nasty

Question 1

where does gastrulation take place

Answer 1

primitive streak

Question 2

what happens during the inwards migration of cells in gastrulation and what do these cells become.

what is important about the part of the primitive streak that the cells use to migrate inwards

Answer 2

they undergo epithelial to mesenchymal transition
these cells become mostly mesoderm and some endoderm.

it dictates what type of tissue the cells will become.
cells that migrate in at hensens node will become axial mesoderm.
cells just below this become paraxial mesoderm
below this they become intermediate mesoderm
and at the bottom they become lateral mesoderm.
PICTURE.

Question 3

what does each of these become...
axial mesoderm
paraxial mesoderm
intermediate mesoderm
lateral plate mesoderm

Answer 3

notocord and prechordal mesoderm

the somites which form the sclerotome (cartilage), syndotome (tendons), myotome (skeletal muscle), endothelial cells and dermatome (dermis)

kidneys and gonads

splanchnic (circulatory system), somatic (body cavity, pelvis, limb bones) and extra embryonic.

Question 4

what is the earliest evidence of segmentation in vertebrates

Answer 4

somites

Question 5

where on the AP axis shows somites

Answer 5

they are only in the anterior.
the somites are lined up down either side of the neural tube. there are clear clefts between each somite.

the posterior cells are all uniform and mesenchymal and are not separated into somites.

Question 6

what does the somite number dictate

how many do humans, chicks and mice have

Answer 6

the number of vertibrae

humans have 33 when they are born, by adulthood they have 24 and 9 fused ones.
the human embryo has 38-44 somites.

the chick has 55 and the mouse has 65.

Question 7

what is the direction of somite formation and what are the basics of what happens

how do they know when to form somites

Answer 7

A to P

at the very posterior of the primitive streak there are new cells forming to extend it.
they will stay unsegmented until it is time for them to become somites.

they know their positional information.
they communicate with the other side of the primitive streak so both opposite somites form together

Question 8

what is the clock and wavefront model

Answer 8

a clock ticks in the pre somatic mesoderm and drives a molecular oscillator that dictates the periodocity of somites.
cells also hit a travelling wavefront, an abrupt change of property occurs leading to the decision to form somites.

Question 9

what is the pattern of expression of C hairy in the chick and what does it cause

Answer 9

the expression oscillates in the pre somatic mesoderm.
one cycle of oscillation takes 90 minutes and a new somite forms each time.

at first there are high levels of C hairy throughout the pre somatic mesoderm. Then it decreases and is only in the top half. and then it goes down to no expression. it increases back up and when it is expressed everywhere again a new cycle starts.

Question 10

what is C hairy called in mice and zebrafish

Answer 10

in mice it is her

in fish it is hes.

Question 11

what is C hairy expression controlled by and what does it code for

Answer 11

notch signalling

it codes for TFs with a short half life which are repressors.

Question 12

what activates the hairy gene

what does the hairy protein do and how long does the action last

Answer 12

notch signalling

the protein represses its own transcription, it has a short half life so it will be degraded quickly and the hairy gene wont be repressed for long.
this means that the protein can be made again which will repress the gene again.
this is a cycle which results in the up down oscillation of the expression of c hairy.

Question 13

what genes other than c hairy show oscillatory expression

how many oscillations are there before a somite forms

Answer 13

wnts, notch and FGF signalling pathways contain them

12

Question 14

which direction does the wavefront move in and what happens when it reaches a cell

Answer 14

it moves P to A.

when it reaches a cell it causes oscillations to stop.

Question 15

what are some gradients in the pre somatic mesoderm

and where is the wavefront positioned.

Answer 15

RA gradient is highest in the somites and gets lower as you go posteriorly.
the somites make an enzyme required for the synthesis of retinoic acid.
FGF8 is made in the posterior and the gradient gets lower as you go up.

the wavefront is positioned where the two low points of the gradients meet in the middle.

Question 16

why cant there be FGF8 and RA in the same place

Answer 16

RA inhibits the production of FGF8.
FGF8 allows production of Cyp26 which is an RA inhibitor.
FGF8 blocks the raldh2 enzyme which is required for RA synthesis.

Question 17

what genes are expressed in the S-1 area

Answer 17

FGF8 will upregulate tbx6.
notch signalling and tbx6 will combine to cause mesp2 signalling.
mesp2 will drive ripply2.
ripply2 will repress mesp2.
This means that mesp2 expression will be restricted to the anterior part of S-1 because there is less FGF8 there.
There is a small band of mesp2 expression in the anterior part of S-1.

Question 18

what happens if you take the anterior S-1 cells and put them in the middle of a presumptive somite

Answer 18

you will see the formation of a cleft in the middle of a somite in the wrong position.
this means that these cells have properties that allow somite boundary formation.

Question 19

what gene is expressed in the somite boundary

what other gene affects this and what happens if you force the expression in the middle of a somite

Answer 19

notch

lunatic fringe
a boundary will form.

Question 20

what happens to somite development if there is a mutation in notch signalling

Answer 20

causes skeleton defects due to lack of periodic segmentation of somites.
in humans a similar mutation causes spondylocostal dysplasia.

Question 21

what is the importance of skeletal muscle

Answer 21

movement and posture
communication
maintaining body temperature
respiration

Question 22

how is a muscle cell made

Answer 22

stem cells are specified and become muscle progenitor cells called myoblasts.
they differenciate into multinucleate cells called myotubes and they mature to become myofibres.
the myotubes form by the fusion of many myoblasts.
during maturation the types of fibres are refined and there is innervation and muscle activity.

Question 23

how to isolate the MyoD gene

Answer 23

start with a fibroblast cell line which can give rise to various cell types including myoblasts
the cells are cultured with 5aza which is a demethylating agent.
the mRNA is extracted and converted to cDNA.
the cDNA is hybridised to the cDNA of fibroblasts not treated with 5aza, but only genes present in both the groups of cDNA will bind.
the different genes will remain single stranded.
he took the cDNA enriched in muscle specific genes and screened them using a myoblast specific probe mRNA, then he isolated the MyoD cDNA.

Question 24

will most of the tissue in the larvae contribute to the adult body

give some of the examples of what will contribute to the adult

Answer 24

no, most of the adult tissue is derived from imaginal discs, small groups of epithelial cells that remain diploid throughout larval development

mouthparts disc, antenna, eye, leg, haltare, wing

Question 25

when are imaginal discs set aside

where are the leg discs, haltare and wing too

Answer 25

during stage 12 when the embryo becomes segmented.

T1,2,3 leg
in T3 dorsal to the leg disc there is the haltare disc
in T2 dorsal to the leg disc there is the wing disc

Question 26

what do imaginal discs start off as and when do they stop dividing

Answer 26

30 cells.

they proliferate until the late third larval instar stage where there will be 50,000 cells, columnar epithelial cells.

Question 27

what is the wing disc split into

which bit of the disc becomes which bit of the wing and what process does this

Answer 27

four quadrants- A and P (front and back)
D and V (top and bottom)

the very middle becomes the wing blade
the surrounding bit becomes the wing hinge
the outer bit becomes the notum

the disc unfolds into the adult wing via a process called eversion.

Question 28

where is wg and ptc expressed in the wing disc

what does scabrous expression show

Answer 28

wg is expressed along the DV split line.
ptc is expressed along the AP split line.

sensory organ precursors on the wing that will become sensory bristles.

Question 29

what is engrailed and what does it do

Answer 29

it is a TF and a segment polarity gene that defines the posterior compartment.
the posterior half of the segment will be expressing engrailed.
the back half of the legs, wings and segments all express engrailed

Question 30

what does engrailed suppress and what does this mean

what does engrailed activate and where can it travel

Answer 30

ci, so ci is only expressed in the anterior half

hh, it can diffuse into the anterior. This is the only place where hh signalling can happen because the posterior has no ci.

Question 31

where is dpp expressed in the wind disc and why

where does it go when it is made and what does this define

Answer 31

in the AP stripe.
this is because dpp is a transcriptional target of hh signalling just like ptc.
the cells which are able to respond to hh will produce dpp and ptc.

it will be made in the stripe and diffuse anteriorly and posteriorly creating a gradient. this will have defined the midline of the disc.

Question 32

what family is dpp in and what does it bind to

what does dpp activate , and what does this then activate

Answer 32

TGFB superfamily.
it binds to type 1 and 2 receptor heterodimers in mammals and thick vein and punt receptors in flies.

a TF called mad. which will turn on omb and sal genes and turn off the brinker gene.

Question 33

what are the dpp conc differences that induce omb and sal expression

Answer 33

high concs of dpp induce sal and omb expression.
low conc of dpp induce omb expression only.

this means that sal is only expressed close to the midline and omb is expressed more widely because it doesnt need as much dpp

Question 34

what happens to dpp distribution when morphotrap is introduced

and what affects does this have downstream

Answer 34

it is a membrane tethered anti GFP that will trap all the dpp protein being made by the midline cells. so it will stay at the midline and will not diffuse away.

mad will only be activated in the midline. and it will not be able to repress brinker in other cells.
the photo made will have a sharp boundary of dpp and mad expression only in the midline.

Question 35

cloning experiments:

creating a group of cells in the posterior that lack engrailed

Answer 35

this means that they cannot make hh, but they can make ci. hh will diffuse to this group of cells which can now respond because they can make ci.
this will allow hh signalling and the production of ptc and dpp.
it has created an ectopic midline in the posterior.

Question 36

cloning experiments:

creating a group of cells in the anterior that are able to make engrailed

Answer 36

this will allow them to make hh, but they cannot make ci.
this group of cells cannot respond to their own hh but it will diffuse and affect the surrounding cells that do have ci expression.
surrounding cells will make dpp and ptc.
this will have created an artificial midline.

Question 37

what happens if we dont express ptc in a group of cells in the anterior

Answer 37

this will stop the regulation of smo.
smo is active and so the hh pathway is active and dpp will be made.
this will mean dpp is made in the group of cells and not just around them like the others.

Question 38

how is polydactyly caused in flies

Answer 38

a ptc clone was induced in the anterior part of the wing.
this will make dpp.
it will cause a mirror image because they all think that they are anterior compartments
it makes a triple anterior wing with one posterior part.

Question 39

what happens if you make a mutant region in the midline of the wing disc that cant express ptc

Answer 39

this means there is nowhere for the hh to bind
so it will keep going past the midline and bind to ptc further away and make dpp there.
the limiter for how far hh can travel is ptc.

Question 40

what are the other words for humerus, ulna + radius and carpals

Answer 40

humerus= stylopod
ulna + radius = zeugopod
carpals and digits = autopod

Question 41

what is a limb bud

what is a limb field and what are they determined by

Answer 41

they are protrusions from the flank of the embryo.

this is where the limb will grow. they are determined by intrinsic properties of the lateral mesoderm.

Question 42

what happens if you take lateral mesoderm that is in the limb field and graft it ectopically

what happens if you take an area of the flank that would not normally form a limb and put it in the place of a limb field

Answer 42

an ectopic limb forms

no limb will form.

Question 43

what TFs specify limb identity

how do they interact

what happens if you force them to swap regions

Answer 43

Tbx5 in the forelimb and Tbx4 in the hindlimb

tbx5 represses tbx4.

this causes a reversal of correct limb identity, but it only works in chicks.
mice still develop the correct limbs.

Question 44

where is pitx1 expressed and what happens if you knock it out

what if you force its expression in the forelimb

what if you dont have pitx1 expression

what gene marks where the forelimb will form

Answer 44

it is expressed in the hindlimb,
KO = loss of hindlimb identity and tbx4 expression

it drives tbx4 expression and hindlimb identity.

forelimb expression is driven

hox6

Question 45

what will a bead soaked in FGF and planted ectopically do

what drives FGF8 and 10 expression and where does this happen
what restricts this

Answer 45

create an ectopic limb.

tbx TFs drive FGF10 transcription in the mesoderm, which drives FGF8 expression in the ectoderm
the expression of FGF10 in the mesoderm is also driven by FGF8 expression in the intermediate mesoderm.

wnt will restrict this to only in the limb field.

Question 46

what three sections are there in the limb bud

Answer 46

the ectoderm covering the limb field thickens and this is the apical ectodermal ridge AER, it is very rich in FGF4 and 8.

the area immeadiately behind this is the progress zone and it is made of mesenchymal cells

a small area in the posterior limb field is the polarising region ZPA which expresses shh.

Question 47

what did saunders do in 1948 and what did he discover

Answer 47

removed the ectoderm from the limb bud at different times during development.
if you remove it after 3 days, the humerus is the only thing to form fully, no digits.
if you remove it after 4 days, the humerus, ulna and radius all form but no digits.
so you need ectoderm to get outgrowth of the limb, and the longer it stays the more of the limb can develop.

Question 48

what is the two signal model of P/D patterning

Answer 48

you initially have progenitors for all parts of the limb.
the cells fate is determined by the signals it receives.

FGF4 and 8 are released from the more distal part of the limb and RA is released from the proximal areas and this created two gradients.

RA drives meis gene expression in the proximal cells, which will form the humerus.
FGF drives the expression of hox genes, hox11 in the middle which makes the ulna and radius and hox13 at the most distal part which makes the digits and carpals.

Question 49

what protein makes RA and what happens if this is knocked out

Answer 49

raldh2, you lose meis expression and proximal limb cell elements.

Question 50

what is the drosophila homologue of hox13 and meis

Answer 50

hox13 = distaless
meis  = homothroax

Question 51

what if you add a second ectopic ZPA to the anterior of the limb bud
and what if the graft was performed earlier

Answer 51

normally chicks have three digits but the graft causes them to have a mirror duplication forming six digits.

if it was performed earlier there was duplication of the ulna.

Question 52

what if you implanted a bead soaked in shh to the anterior of the limb bud

what is formed with altering concs of shh

Answer 52

the shh gradient controls the patterning and it will cause mirror image duplication and six digits will form.

the highest conc of shh causes digit 4
the middle conc causes digit 3
the lowest conc causes digit 2

Question 53

what if there is no shh in the limb bud at all

Answer 53

there is a complete loss of the distal most skeletal elements

Question 54

what ensures that the limb grows as it is patterned

Answer 54

the FGFs are required to maintain the expression of shh in the ZPA
and the ZPA is required to maintain FGF expression

Question 55

how does the ectoderm covering the dorsal part of the limb differ to the ectoderm covering the ventral part

what does KO of lmx1b cause

which fate is the default

Answer 55

wnt7a leads to lmx1b, both expressed dorsally.
engrailed1 is expressed ventrally and this is driven by BMP signalling.

gives signs of a paw forming but no fur, the limb is ventralised.

the ventral fate is the default.
BMP drives engrailed on the ventral side and engrailed will repress wnt7a in the ventral half. this means that it can only be expressed on the dorsal half.

Question 56

where is branching morphogensis needed and what does it help to create

Answer 56

lungs, uretic kidney bud, salivary gland, prostate, mammary gland, pancreas
helps to create a high SA from the branching tree like structure.

Question 57

what interactions cause the kidney to arise

Answer 57

the kidney arises as a result of reciprocal inductive interactions between two structures.
one is the uretic bud which is made of intermediate mesoderm and is epithelial, it sticks out into the metanephric mesenchyme.
it is the interaction between the uretic bud and the metanephric mesenchyme that allows kidney development.
PICTURE

Question 58

what does the metenephric mesenchyme make and where does it go and what does it cause

then what do very high levels cause

Answer 58

the mesenchyme makes GDNF and it travels to the uretic bud and binds to receptor tyrosine kinase called ret.
this causes proliferation and outgrowth of the bud tip cells.
very high GDNF levels start to inhibit proliferation, so the leading edge tip cells stop growing.
the cells on either side will keep growing as they are only exposed to low levels.
this will cause two branches to form.
the two new tips are still surrounded by mesenchyme so this will repeat creating more and more branches.

Question 59

what do the branches of the uretic bud go on to form

what does the metanephric mesenchyme go onto form

Answer 59

collecting ducts of the kidneys

nephric tubes

Question 60

how do the nephric tubes form fromthe mesenchyme

Answer 60

the uretic bud induces the mesenchyme to condense and undergo mesenchymal to epithelial transition and form the renal epithelium and then renal vesicles.

the renal vesicles proliferate and grow into an S shaped body (henle and tubules), one end of this will join to the collecting duct by selective apoptosis and the other end will attract blood vessels and become the bowmanns capsule.

Question 61

what germ layer are the parts of the lungs made from

Answer 61

endoderm and mesoderm
the endoderm becomes the epithelial lining of the airways and undergos branching morphogenesis (same as buds)
the mesoderm will give rise to cartilage, muscle and connective tissue (same as mesenchyme).

Question 62

what is the starting bud called in the lungs that undergoes branching morphogensis

when do they form

Answer 62

respiratory diverticulum and is made of endoderm.

they form during the fourth week.

Question 63

how does the lung bud and the sac of mesoderm communicate

Answer 63

the lung bud has FGF receptors and the mesoderm releases FGF10 onto the bud.
this causes the bud to grow.
high levels of FGF cause expression of BMP4, shh and sprouty and this means that the tip of the bud will become a signalling centre.
the BMP4 stops the tip growing and causes branching.

the shh goes from the tip to the mesoderm and stops FGF10 expression in the area opposite the tip
this promotes the next cycle of branching because the FGF10 will only be released onto the two side tips and not into the centre.

sprouty limits FGF10 so branching is restricted to the tip of the branch.
this is negative feedback because FGF will limit its own expression by activating its inhibitor sprouty.

Question 64

what do the tips of the branches recruit

how much of the alveoli are present at birth

Answer 64

capillaries

only 1/6 and they continue to grow after birth
protracted development