unit 5

Autosomal Dominate

1. Appears in both sexes with equal frequency

2. Does not skip generations/ Affected offspring must have an affected parent, unless they possess a new mutation which you never assume on a pedigree question.

3. When one parent is affected (heterozygous) and the other parent is unaffected, approximately ½ of the offspring will be affected

Autosomal recessive

1. Appears in both sexes with equal frequency

2. Can skip generations / affected offspring can be born to unaffected parents

3. When both parents are heterozygous, approximately ¼ of the offspring will be affected

X-Linked Dominate

1. More females than males are affected

2.  Does not skip generations

3.   Affected sons must have an affected mother

4.   Affected fathers will pass the trait on to all their daughters

5.   Affected mothers (if heterozygous) will pass the trait on to ½ of their sons and ½ of their daughters

X-Linked Recessive

1.     More males than females are affected

2.     Affected sons are usually born to unaffected mother; so the trait skips generations

3.     Approximately ½ of a carrier mother’s sons are affected

4.     It is never passed from father to son without mom being affected

5.     All daughters of affected fathers are carriers

Y-Linked

1.     Only males are affected

2.     It is passed from father to all sons

3.     It does not skip generations

chiasma

the location where the chromosome arms are physically laying over on top of each other during crossing over.

AABB x aabb

100% AaBb

aabb x aabb

100% aabb, parental phenotype

AABB x AABB or AaBB x AABb

both of these that all offspring will have dominant versions of both traits (100% parental phenotype)

AaBb x AaBb

this is a dihybrid cross. If the genes are on separate chromosomes, the phenotype ratio will be 9:3:3:1.

AaBb x aabb

this is a test cross. If the genes are on separate chromosomes, the phenotype ratio is 1:1:1:1.

AaBb x aaBb or AaBb x Aabb

 if the genes are on separate chromosomes, the phenotypic ratio will be 3:3:1:1

 recombinant offspring

dont have the parental phenotype

If A and B are mutually exclusive, then

P ( A or B ) = P ( A ) + P ( B )

If A and B are independent, then:

P ( A and B ) = P ( A ) × P ( B )

In the cross AAbbcc x aaBbCc, what is the chance that the offspring will be AabbCc? 

0.25

For the individual AaBbccDD, what is the chance of it producing the gamete AbcD?

A is a probability of 0.5

b is 0.5

c is 1

D is 1

0.5 x 0.5 x 1 x 1 = 0.25

non-nuclear inheritance

genes that aren’t in the nucleus. 

traits determined by the mitochondrial DNA are

maternally inherited

null hypothesis

“There is no difference between observed and expected outcomes”

degrees of freedom

number of classes — 1

should you ACCEPT or REJECT the null hypothesis

Accept if critical value is under and reject if critical value is over the chi square value