A married couple both has normal color vision, but their daughter has red-green

Question

A married couple both has normal color vision, but their daughter has red-green colorblindness, a X-linked recessive trait. The husband sues the wife for divorce on grounds of infidelity. Can genetics provide evidence supporting his case? Explain. Joe Bob has unattached earlobes like his father, but his mother has attached earlobes Give the genotypes of each of the three. Unattached ear lobes are dominant to attached earlobes. IN guinea pigs, black hair (B) is dominant to white (b). Two black guinea pigs of the same genotype were mated and produced 29 black and 9 white offspring. What would you predict the genotypes o1 the patents to be? Explain. Two individuals are both heterozygous for a single pair of genes (Tt). During the course of there marriage, they produced 8 offspring. Approximately how many (give me a number, not a percentage) of the offspring can be expected to display the dominant phenotype?

Explanation / Answer

Answer:

21). Yes, the red-green colorblind daughter is not the child of the couple as they both have normal color vision.

Explanation:

To produce colorblind daughter, father must be colorblind and mother must either carrier or colorblind.

XC = normal

Xc = colorblindness

Normal father x Normal mother (carrier)

XC Y   x   XC Xc

XC

Xc

XC

XC XC (normal female)

XC Xc (normal, carrier female)

Y

XC Y (normal male)

Xc Y (colorblind male)

22).

Uu= Unattached earlobe, father

uu = Attached earlobe, mother]

Uu= Unattached earlobe, Joe Bob

Explanation:

Unattached = U

Attached = u

Uu (unattached, father) x (attached, mother) uu ---------Parents

u

U

Uu (unattached earlobe)

u

uu (attached earlobe)

23).

Black pig parent genotype = Bb

Explanation:

Bb X Bb---Parents

B

b

B

BB (black)

Bb (black)

b

Bb (black)

bb (white)

24). Dominant phenotype = 6

Explanation:

Tt x Tt ---Parents

T

t

T

TT (dominant)

Tt (dominant)

t

Tt (dominant)

tt (recessive)

Dominant = ¾

Recessive = ¼

Out of 8, dominant progeny = ¾ * 8 = 6

Out of 8, recessive progeny = ¼ * 8 = 2

XC

Xc

XC

XC XC (normal female)

XC Xc (normal, carrier female)

Y

XC Y (normal male)

Xc Y (colorblind male)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.