Question

I marked the correct answers to these questions, but I just want to know how to solve them.

1) In a cross of AaBbCcDdEeFf X AaBbccDdEeFf, what proportion will have the ABCDeF phenotype?

A. 27/64

B. 27/128

C. 27/512

D. 81/512

E. 81/2048 ####

2.) In a cross of two flies +/vg Cy/+ +/se +/ab X +/vg +/+ se/se ab/ab what proportion of the offspring will be mutant in phenotype for all four markers?

A. 0

B. 3/64

C. 1/16

D. 1/32 ####

E. 1/64

3) A cross of two heterozygous individuals produces 74 dominants and 26 recessives. What is the chi-square value for these results?

A. 8/25

B. 16/75

C. 4/75 ####

D. 8/75

E. 12/25

4) Shell orientation in snails is due to a maternal effect gene. A true breeding sinistral (recessive) is crossed to a true breeding Dextral (dominant). The offspring from that cross are self-crossed. What will be the expected ratio of shell types?

A. All sinistral

B. All Dextral ####

C. Half sinistral, half Dextral

D. ¾ Dextral, ¼ sinistral

E. ¾ sinistral, ¼ Dextral

19) In peas, axial (A) flower position is dominant to terminal (a), tall (L) is dominant to short (l), and yellow (Y) is dominant to green (y). If a plant that is heterozygous for all three traits is allowed to self-fertilize, how many of the offspring would be dominant for all three traits?

A) 3/64

B) 9/64

C) 27/64 ####

D) 32/64

E) 64/64

20) The product H substance is needed to express the blood antigen. A mating of IA I B Hh X IA I B hh should produce what ratio of blood types?

a) 1/4 AB, 1/8 A, 1/8 B, 1/2 O ####

b) 3/8 AB, 3/16 A, 3/16 B, 1/4 O

c) 1/4 AB, 3/16 A, 3/16 B, 3/8 O

d) 3/16 AB, 3/8 A, 3/16 B, 1/4 O

e) 1/8 AB, 1/4 A, 1/8 B, 1/2 O

Answer 1

Answer:

1). E). 81/2048

Explanation:

Aa x Aa = A_ (3/4) & aa (1/4)

Bb x Bb = B_ (3/4) & bb (1/4)

Cc x cc = Cc (1/2) & cc (1/2)

Dd x Dd = D_ (3/4) & dd (1/4)

Ee x Ee = E_ (3/4) & ee (1/4)

Ff x Ff = F_(3/4) & ff (1/4)

Proportion of ABCDeF phenotype = A_B_C_D_eeF_ = ¾ * ¾ * ½ & ¾ * ¼* ¾ = 81/2048

2). D. 1/32

Explanation:

+/vg x +/vg = +_ (3/4) & vg/vg (¼)

Cy/+ x ++ = +/+ (1/2) & Cy/+ (1/2)

+/se x se/se = +/se (1/2) & se/se (1/2)

+/ab x ab/ab = +/ab (1/2) & ab/ab (1/2)

Proportion of the offspring will be mutant in phenotype for all four markers = vg/vg Cy/+ se/se ab/ab = ¼ * ½ * ½ * ½ = 1/32

3). C). 4/75

Explanation:

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
Dominant	74	75	-1	1.0000	0.0133
Recessive	26	25	1	1.0000	0.0400
Total	100	100			0.0533

0.0533 = 4/75

4). B. All Dextral

19). C). 27/64

Explanation:

AaLlYy x AaLlYy –Parents

Aa x Aa = A_ (3/4) & aa (1/4)

Ll x Ll = L_ (3/4) & ll (1/4)

Yy x Yy = Y_(3/4) & yy (1/4)

A_L_Y_ = ¾ * ¾ * ¾ = 27/64

20). a) 1/4 AB, 1/8 A, 1/8 B, 1/2 O

Explanation:

IA I B Hh X IA I B hh

IAIB x IAIB = IAIA (1/4) IAIB (1/2) & IBIB (1/4)

Hh x hh = Hh (1/2) & hh (1/2)

A type = IAIA Hh = ¼ * ½ = 1/8

B type = IBIB Hh = ¼ * ½ = 1/8

AB type = IAIB Hh = ½ * ½ = ¼

O type = 1 – ( 1/8 + 1/8 + ¼ ) = ½