Transcript
V\
Sym oiss_ b lack - B Chestnut -b
Phenotype
s
trotting - T pacing -f Mate
x
black, trotter
x
Female
chestnuts_pac^ b
Genotype Gametes
bt 1 BtTt
BbTt
8 bTr
(G enotype) BbTt I BbTt
bt
-
black. trotter (Phenotype)
Female
P2
Genctype Gametes P2 Punnett square
F Phenotypes black, trotter chestnut, trotter
(from F1)
t
i man, brown eye color is due to a dornin^ (B), and blue eye color is due to its allele (b), Right-handedness is due to a dominant gene (t?), and Left-handedness is A man homozygous for both brown eyes and right-handedness blue-eyed, left-handed woman. What is the genoty pe arried a girl With the same genotype, what kinds m of offspring could they produc proportions?
In garden peas, the gene for a smooth speed coat is due to its recessive allele (s). Also, the and a dwarf plant is due to its recessive allele M. Cross heterozygous one for both tallness and smoo th seeds. What kinds of off sprin and in what proportions?
inkled seed ntgene(r), rt
a
they produce
lashi
_ erk hair . They ha l:. iv short eyelashes and dark hai
an w ith short Avolashes nd red hair and one with and children. ypes of parents
inant to blue. A nt to straight and brown eyes is dom ina dom is r hai ly cur n ma In 2.. marries awoman with blue eyes and He r. hai t igh stra and s eye wn man has bro l, too have i_-our have blue eyes and curly hai . outly flair, They have six cniidren. pes of the parents and the offspring oty gen the e Giv r. hai t igh stra brown eyes and u,porn inant allele (B), and chestnut dom a n upo ent end dep is ck bla , in horses inant over its to an allele (T) which i s dom ecessive ("b). T he trotting gait is due ies. Ni w an,) on, different chtomoso ssive pacing allele (t). T hese. g tt er. mated to a homozv ous chestnut tro a) If a homozygous black pacer is ? what will the F 1 caeneration be like Draw a purane o l e i&^divid ills are Gro ssed? it ult res h\ that will he the o. uare: and derive the genotypic rati ive allele dominant gene (E), red to its recess . `rn poultry black color is due to a allele e (C), plain head to its recessive gen nt ina dom a to due is d hea d (e). Creste duce many to a black, plain female. They. pro the (c). A red crested mate bird is mated sted. What were the genotypes of cre ed i f hal and , sted cre ck bla f offspring, hal parents? finds some -haired dachshunds. but he always 5. A man owns a pair of short, red m repeated puppies which are produced. Fro the in r hai ck bla o yom and air long-h black. Could ; 8 long red; 4 short black; 4 long ratings he obtained: 16 short red breeders 1 ratio, thus indicating that both :3: 9:3 ofa le mp exa an be s ult res these es? are heterozygous for the two gen
b_ What phenotypes would you exp ect from the ddVVw
,so
In pine trees, long ne edles are dominant to short and rough bark is do rtin to smooth. Use a punnett square to show the possible geno types of offspring odu d from 2 entirely heterozygous pine trees. .
3. A pea plant heterozygous for height and for yellow seeds is crossed with a dwarf plant having green se eds- Using a p unnett squar e, dete rmine the expected genotypic and phenotypic ratios of the offsprin g.
4. Using the `fraction multiplication" method, dete rm ine the phenotypic ratios of offspring born to a mother with blood type 0 and Rh - and a father with blood type AB and Rh+. ( Rh - is recessive, use rr for it and R for Rh+. The father is heterozygous for Rh type)) 5. Long hair and yellow eyes are dominant in cats, Dete rmine whether or not short -haired, green-eyed kittens could be born from the following parents: mother long hair, yellow eyes father long hair, green eyes
The following question mixes one trait that exibits complete dominance and one trait that exibits incomplete dominance.
6. Roan is a coat color produced when red cattle are bred with white. That is, Red is incompletely dominant to white. Hornless cattle are dominant . at and phenotype as woul d be r expec o ted from a a roan n horno le ss med h
02.049 CONTINUITY
Karyotype p reparation
LIF E SCIENCES ON FILE
02°04
H
0 0 A Blood sample removed from donor 13 Sample suspended in saline; red cells settle out C Catchicine added - stops cell division at metaphase 0 Water added - cells swell and burst E Cells spread onto slide F-G Cellsobserved under micr H Chromosomes photographed € Individual chromosomes cut out J Chromosomes arranged in order of diminishing size
0 2a 050
Bx N XA B^rin6 n 6
7
ofi^A h
A katyotype 3 d karyotype
11
8
zx
AAA )%
12
x xxx
--SDL-UNKAG
Sex ChromosomesX chromosomeY chromosome:
In humans there are 23 pairs of chromosomes (46). Of these, 22 pairs are autosomes and 1 pair, the 23 rd pair, are the sex chromosomes. organ. Morgan worked iat sex chromosomes exist was discovered by fruit flies known as [rophila . H e discovered that when 2 red eyed ents were mated, some individuals in the F 1 generation had white eyes and that the white eyed individuals were always males. To explain this phenomenon, Morgan suggested that the gene for eye color was carried on the sex chromosomes. When genes are carried on the sex genes. Most sex-linked ` chromosomes, they are said to be aex-lbked or x-JmkW genes are carried only on the X chromosome and not the Y. To illustrate:
Parental Cross: Genotypes:
Red eyed male x Red eyed female (red is dominant)
F1 offspring:
In humans, there are many traits carried on the X chromosome. When a defect or mutation occurs in one of the genes on an X chromosome, a female still has a chance of having a normal gene on her other X chromosome, whereas a male who inherits an X that carries a gene defect, will almost always express the s phenotype governed by the mutation. This is humans are also referred to as Male-Disorders.
married a blue e father was blue eyed os wh an m d in r-b lo e () is co 4. A brown eyed, ess). Assume brown ey dn in bl r lo co r fo us go normal vision. eyed carrier ( heterozy blindness recessive to r lo co d an b) e( blu d color blind? er dominant ov ren will be blue eyed an r blind? ild ch eir th of ge ta en rc a) What pe ed and colo e girls will be brown ey ind? b) What percentage of th own eyed and color bl br be ll wi ys bo e th of c) What percentage eyed gene d dominant to the white an d ke lin xse is ) (A e 5. In the fruit fly, red ey notypes of: (r). Give the possible ge a) white eyed males b) red eyed males c) white eyed females d) red eyed females wing crosses: e offspring of the follo th of s pe ty no ge e th e Giv red eyed male a) white eyed female X ale ed female X white eyed m b) heterozygo us red ey e al m ed ed female X red ey c) heterozygous red ey
problgms inked) its which follow a sex-linked (or X-l To solve problems involving tra must be combined with the symbols les alle the for s bol sym the n, inheritance patter omosome(s). ressed as superscripts on the X chr for sex. The allele symbols are exp les and in omosome since it carries so few alle chr Y a on put is bol sym le alle No me w not h sex (:X) linkage, the Y chromoso wit ling dea ns stio que s etic gen st mo l" for a'trait assume that individuals are "norma ays Alw it. tra the for le alle an carry ise. unless the question tells you otherw ce pattern. The 3
ich follow a sex-linked inheritan 't'here are many human traits wh most common traits are: i
ui` o (oniina #^
• eessive) -^ Red/green colorblind ess
-
eWGreen color vision Ability to blood clot Normal visual acuity
Hemophilia Myopia (.nearsightedness)
-linked recessive trait. Show the Kxample.l: Colorblindness is a sex s: genotypes of the following individual Normal male Colorblind male Carrier male
Normal female Colorblind female Carrier female
inant x-linked feather characteristic follows a dom tain cer a , ltry pou In 2: e pk Eu allele (B) produces barred feathers inheritance pattern. The dominant the es non-barred feathers. What is and the recessive allele (b) produc notypes from a cross between a ratio of offspring genotypes and phe non-barred male? heterozygous barred female and a
Parent Phenotypes: Parent Genotypes: Puimett Square:
F1 Genot )es:
h rre_d_f a XX
x
non-barred ttm XY
F1 PhCJIO tVpes:
HemoohWa - a sex-linked disorder. - a blood clotting disorder. - one factor is missing in the "blood clotting cascade" which prevents the proper clotting of blood when a wound occurs. - people with this disorder face the possibility of bleeding to death with the smallest of injuries - even a paper cut! - there are several different kinds of hemophilia - the kind depends on which factor is missing. - historically, most forms of this disorder were found only in males - as females would bleed to death at puberty with the onset of menstruation. - also, ? "mutant" alleles are required in order for a female to inherit the disorder, while males require only one. (Xh`Y = male, h h = female) -this explains why more males have the disorder in any case. - today, many females live with hemophilia past puberty, but they need to routinely have blood transfusions containing the clotting factor that they are missing so that they will not bleed to death (either upon injury or menstruation). today, males with hemophilia also receive transfusions of their missing blood clotting factors in order to reduce the possibility of them bleeding to death upon receiving injuries. - so there are treatments for hemophilia, but there is no cure Pnhlemll A man with hemophiliav-ishes to have a child with a woman who is normal. but whose father also had hemophilia. a) Show the parent genotypes: b) Show the punnett square:
c) What are the chances that a. of their Ch children will be nnormal? Q (1) \\ dt are tl e chances that their bD v ill I)e normal'? a e) What are the chances that their girls will be normal?
°%,
M
ybrid Crosses
-
Summit ry and Short Cuts
B = Black b= white (2 phenottypes) Cron rte:
BB x bb 100%Bb all Ft = Black Fl Cross.
BWxBW 1 BB 2BW 1WW
Bb x Bb I BB 2 Bb I- bb
F2 phenolyxc ratio:
F2 phenotypic ratio:
1 Black: 2 B& W: 1 White
3 black: lwh ite
Dihybrid Cross Short Cut s a peter tzct
