Problem PowerPoint Concepts in Genetics End of Chapter 4

1
Problem PowerPointConcepts in GeneticsEnd of
Chapter 4

• Honors Genetics - Exam

2
Type of Problem

• Epistasis

3
Concepts in Genetics16

• 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.

4
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

5
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

6
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

7
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

8
Offspring Matches?

• A, N
• O,N
• O, MN
• B, M
• B, MN

9
Matches

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

10
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

11
Results

• Mothers father Xrg/Y
• Fathers Father Xrg/Y
• Mother XRGXrg
• Father XRG/Y

12
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

13
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?

14
Based on the son who is colorblind and the blood
type O

• XRGXrg, IAIO ,mother
• Xrg/Y , father

15
Solutions - 25

• The probability of having a female child is ½
• The probabilty that she has normal vision is 1 (
  because the fathers X is normal)
• Probability of type blood is 1/4

16
Problem 10

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

17
Problem 10

• Multiple alleles and rabbit fur
• Himalayan x Himalayan
• Albino
• chca x chca

18
Problem 10

• Full color x albino
• chinchilla
• Ccch x caca cchca

19
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

20
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 red1/2 purple1/3
  white. Regarding radish shape , long is dominant
  to oval in a Mendelian fashion

21
Genotypes

• RR red
• RW purple
• WW white
• O_ long
• Oo oval

22
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)

23
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

24
Epistasis

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

25
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?

26
Answer

• The plants were both
• A_B_ x A_B_

27
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.

28
F1

• Symbol for singed in Fruit flies is sd
• XsdXsd x XY Parental
• F11/2 XsdX This is a female with normal long
  wings
• 1/2 XsdY This is a male with scalloped
  wings that have irregular edges

29
F2 -

• XsdX x XsdY
• Results in
• ¼ XX ( normal female)
• ¼ XsdXsd ( scalloped wing female)
• ¼ XY (normal male)
• ¼ XsdY( male scalloped wings)