Dominance Inheritance ReviewIncomplete Inheritance

Co-dominance Inheritance

SBI3U

• When one (1) allele is stronger (so dominant) than the other allele inheterozygous genotype, resulting in phenotype of the strong (dominant)

Examples:

a) For hair texture trait Curly hair is dominant (C) Straight hair is recessive (c) Therefore, Cc genotype translates to Curly hair phenotype

b) For dimple trait Having dimple is dominant (D) Not having dimple is recessive (d) Therefore, Dd genotype translates to having dimple phenotype

c) For hitchhiker thumb trait Having the hitchhiker thumb is dominant (_____) Not having the hitchhiker thumb is _______ (___) Therefore, ______ genotype translates to _______

Sample problem:Dwarfism, is a dominant trait.If both parents with dwarfism phenotype have normal looking offspring,a) determine the genotype of the parentsb) the percentage of their child looking normalc) the percentage of their grandchildren from the normal offsprings looking dwarfish

given that their life partners are also normal

Answers:a) P: Dd x Dd (because if either one of

the parents has DD, none of theirchildren will look normal)

DD =

Dd =

dd =

142414

= 25% (dwarfism)

= 50% (dwarfism)

= 25% (normal)

c) Zero percent (0%) F1 = dd x dd

b) D d

D DD Dd

d Dd dd

Sample problem:Breast cancer has been tracked to associate with genes BRCA1 & BRCA2, which aretumour suppressors and function to inhibit the growth of tumour (cell cycle genes)It is a recessive trait. That means, if the gene is negatively mutated and theoffsprings carry both mutated alleles, she is more likely to get breast cancer.

a) Determine the phenotype of the genotype variants for this traitb) If the mother is heterozygous for the trait and the father has normal alleles,

how many of their 6 children willi) be carrier of the cancer trait (heterozygous)ii) be homozygous for normal BRCA1iii) develop breast cancer

Answers:

a) BRCA1 + BRCA1 normal; least risk for cancerBRCA1 + brca1 carrier of the cancer traitbrca1 + brca1 develop cancer

b) i) 6

ii) 6

2424

= 3 children

= 3 childreniii) zero

BRCA1BRCA1 BRCA1

BRCA1

BRCA1 brca1BRCA1 BRCA1BRCA1 brca1

BRCA1brca1

• Also known as Blending Inheritance• When two (2) alleles are equally dominant (same strength), they

interact to produce a new, third, phenotype This means, the heterozygous genotype has its own phenotype Three (3) phenotypes are produced from such cross

• Although Mendel did not observe such pattern of inheritance withpeas, there are many examples found in nature.

When theheterozygous

genotype producesa different/new

phenotype

A red hibiscus is crossed with a white hibiscus

What are their genotypes?

P: (HR HR) x (HW HW)

F1: HRHR = ¼ = 25% RedHR HW = ½ = 50% PinkHW HW = ¼ = 25% White

F2: Predict the phenotype ratiofor self-pollinated hybridhibiscus.

In guinea pigs, colour of coat is determined by at least three alleles. Yellow coat is determined by the homozygous genotype YY, White by the homozygous genotype WW, and Cream by the heterozygous genotype YW.

Determine the expected genotype and phenotype ratio of the F1generation which would result from a cross between:a) two cream coloured guinea pigs;

•b) a yellow coated and a cream coated animal

Your Turn:

… SolutionsLet GYGY represent yellow coatLet GWGW represent white coatLet GYGW represent cream coat

a) P generation phenotypes:genotypes:

cream x creamGYGW x GYGW

GY

GY GW

GYGY GY GW

GW GYGW GW GW

F1 : genotypes GYGY = ¼ = 25% (yellow)GYGW = ½ = 50% (cream)

GW GW = ¼ = 25% (white)

Therefore, the expected genotypic ratio is: 1 : 2 : 1the expected phenotypic ratio is: 1 : 2 : 1

b) P generation phenotypes:genotypes:

cream x yellowGYGW x GYGY

GY

GY GY

GYGY GY GY

GW GYGW GYGW

F1 : genotypes GYGY = ½ = 50% (yellow)GYGW = ½ = 50% (cream)

Therefore, the expected genotypic ratio is: 1 : 1the expected phenotypic ratio is: 1 : 1

1. Howdy! My name is Bob Howard, and I own 20 purebred red cows.Something strange happened several months ago. During a violent storm, allof the fences that separate my cattle from my neighbors cattle blew down.During the time that the fences were down, three bulls, one from eachneighbor, had access to my cows. For awhile, I thought that none of the bullsfound my cows, but over the months, I have come to the conclusion that all ofmy cows are expecting calves. One of the bulls is the father. Which bull is it?

A local college professor told me to use a little genetics detective work tofigure out who the father is. He told me to collect information about each ofthe bulls, and to read articles about genetics and Gregor Mendel'sexperiments in genetics. So, I did exactly what he said. I compiled theinformation. Now, I need your help to make sense of the data and to figureout who the father is.

After reading through the information, maybe you can tell me why my redcows had 9 roan calves and 11 red calves. I don’t really understand how thishappened. When you have determined which bull is the father, please tellme the answer.

This is Rocky. He is a 2,200 pound Red bull. The colour of Rocky&’scalves, if mated with a red cow, can be determined by using a Punnett square.His offspring will also be unique in colour compared to the other two bulls.

This is Rufus. He is a 1,920 pound White bull. The colour of Rufus’calves can also be figured out, if he is mated with a red cow, by using aPunnett square. The colour of his calves will also differ from the other twobulls offspring.

This is Ferdinand. He is 2,000 pound Roan bull. The laws of genetics tellus that the offspring he produces will probably be different, in colour, than theother two bulls’ offspring. Using a Punnett square, you can see the genecombinations, for colour, that Ferdinand’s offspring could have if he mateswith a red cow. The colour of Ferdinand’s calves has to do with probability.

Who did it? … a bonus mark for first correct answer posted on Wiki discussion

Co-dominance

• When the dominancy (strength) of the allele changes unpredictably That is, sometimes one allele is dominant; other times it is not This results in the re-appearance of the both parental phenotypes

which causes the offspring phenotype to look patchy!

• The genotypes are not blended and they still obey Mendel’s law ofsegregation, resulting in a mixture of the phenotype.

HaemoglobinThe gene for haemoglobin Hbhas two co-dominant alleles:

i) HbA (the normal gene)ii) HbS (the mutated gene)

Heterozygous: HbA HbS is good for protecting from Malaria

normal redblood cell

sickled redblood cell

Cat coat: orange (OO) x black (oo) = Tortoiseshell

maternal‘O’ allele

paternal‘o’ allele

mosaic adult

http://gslc.genetics.utah.edu/units/basics/blood/types.cfm

Phenotype Genotype Picture Antibodies Notes

Type A

Type B

IAIA or IAi

IBIB or IBi

B – Antibodies

A - Antibodies

Type AB IAIB No antibodies UniversalRecipient

Type O ii A – AntibodiesB – Antibodies

UniversalDonor