Tutorial #3 Question #5

Eye color of the Oriental fruit fly (Bactrocera dorsalis) is determined by a number of wild-type eyes. The F1 were interbred to produce an F2 of 180 wild-type, 59 amethyst eyes (a bright sparkling blue) and 81 yellow eyes.

(a) What modified dihybrid ratio does this represent?(b) What is the name for this dihybrid gene interaction?(c) Give the genotypes for all the flies in the P, F1, and F2

generations.(d) What proportion of the F2

i) wild-type flies are homozygous?ii) yellow flies are homozygous?iii) amythest flies are homozygous?

(a) What modified dihybrid ratio does this represent?

F2 progeny:

180 wt : 59 amethyst : 81 yellow

(b) What is the name for this dihybrid gene interaction?

18020

5920

8120

9 3 4: :

Divide by the least common denominator

9 A_ B_ 9 wt

3 A_ bb 3 amethyst

3 aa B_4 yellow

1 aa bb

Recessive epistasis:homozygous recessive phenotype of

a gene masks the phenotype of a 2nd gene

(c) Give the genotypes for all the flies in the P, F1, and F2 generations.

wt eyes × yellow eyes

wt eyes

9 wt eyes : 3 amethyst eyes : 4 yellow eyes

P:

F1:

F2:

Aa Bb

AA BB aa bb

A_ B_ A_ bb aa B_aa bb

(d) What proportion of the F2

i) wild-type flies are homozygous?

ii) yellow flies are homozygous?

iii) amythest flies are homozygous?

AA BB

1/3

2/3

1/3

2/31/9 AA BB PRODUCT RULE: the

probability of 2 or more independent events occurring together

aa BB

AA Bb Aa BB Aa Bb

aa Bb

BB

Bb1

bb

1/4

2/4

1/4

aa bb

1/4 aa BB

1/4 aa bb

AA

Aa

BB

Bb

BB

Bb

aa

2/4

SUM RULE: the probability of any 1 of 2 or more mutually exclusive events occurring

AA bb Aa bb

1/3

1 1/3 AA bb

2/3

BB

Bb

bb

BB

Bb

bb

AA

Aa

Tutorial #3 Question #6

Prune (pn) is a recessive eye-color mutant located on the X chromosome of D. melanogaster. An autosomal dominant gene, Prune-killer (K), causes death of all prune larvae. If a cross is made between a pn/pn+ ; K/K+ female and a pn+/Y ; K/K+ male, what will be the sex ratio of the progeny and what will be the frequencies of wild-type and prune phenotypes among the adult progeny?

pn/pn+ ; K/K+ × pn+/Y ; K/K+

pn+/pn+

pn+/Y

pn/Y

pn/pn+

1/4

1/4

1/4

1/4

K+/K+

K+/K

K/K

K+/K+

K+/K

K/K

K+/K+

K+/K

K/K

K+/K+

K+/K

K/K

1/4

2/4

1/4

1/4

2/4

1/4

1/4

2/4

1/4

1/4

2/4

1/4

pn+/pn+ ; K+/K+

pn+/pn+ ; K+/K

pn+/pn+ ; K/K

pn/pn+ ; K+/K+

pn/pn+ ; K+/K

pn/pn+ ; K/K

pn+/Y ; K+/K+

pn+/Y ; K+/K

pn+/Y ; K/K

pn/Y ; K+/K+

pn/Y ; K+/K

pn/Y ; K/K

1/16

2/16

1/16

1/16

2/16

1/16

1/16

2/16

1/16

1/16

2/16

1/16

8/16 wt females

4/16 wt males

3/16 males die

1/16 prune males

Sex ratio = 8 females : 5 males Phenotypic ratio = 12/13 wt : 1/13 prune

Tutorial #3 Question #11

Fat mice can be produced by 2 independently assorting genes. The recessive genotype o/o produces a fat sterile mouse called “obese”. Its dominant allele produces normal growth. The recessive genotype a/a also produces a fat sterile mouse called “adipose” and its dominant allele produces normal growth. What phenotypic proportions of fat vs. normal would be expected among the F1 and F2 from parents of the genotype O/o ; A/a?

O/o ; A/a × O/o ; A/a

9 O_ ; A_ Normal

3 O_ ; aa Adipose + sterile

3 oo ; A_ Obese + sterile

1 oo ; aa Adipose + obese + sterile

9 normal : 7 fat + sterile

P:

F1:

F2:

F1 intercross only involves the normal O_ ; A_ progeny:

OO ; AAOO ; Aa

Oo ; AA

Oo ; Aa

1/92/9

2/9

4/9

1/9 OO ; AA 2/9 OO ; Aa 2/9 Oo ; AA 4/9 Oo ; Aa

1/9 OO ; AA 1/81 × 0 2/81 × 0 2/81 × 0 4/81 × 0

2/9 OO ; Aa 2/81 × 0 4/81 × 1/4 4/81 × 0 8/81 × 1/4

2/9 Oo ; AA 2/81 × 0 4/81 × 0 4/81 × 1/4 8/81 × 1/4

4/9 Oo ; Aa 4/81 × 0 8/81 × 1/4 8/81 × 1/4 16/81 × 7/16

Proportion of fat mice = 1/81 + 2/81 + 1/81 + 2/81 + 2/81 + 2/81 + 7/81 = 17/81

Proportion of normal mice = 1 - 17/81 = 64/81

Chapter 3 Question #30

In cats, curled ears result from an allele (Cu) that is dominant over an allele (cu) for normal ears. Black colour results from an independently assorting allele (G) that is dominant over an allele for gray (g). A gray cat homozygous for curled ears is mated with a homozygous black cat with normal ears. All the F1 cats are black and have curled ears.

(a) If 2 of the F1 cats mate, what phenotypes and proportions are expected in the F2?

(b) An F1 cat mates with a stray cat that is gray and possesses normal ears. What phenotypes and proportions of progeny are expected from this cross?

(a) If 2 of the F1 cats mate, what phenotypes and proportions are expected in the F2?

homozygous curled gray × normal homozygous black

All curled black

9 curled black : 3 normal black : 3 curled gray : 1 normal gray

P:

F1:

F2:

Cu/cu G/g

Cu/Cu g/g cu/cu G/G

Cu/_ G/_ cu/cu G/_ Cu/_ g/g cu/cu g/g

(b) An F1 cat mates with a stray cat that is gray and possesses normal ears. What phenotypes and proportions of progeny are expected from this cross?

heterozygous curled black × normal gray stray

1 curled black : 1 curled gray : 1 normal black : 1 normal gray

P:

F1:

Cu/cu G/g cu/cu g/g

1 Cu/cu G/g 1 Cu/cu g/g 1 cu/cu G/g 1 cu/cu g/g: : :

Chapter 3 Question #31

The following 2 genotypes are crossed :

Aa Bb Cc dd Ee × Aa Bb Cc Dd Ee

What will the proportion of the following genotypes be among the progeny of this cross?

(a) Aa Bb Cc Dd Ee

(b) Aa bb Cc dd ee

(c) aa bb cc dd ee

(d) AA BB CC DD EE

P = ½ × ½ × ½ × ½ × ½ = 1/32

P = ½ × ½ × ½ × ½ × ¼ = 1/64

P = ¼ × ½ × ¼ × ½ × ¼ = 1/256

P = 0

Chapter 3 Question #38

In the California poppy, an allele for yellow flowers (C) is dominant over an allele for white flowers (c). At an independently assorting locus, an allele for entire petals (F) is dominant over an allele for fringed petals (f). A plant that is homozygous for yellow and entire petals is crossed with a plant that is white and fringed. A resulting F1 plant is then crossed with a plant that is white and fringed, and the following progeny are produced: 54 yellow and entire, 58 yellow and fringed, 53 white and entire, and 10 white and fringed.

(a) Use a chi-square test to compare the observed numbers with those expected for the cross.

(b) What conclusion can you make from the results of the chi-square test?

(c) Suggest an explanation for the results.

homozygous yellow entire × white fringed

yellow entire petals × white fringed

54 yellow entire : 58 yellow fringed : 53 white entire : 10 white fringed

Ho : There is no significant difference between the observed and expected results. OR

The observed results occur in a 1 : 1 : 1 : 1 ratio, according to Mendelianindependent assortment.

Ha : The observed results are significantly different from the expected results.OR

The observed results do no occur in a 1 : 1 : 1 : 1 ratio, according to Mendelian independent assortment.

P:

F1:

CC FF cc ff

¼ Cc Ff ¼ Cc ff ¼ cc Ff ¼ cc ff

Cc Ff cc ff

F2:

Observed Expected (O-E)2/E

Cc Ff 54 43.75 (54 – 43.75)2/43.75 = 2.40

Cc ff 58 43.75 (58 – 43.75)2/43.75 = 4.64

cc Ff 53 43.75 (53 – 43.75)2/43.75 = 1.96

cc ff 10 43.75 (10 – 43.75)2/43.75 = 26.0

Total 175 175 2 = 35

df = 4 – 1 = 3

P << 0.005

Ho : There is no significant difference between the observed and expected results.

OR

The observed results occur in a 1 : 1 : 1 : 1 ratio, according to Mendelian independent assortment.

Ha : The observed results are significantly different from the expected results.

OR

The observed results do no occur in a 1 : 1 : 1 : 1 ratio, according to Mendelian independent assortment.

Possible explanation? Poppies are sublethal when homozygous.

Chapter 5 Question #15

Assume that long ear lobes in humans are an autosomal dominant trait that exhibits 30% penetrance. A person who is heterozygous for long ear lobes mates with a person who is homozygous for normal ear lobes. What is the probability that their first child will have long ear lobes?

Heterozygous long ear lobes × homozygous normal ear lobes

long ear lobes × normal ear lobes

P:

F1:

L/l l/l

½ L/l ½ l/l

Penetrance (long ear lobes) = 30% = 0.3

P(1st child is L/_) = 0.5 × 0.3 = 0.15 = 15%

Chapter 5 Questions #18

As discussed in the introduction to this chapter, Cuénot studied the genetic basis of yellow cat color in mice. He carried out a number of crosses between 2 yellow mice and obtained what he thought was a 3:1 ratio of yellow to gray mice in the progeny. The following table give Cuénot’s actual results, along with the results of a much larger series of crosses carried out by Castle and Little (W.E. Castle and C.C. Little. 1910. Science 32:858–870).

Investigators YellowProgeny

Non-yellow progeny

Total progeny

Cuénot 263 100 363

Castle and Little 800 432 1235

Both Combined 1063 535 1598

Observed Expected (O-E)2/E

Yellow 263 272.25 (263 – 272.25)2/272.25 = 0.31

Non-yellow 100 90.75 (100 – 90.75)2/90.75 = 0.94

Total 363 363 2 = 1.25

df = 2 – 1 = 1

a) Using chi-square test, determine whether Cuénot’s results are significantly different from the 3:1 ratio that he thought he observed. Are they different from a 2:1 ratio?

Test 3:1 ratio:

Ho: Cuénot’s results are not significantly different from a 3:1 ratio.Ha: Cuénot’s results are significantly different from a 3:1 ratio.

0.5 > P > 0.1Therefore, we accept the null hypothesis that Cuénot’s results do not differ significantly from a 3:1 ratio.

Observed Expected (O-E)2/E

Yellow 263 242 (263 – 242)2/242 = 1.8

Non-yellow 100 121 (100 – 121)2/121 = 3.6

Total 363 363 2 = 5.4

df = 2 – 1 = 1

Test 2:1 ratio:

Ho: Cuénot’s results are not significantly different from a 2:1 ratio.Ha: Cuénot’s results are significantly different from a 2:1 ratio.

0.025 > P > 0.01Therefore, we reject the null hypothesis that Cuénot’sresults do not differ significantly from a 2:1 ratio.

Expressivity vs. Norm of Reaction

Expressivity: degree to which a trait is expressed.

Norm of reaction: range of phenotypes produced by a particular genotype in different environmental conditions.

Calculating Allelic Frequency Changes due to Selection

Using these values …

You could also enter your values at the beginning and calculate for WAA.