Paediatric Exams: A Survival Guide PDF
Preview Paediatric Exams: A Survival Guide
aedi-atric xams Survival Gu ide .. . ~~:i~<~_-> aediatric y.PAss~ ~. - ..: ·.'_.a_,•. ·' -. . . •' •····.• -I?AEDJATRfCS> - xams ~,,,__v'...(.<'<- ' Survival Guide ul Gaon Mass MRcP(uK) MRcPcH ltltdlatric Consultant, Israel ''reword by rofessor Anthony Milner . . MD FRCP FRCPCH DCH lllfofessor of Neonatology, Guy's, King's and Thomas' School of Medicine, London, UK :ontents 1. Genetics 1 2. Cardiology 13 3. Respiratory system 37 4. Nephrology 67 5. Neurology 93 6. Gastroenterology 135 7. Endocrinology 1 69 8. Haematology 199 9. Immunology 235 10. Metabolic diseases 247 11. Rheumatology 269 12. Dermatology 285 13. Vision and hearing 303 14. Paediatric syndromes 315 15. Paediatric development 331 16. The long case 339 Bibliography and recommended further reading 343 Index 345 1 Genetics Genetics questions appear in the examination mainly as pedigrees in the data interpretation section. A broader knowledge of basic genetic issues is, however, expected and this reflects the· general importa~ce of this topic in paediatrics. CLASSIFICAnON ·OF· CONGENITAL: ABNORMAUnES- .. · ·'t:~'· : :: A common problem faced by paediatricians is the child with structural defects. A nom~nclature has arisen to describe the genesis of s,_.ch deformities. ·· · - Malforrruztion: A structural defect resulting from abnormal development; - Deformation: This is where mechanical outside forces have altered the pre- viously normal shape of body parts. Examples include oligohydramnios l~ading to pulmonary hypoplasia and talipes. - Disruption: An extrinsic force alters a normal fetus usually by a destructive process (mechanical, vascular or infectious). Examples include amniotic bands leading to amputations, and congenital rubella. - Sequenu: When a single error results in a chain or cascade of subsequent events resulting in a series of abnormalities. Examples include the posteri orly placed tongue in Pierre-Robin syndrome that results in micrognathia and a cleft palate. - Malformntion syndrome: This occurs when there are multiple structural defects present that cannot be explained by a sequence (see above). Examples of this are polydactyly and exomphalos.ln a large number of cases the cause is not known. Other causes in~lude genetic causes (single/multi teratoiens ple gene effect/chromosomal defects/multifactorial), such as viruses (le~ding to congenital infection), drugs, chemicals and irradiation: - Association: This refers to cases where there are a number of malforma tions present that occur more often together than would be expected by chance alone. for example VATER/VACTERL and CHARGE syndromes 1 (see Ch.~pter 1-J). In Onl·-tenth of ,,1[ infants with a congenital m,llforma· tion there will generally be another malformation associated with it. lf you find one malformation, ,1Iways look for others. ll.EVELOPMENT AND TERATOGENESIS - ~ "'" . . . .. . ' ~ ·-·'i The effect of a particular teratogen on the fetus will depend on the nature of the teratogen and the fetal age, in addition to other factors. Weeks 1-3 Embryonic stage. At this stage teratogens c1ct in an all or nothing fashion - either killing the fetus or not affecting it at all. Weeks 3-10 Organogenesis. This is when organ sy:>tems are the most susceptible to damage. Weeks 16-40 Fetal growth and maturation. At this stage unlikely to be teratogenic but may interfere with growth ~nd physiological functioning of the normally formed fetal tissues and organ systems. Genetic pedigrees routinely come up in the data interpretation part of the examination and you ~hould have a good system for dealing with these questions. A certain amount of assumed knowledge is expected here such as the theory behind the different types of inheritance (found in most paediatric textbooks). You must know the commonly used symb(!ls !or.g enetic pedigrees (Fig. 1.1 ). Essential genetic pedigree knowledge Autosomal recessive (ARl conditions - The affected individual has two heterozygote (carrier) parents. There is usually no antecedent history; the disease appears out of the blue. - 1f both parents are heterozygotes then half the offspring will be carriers, a quarter will be affected and a quarter will be nonnal. -The spontaneous mutation rate for recessive conditions is extremely low. -Autosomal recessive conditions and their genes are very rare but the same recessh·e genes are more likely to be present in the gene pool of families or smaller races that do not intermarry. Thus consanguineous relations are more likely to gh·e rise to affected individuals.. This is because heterozygotes for a particular condition are more likely to come in contact with each other, producing the abnonnality (homozygous indi,·iduals). 2 -The male:female (!vt:F) ratio is 1:1. + Male 0 Normal mating D-0 D-0 0 Consanguineous mating Female • db Monozygotic twins Q Affected male (I) ::::s Do e DizygotiC tw1ns (I) !:t. Affected female 0 en IJ Heterozygote male for AR conditiOo Numbering of a genetic pedigree I· Heterozygote female () .tJ u Proband (affected person) "' 0 Deceased Proband is 111-1 lg. 1.1 A summary of the main symbols used in genetic pedigrees. E;romptes include: cystic fibrosis (carrier rate 1/25, one-quarter of children will be affected so that the incidence in the population is 1 /2500), sickle .cell disease, beta thalassaemia, most inborn errors of metabolism and spinal muscular atrophy. . \utosomal doJnant (AD) inheritance - The condition is inherited from one affected parent. ..... Half of the offspring will be affected and half will be normaL The spontaneous mutation rate is relatively high (compared to autosomal 1- recessive conditions). ~Male:female ratio is 1:1. Compa~ed to AR d~ease, AD disease displays a great deal of variation in expreSSIOn. Examples include: hereditary spherocytosis, myotonic dystrophy, retinoblastoma, ruberous sclerosis, Fried reich ataxia and polycystic kidneys. ~ ·linked recessive conditions -Males are affected. This is because the abnormal gene is on an X chromosome. Females ha\·e an extra X chromosome to protect them and therefore only become carriers. whereas males with only another 'empty' Y chromosome are affected. During cell division, however, a process called lyonisation can occur where there is a random inactivation of one of the X chromosomes in all the female cells. and thus some heterozygote females may have some of the features of the fully expressed X-linked condition ..A n example of this is haemophilia A where female heterozygotes may have prolonged clotting times. Another example is the raised creatine phosphokinase (CPK) levels found in heterozygote females in Duchenne muscular dystrophy. The lyonisation process that results in an inactin? X chromosome is also responsible for the Barr body seen in ceHs as a densely stained mas•: of chromatin within the nuclei (sample cells are most com:eniently taken from a buccal smear). As a rule the number of B&n bodies seen will be - one less than the total number of X chromosomes. Thus normal females co will have one Barr body and girls with Turner syndrome will ha\·e none .2: - A female carrier parent will result in· half the male offspring being c: affected and half the female offspring being carriers. e:n:J - An affected male parent will result in all the female offspring being carriers and all the male offspring being normal since they inherit the <( norma! Y chromosome from the father. E co >< X-iinked dominant conditions w - n.ese conditions a1·e extremely rare. According to the p€ digrees affat~d (.) 'i: males wiU have affected daughters and norrtlal sons (d. AD inh.:;;itanct), , .!!! Affected fumales pass the condition to hall of their offspring independent ol their ::;ex. MaieS f)ass the condition on to aU their daughters dfld none of 4!-.:lr Gco> sons. L"' p~ctice Cl number of these conditions appear io be lethal to m,1lt...1 D. (v.·ho may be stillborn) and thus only femaies a.re affected. You .should knv~v a co:;p re oi examples such as X-linked hypophosphataemic-rickets and incon tinentia pigmenti (lethal to males). U$ehli ruies in determining ii"~n~-ri~nee patien-• i.-. genetic pec:l!Q.g~~&Z» . 1. Count the m:1.rnbers of maies and fema!e:s affected and determine the M:F ratio. - If th~ ratio ~s 1:1 then the condition is likely to beAR or AD. Smaller nu:nb~rs of affected individuals will usually be present' in a gent: tic pec:H~ of AR inheritance. - If females are affected more common!y !han males t~.i:nk of X-li'."lked dom.inim~ inheritance. - If no furr>a!res are affected think of X-llnked recessive inheritance . . :i! 1f a."' aff«wd male gives rise to an affe:ci·~ mal~ trunk of AD inh~rit~'"lc. (AR <;;;s.;, possible but extremely rare). This is in contrast to X-linked dominant inheritance (affected males never give rise to an affected rr.ale). Sometimes it is difficult to disr.nguish betwe-i?n AD and X-Iin::...:..d dominant ill.l)eri tance and therefore large pedigrees are m!<:es...aty to de:nonstrate male to male transmission (and thus AD inheritanc~). 3. Remember that most indi\•iduals affected by AR ccnditions are w.•1a!iy sterile or die early. &ies , In these examples (as is conunon in the examination) only affected indivtdu als will be indicated (shadecl in) and carriers wili not be shown. Work ttu:ough 4 these examples '(Fig. 1.2) using the rules above. + example~ r-; :.'l :...J -(? ' 6 [5 ~ G) • I (!) I -, If- --'--1 :::J 0 ~ 0 0 (n~1) en ..r:-~ u - . 1-.2a This is an example of X·linked dominant inheritance. If asked fo:r ,;\n xample, say X·linked dominant hypophosphataemic ricitets rath«r than incontinentia igmer.ti since the latter is lethal to ma!es. Exampl&2 + 6 0r r0m 6:---+-tJ--,6 Jllg. 1.2b This ~ an -example.o f an X ·Hnlt~ rec.."Ssiva distwl9r. Example3 6 Fig. 1..2c This is an example of Af< in'-~ritancE. Example4 5 Another commonly asked question includes the following: A couple come for genetic counselling about the risk that their child will ha\·e cystic fibrosis. The husband's sister has it, but the husband is not affected and there is no historv in the wife's familv of the disease. A11swt'r: Cystic fibrosis is inherited as an aut~somal recessive condi· tion. Therefore since the sister is affected. both her parents must be car - riers. Since the brother is not affected therP. is a two-thirds chance that co > the brother is a carrier. We know that the carrier risk in the general pop ·~ ulation is 1/25. Thus we assume that this is also the wife's risk of being ::::J a carrier. C/) Therefore the risk of their child being affected is 2/3 (father's risk of <.. being a earner) multiplied by 1/25 (mother's risk of being a carrier) E multiplied-by 1/4 (risk of child being affected with two carrier parents). co .n -You should know about the rare diseases that demonstrate mitochondrial inheritance. Here we see that some mitochondrial .0 proteins are encoded. for by the mitochondrial chromosome (rather :5 than the nuclear genes). Since mitochondria are derived from the .~ mother then these diseases demonstrate maternal inheritance. Male "C , (1) and female offspring are equally likely to develop the condition. co 0. Examples include MELAS and MERRF (see Chapter 10). CHROMOSOME ANALYSIS ., :.. Thls is another popular data interpretation question. In classic ch,romosomal . analysis one finds that there are 23 pairs of chromosomes (22 autosomes and a pair of sex chromosomes). They are sorted by size, position of the centromere and banding pattern so that the autosomes are labelled from 1-22 and the sex chromosomes are called X andY {see Fig. 1.3). In the top left hand side are the large chromosomes with median placed centromeres. ·As one passes from left to right the chromosomes become smaller in size and the centromere moves distally forming progressively more acrocentric chromosomes. The short arm of the chromosome is called the p arm and the long arm is ..::all~d the q arm. An older nomenclature divided similar look ing chromosomes into seven groups designated A-G + the sex chromosomes Patau classification. Genetic testing is most conveniently carried out using peripheral blood lymphocytes. but almost any tissue may be used. The chromosomes are arrested in metaphase and stained. In order to diag nose tJ:te problem you must label the chromosomes from 1-22 and the sex chromosomes. 6 xamples Example 1 .";.·.. ' .#.... .: ·.:~. •- .v !4w• < .. .•..".. =. " "..'. .. ..: !; .;" :-"=.- .~. .; ' .•,.~. . .;. .' .. .·:••". - .~, "' .. ' 1~ \2 \- 3) ' 4 I I -s ' \.; 6 I \•; ;; !':: ,•3~; •:-~, ..•'9 I~. . '';. .. :~; ' !'-'= . 1~•~ ~.',!. I ,.~] 7 10 12 ~:~ ; !;§!; ~!.iI. S!'!. . . s& :'t. ' i i"";:"r.",:-:; !.~I.!.{!. :'!•-:1~! •~"-• I ' !..l. ' ~•""1 "'•3"'! ~~•;1 :;•4:.•.. R".. 1 .55'I ~"1''~.63 i1 7 1& 19 20 21 22 ~ Cl X y fig. 1.3a If there is one lone X chromosome then tht! individual has Turner syndrome. Example 2 . ff:. :;, :-•;' .;.. :.;r- -;~~ " :.,.:... , . ......, 5.. .... . ;... 7 ;j' , '~ •J "' •....'. "'" "• ~I "' ~' 2 3 4 5 s . ~.. .;.; j../ ....... ;i :i • • i !! • •--• •~: '! 7 8 9 10 11 •!w;I j ::':!•.!. Q~.-. i:= : !'IjlI i:: _1 j I'0 !.. •~a •"J 13 14 15 16 17 "•.$. '•""1J •".:~r "~;• . iA ..3=, ...:; l !•;
