# Problem PowerPoint Concepts in Genetics End of Chapter 4

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Problem PowerPoint Concepts in Genetics End of Chapter 4
Honors Genetics - Exam

Type of Problem Epistasis

Concepts in Genetics 16 Pigment in mouse fur is only produced when the C allele is present. Individuals of the cc genotype are white. If c olor is present it may be determined by the A,a alleles AA, results in agouti color, while aa results n black fur.

16(a) What F1 and F2 genotypic and phenotypic ratios are obtained from a cross between AACC and aacc mice F1 = AaCc = agouti F2 = 9 agouti,4 white, 3 black

16(b) 1. How do you get 8 Agouti and 8 white AACc x AAcc
How do you get 9 agouti, 10 black aaCC x AaCC How do you get 4 agouti, 5 black, and 10 white AaCc x aacc

Problem 23. Five human matings( 1-5) including both maternal and paternal phenotypes for ABO and MN blood group anitgen status, are shown in the following table:

Parental Phenotypes ( 1) A, M x A, B (2) B,M x B, M (3) O,N x B, N
(4) AB,M x O,N (5) AB, MN x AB,MN

Offspring – Matches? A, N O,N O, MN B, M B, MN

Matches ( 1) = c (2) = d ( 3) = b ( 4) = e (5) = a

Color Vision- Problem 24 RG – normal vision Rg- color blind
A husband and wife have normal vision , although both of their fathers are red-green color blind, an inherited X linked recessive condition. What is the probability that their first child will be

Results Mother’s father: Xrg/Y Father’s Father: Xrg/Y Mother: XRGXrg

Explanation Notice that the mother must be heterozygous for the rg allele( being normal and having inherited an Xrg from her father and the father because he has normal vision, must be XRG

Problems - 25 In humans, the ABO blood type is under the control of autosomal alleles. Color blindness is a recessive X-linked trait. It two parents who are both type S and have normal vision produce a son who is color blind and is type O, what is the probability that their next child will be female with normal vision?

Based on the son who is colorblind and the blood type O
XRGXrg, IAIO ,mother Xrg/Y , father

Solutions - 25 The probability of having a female child is ½
The probabilty that she has normal vision is 1 ( because the father’s X is normal) Probability of type blood is 1/4

Problem 10 Fur color C ( full color) cch ( chinchilla)
ch ( himalayan ) ca ( albino )

Problem 10 Multiple alleles and rabbit fur Himalayan x Himalayan
Albino chca x chca

Problem 10 Full color x albino chinchilla Ccch x caca cchca

Problem 1 – Incomplete dominance
In shorthorn cattle, coat color may be red, white, or roan. Roan is an intermediate phenothype expressed as a mixture of re and white hairs Red x white roan RR x WW RW

Mixed Problem – Incomplete dominance and dominant and recessive
In radishes, flower color may be red, pruple or white. The edible protion of the raidsh may be long or oval. When only flower color is studied not dominance is evident. Red x White yields all Purple. If these F1 purples are interbred, the F2 generation consists of ¼ red:1/2 purple:1/3 white. Regarding radish shape , long is dominant to oval in a Mendelian fashion

Genotypes RR = red RW = purple WW = white O_ = long Oo = oval

Problem based on radish facts
If a red X unknown produced 204 red and 198 purple what is the genotype of the unknown plant? What would the cross of a RWOo plant x RW oo plant produce( just phenotypes)

Problem 17 In some plants, a red pigment cyanidin is synthesized by a colorless precursor. The addition of a hydroxyl group ( OH) to the molecule causes it to become purple

Epistasis Colorless is aa Purple is A_ B_ Red is A_bb
In a cross between two purple plants the following results were obtained

Results 94 purple 31 red 43 white
This is approximately what epistatic ratio? What does this tell you about the genotype of the original purple plants?

Answer The plants were both A_B_ x A_B_

Problem 26 The X linked recessive mutation, scalloped causes irregular wing margins in Drosophila melanogaster, the fruit fly. What occurs if a scalloped female is crossed with a normal male in the F1.

F1 Symbol for singed in Fruit flies is sd XsdXsd x X+Y – Parental
F1=1/2 XsdX+ This is a female with normal long wings 1/2 XsdY – This is a male with scalloped wings that have irregular edges

F2 - XsdX+ x XsdY Results in ¼ X+X+ ( normal female)
¼ XsdXsd ( scalloped wing female) ¼ X+Y (normal male) ¼ XsdY( male scalloped wings)

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