
♂TY
♂Ty
♂tY
♂ty

Genotype of off-spring: You can represent %, ratio (1:2) or fraction (½ ), just be consistent

Phenotype of offspring: You can represent %, ratio (1:2) or fraction (½ ), just be consistent

Solutions to Genetics Problems:

1. Genotype Gametes

WW W, W

Ww W, w

ww	w, w

Ww X ww	w	w

W	Ww	Ww

w	ww	ww

Genotype: 50% heterozygous, 50% homozygous recessive; Phenotype: 50% white, 50% yellow

W X Ww	W	w

W	WW	Ww

W	WW	Ww

Genotype: 50% homozygous dominant, 50% heterozygous; Phenotype: 100% white

WW X ww	w	w

W	Ww	Ww

W	Ww	Ww

Genotype: 100% heterozygous; Phenotype: 100% white

Ww X Ww	W	w

W	WW	Ww

w	Ww	ww

Genotype: 25% homozygous dominant, 50% heterozygous, 25% homozygous recessive

Phenotype:	75% white, 25% yellow

2.	A. Genotype of woman:	XH XH	Genotype of man:	XH Y

XH	XH

XH	XH XH	XH XH

Y	XH Y	XH Y

0% of children with hemophilia

B.Genotype of man with hemophilia: XhY

XH XH

Xh	XHXh	XHXh

Y	XHY	XHY

0% children will have hemophilia

100% of daughters will be carriers

3.	Genotype of man = Bb	Genotype of woman = bb

b	b

B	Bb	Bb

b	bb	bb

Genotype:	50% heterozygous, 50% homozygous recessive

Phenotype:	50% brown, 50% blue

4. Father: BB	Mother: bb	Children: Bb

5.

Blood Type – Phenotype Genotypes
A + IAIARR, IAIARr, IAiRR IAiRr
A- IAIArr, IAirr
B+ IBIBRR, IBIBRr, IBiRR IBiRr
B- IBIBrr, IBirr
AB+ IAIBRR, IAIBRr
AB- IAIBrr
O+ iiRR, iiRr
O- iirr

IA	IB

i	IA i	IBi

i	IA i	IBi

Genotype: 100% heterozygous (50% heterozygous for type A and 50% heterozygous for type B)

Phenotype: 50% type A blood, 50% type B blood

6.	Phenotype	Genotype

Normal hemoglobin	SS

Sickle cell anemia	ss

Sickle cell trait	Ss

S	s

S	SS	Ss

s	Ss	ss

Genotype:	25% homozygous dominant

50% heterozygous

25% homozygous recessive

Phenotype:	25% normal hemoglobin

50% carriers of sickle trait

25% sickle cell anemia