GeneticsJunior IB Biochemical Biology

OR

I. Mendel’s Law of Segregation

G1 of Interphase

S of Interphase

(chromosomes replicated)

• These chromosomes are homologous

• One came from the mother and one from the father

R r

R R r r

IB Review Book p23Locus: the position of a gene on a chromo-some

• The R and the r are “alleles”• Alleles are alternate forms of the same gene• Genes are pieces of DNA that code for proteins

(one gene = one protein)• This individual is HETEROZYGOUS for “R”• The diploid (2n) number is 2• The haploid (n) number is 1

R r

R R r r

•Mendel’s Law of Segregation – An individual’s two homologous chromosomes separate during anaphase I.This assures that… 1. Each gamete will carry ONE version of every single gene 2. When two gametes join, they will create a new individual with exactly TWO copies of every single gene

• r R r R• Alleles separate from each other during

gamete formation

Fertilization:

R R R r

r R r r

F1 (1st Filial) = offspring of P1

R r

R

r

P1 R r x R rThese are the two possible gametes that the father could produce – there are only 2 possible sperm: R and r, and there is an equal chance of each…

Thus, a Punnett square gives every possible offspring that could be produced between two parents

• Phenotypic Ratio (looks):

• 3 round : 1 wrinkled

• Genotypic Ratio (homo- or hetero-)

• 1 homozygous dominant : 2 heterozygous : 1 homozygous recessive

R r

R RR Rr

r Rr rr

If R is a protein that makes a seed round and r is a protein that makes a seed wrinkled…

II. Mendel’s Law of Independent Assortment• Alleles segregate in a random fashion (no

pattern, no linkage)

T R T r t R t r

T t

R r

In other words: because of the way chromosomes randomly line up during metaphase I, “T” has the same chance of ending up with “R” as it does with “r”

4 possible combinations in gametes

A. Dihybrid Cross…… of unlinked, autosomal genes

TtRr x TtRr

T R T r t R t r

T R TTRR TTRr TtRR TtRr

T r TTRr TTrr TtRr Ttrr

t R TtRR TtRr ttRR ttRr

t r TtRr Ttrr ttRr ttrr

Which are recombinants (new combinations)?

Phenotypic ratio: 9 Tall, Round : 3 Tall, wrinkled

9:3:3:1 3 short, Round : 1 short, wrinkled

B. Dihybrid x pure recessive

TtRr x ttrr

T R T r t R t r

t r TtRr Ttrr ttRr ttrr

1 : 1 : 1 : 1 Tall Tall short short

Round wrinkled Round wrinkled

*Recombinant (does not have parental combinations)

C. Parents: AABbccDdEe x aabbCcDdEe

• Chance of having a child: AabbccDDEe

• Aa bb cc DD Ee1/1 x ½ x ½ x ¼ x 1/2 = 1/32

A A B ba Aa Aa b Bb bba Aa Aa b Bb bb

C. Parents: AABbccDdEe x aabbCcDdEe

• Chance of having a child: AaBbCcddee

• Aa Bb Cc dd ee

1/1 x ½ x ½ x ¼ x 1/4 = 1/64

• Aa bb cc Dd Ee

• 1/1 x ½ x ½ x ½ x ½ = 1/16

Pedigree• Sickle Cell Anemia

– HbS – Normal Dominant– Hbs – Sickle Cell

male female

HbSHbs HbSHbs

HbSHbs HbsHbs HbSHbS

carrier sickle cell normal

Dominant or Recessive?

III. Multiple Alleles and Codominance (as illustrated by blood groups…)

• 3 alleles: IA, IB, and i

• Phenotype Genotype

O ii

A IAIA or IAi

B IBIB or IBi

AB IAIB

TYPE CELLS Rh

O + or -

A + or -

B + or -

AB + or -

Rh factor… ‘+’ have it or ‘-’ don’t

A A A A A A

B B

B B B A A B A B A

A-antigen that can be recognized and destroyed by foreign bodies that will produce antibodies against it

• Rh factor works the same way – if you are Rh+ it means you have Rh antigens on the surface of your blood cells

• If you are Rh- then you have no antigens, and will produce antibodies in the presence of Rh+ blood

• Universal Donor:

Type O-

• Universal Recipient:

Type AB+

CHINA U.S.

Type AB Type O

Problem #1:Mother is type A and father is type A.

What are all possible children?

IA i A A

IA IAIA IAi A O

i IAi ii

Problem #2:

Mrs. Brown AB Mrs. Smith A

Mr. Brown B Mr. Smith B

Babies: Baby 1 is type O… Baby 2 is type A

IV. Testcross• Sheep: White is dominant

Black is recessive• White male• $1,000.00 per vial of sperm from a pure white male.

How do we know if a male is homozygous dominant or not?

• Cross with a homozygous recessiveW? x ww …if it is pure, all offspring will

W ? be white (you need to have a lot w Ww ww of offspring to be sure) w Ww ww 2 white : 2 black

(not a pure homozygous dominant)*What are the chances that a heterozygous male crossed

with a black female would give rise to 4 white offspring in a row?

*A: ½ x ½ x ½ x ½ = 1/16th chance

V. Sex-linked traits (ex: hemophilia & colorblindness)

• Found on X chromosome but not on Y

NORMAL

CARRIER

COLORBLIND

NORMAL

COLORBLIND

B B

B b

b b

B

b

Ex 1: Cross colorblind male with carrier female

• XbY x XBXb

Xb Y

XB XBXb XBY

Xb XbXb XbY

– Of the offspring, what are the chances of having a normal son?– 1/4– What are the chances of having a colorblind son?– 1/4

Ex 2: Cross a hemophilia-carrier female with a normal male…

XHXh x XHY

XH Xh

XH XHXH XHXh

Y XHY XhY

• Chances of a child being a hemophiliac?

• ¼

Ex 3: Colorblindness (sex-linked)

P1 (parents)

F1

Parents must be…

DAD: XBY

MOM: XBXb

Note: If a normal mother has a son who is colorblind, then she MUST be heterozygous (a carrier)

Genetics review…

•This is a karyotype (all of the chromosomes in the human body)•Each chromosome has genes on it (to code for proteins)•There are multiple versions of each gene (example: T = tall, t = short)•These multiple versions are called “alleles”•You always have two alleles, because you have two versions of each chromosome (one from mom, one from dad)

B bR Rq q

T t

Example: This individual’s genotype would be BbRRqqTt

VI. Modifying Mendel with linkage and crossing over• TtBb x ttbb

A. If no linkage, you would expect to get 1TtBb : 1Ttbb : 1ttBb : 1ttbb ratio

B.Suppose there is the following linkage:

T B t b

t b t b• With this linkage,

you would expect: TB tb

tb TB tb

tb tb

tb TB tb

tb tb a 1 : 1 ratio!

C. But… what if the actual ratio of the cross is 7:1:1:7?• There must be linkage (we are close to 7:7) with some

crossing-over occurring (a source of recombination)

T B t b

t b t b

T b t Bt b T b t B t b t bt b T b t B t b t b

This happened 2 out of 16 times!

1 : 1

7 : 1 : 1 : 7TB Tb tB tb

tb tb tb tb

#’s like this imply crossing over AND linkage

2/16 14/16

With a cross of a double heterozygote (TtRr) and a double homozygous recessive (ttrr), REMEMBER: WITHOUT linkage we expect 1:1:1:1, and WITH linkage we expect 1:1

• You observe a ratio of 84 : 79 : 88 : 81. Do you think there is linkage?

• NO• You observe a ratio of 118 : 121. Do you think there is

linkage?• YES• You observe a ratio of 131 : 12 : 9 : 137. What is going

on?• Linkage with crossing over! This is basically a 1:1 ratio,

with 21 (12 + 9 = 21) “out of place” results mixed in.

VII. Analysis of COV data to construct gene maps

• If ratio is 7:1:1:7, then 2/16 were recombinants

• 2 / 16 = .125

COV = 12.5%

= 12.5 cM (centi-Morgans apart)

• 1% COV = 1 centi-Morgan

Problem #1: pure white (r), smooth leaves (p) x pure red (R), pointed (P) leaves

• F1 are all red, pointed• A test-cross is done… RrPp x rrpp (to detect linkage)

40 white, smooth36 red, pointed10 white, pointed14 red, smooth100

• Q1: Is there linkage?• Q2: How far apart are these 2 genes (R & P)?

R P ?

A1: The RrPp x rrpp cross will determine if there is linkage…

• If no linkage…

RP Rp rP rp

rp RrPp Rrpp rrPp rrpp

1 : 1 : 1 : 1• If linkage…

RP rp

rp RrPp rrpp

1 : 1• If both linkage & crossing over…

something like 7 : 1 : 1 : 7

• (40:36:10:14 implies that there IS linkage and crossing over)

A2…distance apart

• 24/100 = 0.24

COV = 24%

= 24 cM

R P

24 cM

Problem 2: Assume Gene T and B are 12.5 cM apart and another gene A is linked with T & B. A has a COV with T of 5 cM and a COV with B of 7.5 cM. Draw a map of TBA.

T A B

12.5 cM5 cM

7.5 cM

Problem #3: COV of R & T = 6.0 COV of S & T = 13.5 COV of R & S = 19.5

R T S

6.0

13.5

19.5

R /r T/t B/b E/e

5 17 2

VIII. A. Modifying Mendel with Polygenic Inheritance• Many genes control one trait

• Producing continuous variation

• NOT producing a ratio

• Ex1: Human skin color

• Ex 2: Human height

• Ex 3: Hair color

B. Codominance (and Incomplete Dominance)

• Codominance: both alleles are dominant, and so both are represented (ex: blood type… IAIB is “type AB”

• Incomplete Dominance: neither is completely dominance, and so there is a blend (if RED snapdragons crossed with WHITE snapdragons give PINK flowers)

C. Mendelian ratios can also be modified by “gene interaction”

• Studied by Bateson

• Ex: Epistasis (when genes act on one another)

CcPp x CcPp purple flowers purple flowers• Mendelian genetics expects 9:3:3:1• Actual ratio is 9 purple : 7 white. WHY?

Purple : WhiteDom; Dom one or no dominant genes

C_P_ Ccpp, ccPp, ccpp

IX. Nondisjunction – errors in meiosis

• Causes extra chromosomes or too few chromosomes• Chance occurrence• Failure of chromosomes to separate• Tested for by amniocentesis or CVS (chorionic villi sampling)

A. In Sex chromosomes…

spermatogenesis oogenesis

Interphase XY XX

Anaphase I X-X Y-Y X-X X-X

Prophase II X-X Y-Y X-X X-X

Anaphase II X-X Y Y X-X X X

• XX = normal female

• XY = normal male

• X = sterile female (called “Turner syndrome”)… chromosome # is 45

• XXY = sterile male (called “Klinefelter’s syndrome”)… chromosome # is 47

B. Nondisjunction in autosomes ex: Down’s syndrome

• 95% due to maternal age egg sperm

#21 #21 #21

Trisomy-21

is Down’s #21 #21 #21

ZYGOTE

Terms:• Karyotype –

photograph of an individual’s chromosomes:

• Autosome – non-sex chromosome

• Linkage group – group of genes who have their loci on the same chromosome

aB

d

Ab C c D

• As far as these chromosomes are concerned, what is the individual’s genotype?

• AaBbCcDd• Do genes B and C assort independently (randomly)?• Yes• Which genes are in the same linkage group?• A, B, and D• Set up a Punnett square for the cross between this individual and

an individual with the genotype bc (we are only concerned with the B and C genes).

BC Bc bC bc bc

• Set up a Punnett square for the cross between this individual and an individual with the genotype abd (we are only concerned with the A, B, and D genes).

aBd AbD abd

T = tall B = brown R = round M = loud Z = hairy

t = short b = white r = wrinkled m = quiet z = bald

• TtBb x ttbb

45 tall,brown:3 short,brown:5 tall,white:47 short, white• BbRr x bbrr

27 brown,round:26 brown,wrinkled:22 white,round:25 white, wrinkled

• MmTt x mmtt

41 loud,tall:9 loud,short:5 quiet,tall:45 quiet, short• TtZz x ttzz

21 tall,hairy:28 tall,bald:30 short,hairy:21 short,bald• BbMm x BbMm

47 brown,loud:4 brown,quiet:1 white,loud:48 white,quiet

Determine which are linked, and map the chromosome.

T B M8

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