Genetics Test 2 Flashcards

Question

Genetic heterogeneity

Answer 1

at least two alleles create similar or same phenotype divided into: locus heterogeneity, allelic heterogeneity, and cliincal heterogeneity

Answer 2

alleles from two genes cause a phenotype/ mutation at a different gene causes a similar phenotype e.g. epistasis Myo7A, Harmonin, Cadherin 23 all disrupt Steriocilia in retina causing Usher Syndrome 1. For example, retinitis pigmentosa has autosomal dominant, autosomal recessive, and X-linked origins. However, only one mutant locus is needed for the phenotype to manifest.

Answer 3

allelomorphy-alleles at same location (gene) cause phenotype different mutations at the same locus causes a similar phenotype. For example, β-thalassemia may be caused by several different mutations in the β-globin gene.

Answer 4

penetrance and expressivity mutation within the same gene causes a different phenotype.

Answer 5

Common Recessive=Dom.

Answer 6

Incomplete Penetrance

Answer 7

Variable Expressivity

Answer 8

Paternal Imprinted Auto Dom

Answer 9

Maternal Imprinted Auto Dom

Answer 10

X Linked, Dom Hemis Abort

Answer 11

X Linked Inbred “M to M”

Answer 12

Novel Mutation

Answer 13

• **The Westermarck Effect**-not liking people you were raised in close proximity to. BUT, 1st and 2nd cousins tend not to experience WE. • Women tend to prefer males with alleles in common to their f**ather’s Human Leukocyte Antigen/MHC.** Men with 2-7 alleles in common are preferred. **• 2nd order and above relatives**…We tend not be raised in close proximity/and they resemble us in ways that may contribute to mate choice.

Answer 14

(1/2)^n x (1+F) n= number of individuals involved in the path F= inbreeding coefficient for the common ancestor http://pawpeds.com/pawacademy/genetics/genetics/inbreeding.html

Answer 15

•To many genes we stop using Mendel and look at Means, Variance and Standard Deviation of “Normal Distribution” (a graph of Observations)

Answer 16

= A Groups Total Average Distance from Mean measures how far a set of numbers is spread out. (A variance of zero indicates that all the values are identical.)

Answer 17

= Range we find 68% (called 1 SD) of data points, 95% (is 2SDs), 99.7% (is 3SDs) A measure of the dispersion of a set of data from its mean. The more spread apart the data, the higher the deviation.

Answer 18

``` _Sum of (each value-mean)^2_ # of values – 1 ``` \*Answer in X units squared

Answer 19

Gene Variance/Total Variance

Answer 20

Square Root of Variance \*Answer in X units –Shows us the width of the curve at around the mean where 68%, 95% and 99.7% of data is included

Answer 21

Child 1 of Diseased Parents is more likely to acquire a disease than Child 2 of Undiseased Parents of a population if the disease is at all genetic. Less overlap of graphs = greater genetic influence on the disease = lesser environmental influence Falconer postulated the existence of a threshold. Embryos whose susceptibility exceeds a critical threshold value develop cleft palate; those whose susceptibility is below the threshold, even if only just below, avoid cleft palate. (polygenic)

Answer 22

polygenic, continuous Variation like height, skin color or Crop Yield

Answer 23

The trait is there or it’s not

Answer 24

–Large Population •Prevents Genetic Drift-change in allele freq. due to: –Bottle Neck Effect eg. Killing off of homozygous people –Founder Effect eg. Pickett’s infamous African Diaspora –Genetic drift leads to “Fixation” allele freq.=100% –Random Mating –No Mutations –No Natural Selection –No immigration into population (No Genetic Flow)

Answer 25

•Hardy Weinberg Distribution- Occurs when alleles freq is constant = No occuring evolution •Failure to meet regulation means allele freq has changed and evolution has occurred.

Answer 26

**p + q = 1** –p = % Dominant alleles (A) in population –q = % of recessive alleles (a) in population **p2 + 2pq + q2 = 1** –p2 = # AA individuals –2pq = # Aa individuals –q2 = # aa individuals

Answer 27

**Practice work/answer** * q^2 = 64/100 = .64 * q = .80 * p = -q – 1 = .80 -1 = .20 * Dominant allele freq = .20 * Brown cows = 2pq * 2 x .20 x .80 = 32 brown cows

Answer 28

A and B are In Coupling Orientation

Answer 29

a and b are In Coupling Orientation

Answer 30

A and B are **In Repulsion** to a and b

Answer 31

A is **‘In Cis’** to B, a is ‘**In Cis’** to b

Answer 32

A is ‘**In Trans’** to b AND to a B is **‘In Trans’ t**o a AND to b

Answer 33

-The Exchange of Genetic Material By Homologous Chromosomes During Meiotic Prophase. -Changes the Cis-Trans Orientation

Answer 34

Recombinant Chromosomes are **Non-Parenta**l, Because the Parent Does Not Have This **Cis-Trans Orientation.**

Answer 35

**“Non-sister”** recombination is genetically detectable

Answer 36

**“Sister-sister”** recombination is NOT detectable

Answer 37

**F1 X F1** 9:3:3:1 Dihybrid Ratio Weaker linked genes recombine more Decreasing parental haplotypes= Less Parental and more Recombinant Gamete Frequencies More like **dihybrid cross** (cross between F1 offspring (first-generation offspring) of two individuals that differ in two traits of particular interest. For example, BB × bb)

Answer 38

Ratio Moves Toward 3:1 Stronger linked genes do not recombine as often, keeping Parental haplotypes together = More Parental and Less Recombinant Gamete Frequencies More like **Monohybrid cross** (mating between individuals who have different alleles at one genetic locus of interest)

Answer 39

Measuring Genetic Linkage Violates **Mendel’s Law of Independent Assortment** because **Multiple genes move as one unit (like single gene)**

Answer 40

•Recombination Fraction (θ) -- recombs/total chances to recomb \*\*\*Two are basically the same thing in the end: •θ = r = total of both recombinants

Answer 41

•Frequency of Gametic Recombination (r) – recomb gametes/total gametes \*\*\*Two are basically the same thing in the end: •θ = r = total of both recombinants

Answer 42

λ1 + λ2 = r (total recombinants) –assuming recombinants are equal (they should be)

Answer 43

= r –= ½ r X ½ r = chances receiving two recombinants

Answer 44

* Each couple makes only few babies * Cis-Trans orientation hard to determine from pedigree * Diseases are usually rare * Overall we do not have much data to work with

Answer 45

= How we decide a rare disease is linked or not linked The LOD score compares the likelihood of obtaining the test data if the two loci are indeed linked, to the likelihood of observing the same data purely by chance. Positive LOD scores favor the presence of linkage, whereas negative LOD scores indicate that linkage is less likely. Likely linked \> 3........-2 \> Likely Independently Assorted \*\*\*Requires we know Cis-Trans Orientations (Phase Known)

Answer 46

•Because LODs are Logs they are **additive** –This way we can add data from different families and create a statistical pool of data for a disease. –When the pool total becomes **\>3** we can say the disease is **linked** –When **less than -2** we can say it’s **independently assorted**

Answer 47

Recombination is not uniformly distributed along a chromosome. But usually broadly enough distributed. However there is a tendency for linked genes to be inherited as haplotypes and so hotspots can tend to cluster alleles of linked genes

Answer 48

Mapping using many molecular markers and the disease allele can provide higher resolution. Many markers have been mapped against each other and placed on the sequence map…so they form a useful marker framework to position new loci. Three point mapping is very powerful because in humans flanking double recombinants on a dense marker map are extremely rare.

Answer 49

Rare recombinants in small families can be very powerful tools despite low contribution to Z. With dense marker maps a recombination breakpoint can provide the “maximal physical extent” of a genetic locus.

Genetics Test 2 Flashcards

(83 cards)